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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *27* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *27 * 0.0293 in² = 0.7911 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.7911}{0.814} \approx 0.97$$**1-1/2 inch EMT conduit is 97% full with 27 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 27 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 27 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 27 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 27 \times 0.0293 = 0.7911 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which is **not sufficient** for our total area of 0.7911 in².\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is sufficient** for our total area.\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.814 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 27 #14 RHW-2 conductors, you should use at least a 1-1/2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch EMT conduit will be 97% full with 27 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *25* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *25 * 0.0293 in² = 0.7325 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.7325}{0.814} \approx 0.90$$**1-1/2 inch EMT conduit is 90% full with 25 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 25 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 25 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 25 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 25 \times 0.0293 = 0.7325 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which is **not sufficient** for our total area of 0.7325 in².\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is sufficient** for our total area.\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.814 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 25 #14 RHW-2 conductors, you should use at least a 1-1/2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch EMT conduit will be 90% full with 25 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *10* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *10 * 0.0293 in² = 0.2930 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**3/4 inch EMT:** *0.213 in²* \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.2930}{0.346} \approx 0.85$$**1 inch EMT conduit is 85% full with 10 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 10 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 10 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 10 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 10 \times 0.0293 = 0.2930 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 3/4 inch EMT conduit has a fill capacity of 0.213 in², which is **not sufficient** for our total area of 0.2930 in².\n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which **is sufficient** for our total area.\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.346 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 10 #14 RHW-2 conductors, you should use at least a 1 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch EMT conduit will be 85% full with 10 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *39* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *39 * 0.0293 in² = 1.1427 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{1.1427}{1.342} \approx 0.85$$**2 inch EMT conduit is 85% full with 39 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 39 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 39 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 39 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 39 \times 0.0293 = 1.1427 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 1.1427 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 39 #14 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 85% full with 39 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *30* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *30 * 0.0293 in² = 0.8790 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.8790}{1.342} \approx 0.65$$**2 inch EMT conduit is 65% full with 30 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 30 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 30 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 30 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 30 \times 0.0293 = 0.8790 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 0.8790 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 30 #14 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 65% full with 30 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *29* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *29 * 0.0293 in² = 0.8497 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.8497}{1.342} \approx 0.63$$**2 inch EMT conduit is 63% full with 29 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 29 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 29 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 29 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 29 \times 0.0293 = 0.8497 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 0.8497 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 29 #14 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 63% full with 29 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *14* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *14 * 0.0293 in² = 0.4102 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.4102}{0.598} \approx 0.69$$**1-1/4 inch EMT conduit is 69% full with 14 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 14 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 14 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 14 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 14 \times 0.0293 = 0.4102 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which is **not sufficient** for our total area of 0.4102 in².\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is sufficient** for our total area.\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.598 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 14 #14 RHW-2 conductors, you should use at least a 1-1/4 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch EMT conduit will be 69% full with 14 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *21* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *21 * 0.0293 in² = 0.6153 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.6153}{0.814} \approx 0.76$$**1-1/2 inch EMT conduit is 76% full with 21 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 21 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 21 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 21 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 21 \times 0.0293 = 0.6153 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which is **not sufficient** for our total area of 0.6153 in².\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is sufficient** for our total area.\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.814 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 21 #14 RHW-2 conductors, you should use at least a 1-1/2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch EMT conduit will be 76% full with 21 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *15* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *15 * 0.0293 in² = 0.4395 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.4395}{0.598} \approx 0.73$$**1-1/4 inch EMT conduit is 73% full with 15 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 15 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 15 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 15 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 15 \times 0.0293 = 0.4395 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which is **not sufficient** for our total area of 0.4395 in².\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is sufficient** for our total area.\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.598 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 15 #14 RHW-2 conductors, you should use at least a 1-1/4 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch EMT conduit will be 73% full with 15 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *23* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *23 * 0.0293 in² = 0.6739 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.6739}{0.814} \approx 0.83$$**1-1/2 inch EMT conduit is 83% full with 23 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 23 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 23 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 23 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 23 \times 0.0293 = 0.6739 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which is **not sufficient** for our total area of 0.6739 in².\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is sufficient** for our total area.\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.814 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 23 #14 RHW-2 conductors, you should use at least a 1-1/2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch EMT conduit will be 83% full with 23 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *38* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *38 * 0.0293 in² = 1.1134 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{1.1134}{1.342} \approx 0.83$$**2 inch EMT conduit is 83% full with 38 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 38 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 38 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 38 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 38 \times 0.0293 = 1.1134 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 1.1134 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 38 #14 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 83% full with 38 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *26* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *26 * 0.0293 in² = 0.7618 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.7618}{0.814} \approx 0.94$$**1-1/2 inch EMT conduit is 94% full with 26 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 26 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 26 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 26 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 26 \times 0.0293 = 0.7618 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which is **not sufficient** for our total area of 0.7618 in².\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is sufficient** for our total area.\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.814 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 26 #14 RHW-2 conductors, you should use at least a 1-1/2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch EMT conduit will be 94% full with 26 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *40* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *40 * 0.0293 in² = 1.1720 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{1.1720}{1.316} \approx 0.89$$**2 inch ENT conduit is 89% full with 40 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 40 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 40 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 40 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 40 \times 0.0293 = 1.1720 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which is **not sufficient** for our total area of 1.1720 in².\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.316 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 40 #14 RHW-2 conductors, you should use at least a 2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch ENT conduit will be 89% full with 40 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *21* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *21 * 0.0293 in² = 0.6153 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.6153}{0.794} \approx 0.77$$**1-1/2 inch ENT conduit is 77% full with 21 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 21 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 21 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 21 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 21 \times 0.0293 = 0.6153 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which is **not sufficient** for our total area of 0.6153 in².\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is sufficient** for our total area.\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.794 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 21 #14 RHW-2 conductors, you should use at least a 1-1/2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch ENT conduit will be 77% full with 21 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *13* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *13 * 0.0293 in² = 0.3809 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1 inch ENT:** *0.333 in²* \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.3809}{0.581} \approx 0.66$$**1-1/4 inch ENT conduit is 66% full with 13 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 13 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 13 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 13 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 13 \times 0.0293 = 0.3809 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1 inch ENT conduit has a fill capacity of 0.333 in², which is **not sufficient** for our total area of 0.3809 in².\n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which **is sufficient** for our total area.\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.581 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 13 #14 RHW-2 conductors, you should use at least a 1-1/4 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch ENT conduit will be 66% full with 13 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *31* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *31 * 0.0293 in² = 0.9083 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.9083}{1.316} \approx 0.69$$**2 inch ENT conduit is 69% full with 31 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 31 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 31 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 31 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 31 \times 0.0293 = 0.9083 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which is **not sufficient** for our total area of 0.9083 in².\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.316 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 31 #14 RHW-2 conductors, you should use at least a 2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch ENT conduit will be 69% full with 31 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *26* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *26 * 0.0293 in² = 0.7618 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.7618}{0.794} \approx 0.96$$**1-1/2 inch ENT conduit is 96% full with 26 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 26 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 26 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 26 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 26 \times 0.0293 = 0.7618 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which is **not sufficient** for our total area of 0.7618 in².\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is sufficient** for our total area.\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.794 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 26 #14 RHW-2 conductors, you should use at least a 1-1/2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch ENT conduit will be 96% full with 26 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *9* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *9 * 0.0293 in² = 0.2637 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**3/4 inch ENT:** *0.203 in²* \n**1 inch ENT:** *0.333 in²* \n**1-1/4 inch ENT:** *0.581 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.2637}{0.333} \approx 0.79$$**1 inch ENT conduit is 79% full with 9 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 9 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 9 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 9 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 9 \times 0.0293 = 0.2637 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 3/4 inch ENT conduit has a fill capacity of 0.203 in², which is **not sufficient** for our total area of 0.2637 in².\n - The 1 inch ENT conduit has a fill capacity of 0.333 in², which **is sufficient** for our total area.\n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.333 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 9 #14 RHW-2 conductors, you should use at least a 1 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch ENT conduit will be 79% full with 9 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *37* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *37 * 0.0293 in² = 1.0841 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{1.0841}{1.316} \approx 0.82$$**2 inch ENT conduit is 82% full with 37 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 37 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 37 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 37 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 37 \times 0.0293 = 1.0841 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which is **not sufficient** for our total area of 1.0841 in².\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.316 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 37 #14 RHW-2 conductors, you should use at least a 2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch ENT conduit will be 82% full with 37 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *29* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *29 * 0.0293 in² = 0.8497 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.8497}{1.316} \approx 0.65$$**2 inch ENT conduit is 65% full with 29 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 29 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 29 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 29 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 29 \times 0.0293 = 0.8497 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which is **not sufficient** for our total area of 0.8497 in².\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.316 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 29 #14 RHW-2 conductors, you should use at least a 2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch ENT conduit will be 65% full with 29 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *17* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *17 * 0.0293 in² = 0.4981 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1 inch ENT:** *0.333 in²* \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.4981}{0.581} \approx 0.86$$**1-1/4 inch ENT conduit is 86% full with 17 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 17 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 17 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 17 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 17 \times 0.0293 = 0.4981 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1 inch ENT conduit has a fill capacity of 0.333 in², which is **not sufficient** for our total area of 0.4981 in².\n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which **is sufficient** for our total area.\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.581 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 17 #14 RHW-2 conductors, you should use at least a 1-1/4 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch ENT conduit will be 86% full with 17 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *8* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *8 * 0.0293 in² = 0.2344 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**3/4 inch FMC:** *0.213 in²* \n**1 inch FMC:** *0.327 in²* \n**1-1/4 inch FMC:** *0.511 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.2344}{0.327} \approx 0.72$$**1 inch FMC conduit is 72% full with 8 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 8 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 8 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 8 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 8 \times 0.0293 = 0.2344 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 3/4 inch FMC conduit has a fill capacity of 0.213 in², which is **not sufficient** for our total area of 0.2344 in².\n - The 1 inch FMC conduit has a fill capacity of 0.327 in², which **is sufficient** for our total area.\n - The 1-1/4 inch FMC conduit has a fill capacity of 0.511 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.327 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 8 #14 RHW-2 conductors, you should use at least a 1 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch FMC conduit will be 72% full with 8 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *35* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *35 * 0.0293 in² = 1.0255 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**2-1/2 inch FMC:** *1.963 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{1.0255}{1.307} \approx 0.78$$**2 inch FMC conduit is 78% full with 35 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 35 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 35 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 35 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 35 \times 0.0293 = 1.0255 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which is **not sufficient** for our total area of 1.0255 in².\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is sufficient** for our total area.\n - The 2-1/2 inch FMC conduit has a fill capacity of 1.963 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.307 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 35 #14 RHW-2 conductors, you should use at least a 2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch FMC conduit will be 78% full with 35 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *33* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *33 * 0.0293 in² = 0.9669 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**2-1/2 inch FMC:** *1.963 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.9669}{1.307} \approx 0.74$$**2 inch FMC conduit is 74% full with 33 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 33 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 33 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 33 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 33 \times 0.0293 = 0.9669 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which is **not sufficient** for our total area of 0.9669 in².\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is sufficient** for our total area.\n - The 2-1/2 inch FMC conduit has a fill capacity of 1.963 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.307 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 33 #14 RHW-2 conductors, you should use at least a 2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch FMC conduit will be 74% full with 33 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *24* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *24 * 0.0293 in² = 0.7032 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/4 inch FMC:** *0.511 in²* \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.7032}{0.743} \approx 0.95$$**1-1/2 inch FMC conduit is 95% full with 24 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 24 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 24 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 24 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 24 \times 0.0293 = 0.7032 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/4 inch FMC conduit has a fill capacity of 0.511 in², which is **not sufficient** for our total area of 0.7032 in².\n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which **is sufficient** for our total area.\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.743 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 24 #14 RHW-2 conductors, you should use at least a 1-1/2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch FMC conduit will be 95% full with 24 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *26* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *26 * 0.0293 in² = 0.7618 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**2-1/2 inch FMC:** *1.963 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.7618}{1.307} \approx 0.58$$**2 inch FMC conduit is 58% full with 26 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 26 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 26 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 26 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 26 \times 0.0293 = 0.7618 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which is **not sufficient** for our total area of 0.7618 in².\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is sufficient** for our total area.\n - The 2-1/2 inch FMC conduit has a fill capacity of 1.963 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.307 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 26 #14 RHW-2 conductors, you should use at least a 2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch FMC conduit will be 58% full with 26 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *6* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *6 * 0.0293 in² = 0.1758 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1/2 inch FMC:** *0.127 in²* \n**3/4 inch FMC:** *0.213 in²* \n**1 inch FMC:** *0.327 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.1758}{0.213} \approx 0.83$$**3/4 inch FMC conduit is 83% full with 6 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 6 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 6 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 6 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 6 \times 0.0293 = 0.1758 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1/2 inch FMC conduit has a fill capacity of 0.127 in², which is **not sufficient** for our total area of 0.1758 in².\n - The 3/4 inch FMC conduit has a fill capacity of 0.213 in², which **is sufficient** for our total area.\n - The 1 inch FMC conduit has a fill capacity of 0.327 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 3/4 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.213 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 6 #14 RHW-2 conductors, you should use at least a 3/4 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 3/4 inch FMC conduit will be 83% full with 6 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *18* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *18 * 0.0293 in² = 0.5274 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/4 inch FMC:** *0.511 in²* \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5274}{0.743} \approx 0.71$$**1-1/2 inch FMC conduit is 71% full with 18 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 18 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 18 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 18 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 18 \times 0.0293 = 0.5274 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/4 inch FMC conduit has a fill capacity of 0.511 in², which is **not sufficient** for our total area of 0.5274 in².\n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which **is sufficient** for our total area.\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.743 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 18 #14 RHW-2 conductors, you should use at least a 1-1/2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch FMC conduit will be 71% full with 18 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *25* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *25 * 0.0293 in² = 0.7325 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/4 inch FMC:** *0.511 in²* \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.7325}{0.743} \approx 0.99$$**1-1/2 inch FMC conduit is 99% full with 25 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 25 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 25 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 25 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 25 \times 0.0293 = 0.7325 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/4 inch FMC conduit has a fill capacity of 0.511 in², which is **not sufficient** for our total area of 0.7325 in².\n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which **is sufficient** for our total area.\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.743 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 25 #14 RHW-2 conductors, you should use at least a 1-1/2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch FMC conduit will be 99% full with 25 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *33* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *33 * 0.0293 in² = 0.9669 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**2-1/2 inch FMC:** *1.963 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.9669}{1.307} \approx 0.74$$**2 inch FMC conduit is 74% full with 33 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 33 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 33 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 33 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 33 \times 0.0293 = 0.9669 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which is **not sufficient** for our total area of 0.9669 in².\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is sufficient** for our total area.\n - The 2-1/2 inch FMC conduit has a fill capacity of 1.963 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.307 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 33 #14 RHW-2 conductors, you should use at least a 2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch FMC conduit will be 74% full with 33 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *17* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *17 * 0.0293 in² = 0.4981 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1 inch FMC:** *0.327 in²* \n**1-1/4 inch FMC:** *0.511 in²* \n**1-1/2 inch FMC:** *0.743 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.4981}{0.511} \approx 0.97$$**1-1/4 inch FMC conduit is 97% full with 17 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 17 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 17 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 17 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 17 \times 0.0293 = 0.4981 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1 inch FMC conduit has a fill capacity of 0.327 in², which is **not sufficient** for our total area of 0.4981 in².\n - The 1-1/4 inch FMC conduit has a fill capacity of 0.511 in², which **is sufficient** for our total area.\n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.511 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 17 #14 RHW-2 conductors, you should use at least a 1-1/4 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch FMC conduit will be 97% full with 17 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *38* \n**Conduit Type:** *FMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *38 * 0.0293 in² = 1.1134 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)):</strong></u> \n**1-1/2 inch FMC:** *0.743 in²* \n**2 inch FMC:** *1.307 in²* \n**2-1/2 inch FMC:** *1.963 in²* \n**Link:** [Chapter 9 Table 4(FMC)](https://2023.antflix.net#table-4(FMC)) \n \n$$\text{Space occupied by conductors} = \frac{1.1134}{1.307} \approx 0.85$$**2 inch FMC conduit is 85% full with 38 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size FMC conduit is needed to run 38 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 38 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 38 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 38 \times 0.0293 = 1.1134 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)), we can see the following: \n - The 1-1/2 inch FMC conduit has a fill capacity of 0.743 in², which is **not sufficient** for our total area of 1.1134 in².\n - The 2 inch FMC conduit has a fill capacity of 1.307 in², which **is sufficient** for our total area.\n - The 2-1/2 inch FMC conduit has a fill capacity of 1.963 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch FMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.307 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(FMC)).\n\n**Conclusion** \nTo safely install 38 #14 RHW-2 conductors, you should use at least a 2 inch FMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch FMC conduit will be 85% full with 38 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *40* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *40 * 0.0293 in² = 1.1720 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1-1/2 inch IMC:** *0.89 in²* \n**2 inch IMC:** *1.452 in²* \n**2-1/2 inch IMC:** *2.054 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{1.1720}{1.452} \approx 0.81$$**2 inch IMC conduit is 81% full with 40 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 40 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 40 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 40 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 40 \times 0.0293 = 1.1720 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which is **not sufficient** for our total area of 1.1720 in².\n - The 2 inch IMC conduit has a fill capacity of 1.452 in², which **is sufficient** for our total area.\n - The 2-1/2 inch IMC conduit has a fill capacity of 2.054 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.452 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 40 #14 RHW-2 conductors, you should use at least a 2 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch IMC conduit will be 81% full with 40 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *19* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *19 * 0.0293 in² = 0.5567 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**1-1/2 inch IMC:** *0.89 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5567}{0.659} \approx 0.84$$**1-1/4 inch IMC conduit is 84% full with 19 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 19 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 19 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 19 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 19 \times 0.0293 = 0.5567 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which is **not sufficient** for our total area of 0.5567 in².\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is sufficient** for our total area.\n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.659 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 19 #14 RHW-2 conductors, you should use at least a 1-1/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch IMC conduit will be 84% full with 19 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *10* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *10 * 0.0293 in² = 0.2930 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**3/4 inch IMC:** *0.235 in²* \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.2930}{0.384} \approx 0.76$$**1 inch IMC conduit is 76% full with 10 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 10 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 10 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 10 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 10 \times 0.0293 = 0.2930 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 3/4 inch IMC conduit has a fill capacity of 0.235 in², which is **not sufficient** for our total area of 0.2930 in².\n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which **is sufficient** for our total area.\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.384 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 10 #14 RHW-2 conductors, you should use at least a 1 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch IMC conduit will be 76% full with 10 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *14* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *14 * 0.0293 in² = 0.4102 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**1-1/2 inch IMC:** *0.89 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.4102}{0.659} \approx 0.62$$**1-1/4 inch IMC conduit is 62% full with 14 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 14 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 14 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 14 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 14 \times 0.0293 = 0.4102 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which is **not sufficient** for our total area of 0.4102 in².\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is sufficient** for our total area.\n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.659 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 14 #14 RHW-2 conductors, you should use at least a 1-1/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch IMC conduit will be 62% full with 14 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *19* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *19 * 0.0293 in² = 0.5567 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**1-1/2 inch IMC:** *0.89 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5567}{0.659} \approx 0.84$$**1-1/4 inch IMC conduit is 84% full with 19 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 19 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 19 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 19 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 19 \times 0.0293 = 0.5567 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which is **not sufficient** for our total area of 0.5567 in².\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is sufficient** for our total area.\n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.659 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 19 #14 RHW-2 conductors, you should use at least a 1-1/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch IMC conduit will be 84% full with 19 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *8* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *8 * 0.0293 in² = 0.2344 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1/2 inch IMC:** *0.137 in²* \n**3/4 inch IMC:** *0.235 in²* \n**1 inch IMC:** *0.384 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.2344}{0.235} \approx 1.00$$**3/4 inch IMC conduit is 100% full with 8 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 8 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 8 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 8 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 8 \times 0.0293 = 0.2344 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1/2 inch IMC conduit has a fill capacity of 0.137 in², which is **not sufficient** for our total area of 0.2344 in².\n - The 3/4 inch IMC conduit has a fill capacity of 0.235 in², which **is sufficient** for our total area.\n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 3/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.235 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 8 #14 RHW-2 conductors, you should use at least a 3/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 3/4 inch IMC conduit will be 100% full with 8 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *32* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *32 * 0.0293 in² = 0.9376 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1-1/2 inch IMC:** *0.89 in²* \n**2 inch IMC:** *1.452 in²* \n**2-1/2 inch IMC:** *2.054 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.9376}{1.452} \approx 0.65$$**2 inch IMC conduit is 65% full with 32 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 32 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 32 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 32 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 32 \times 0.0293 = 0.9376 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which is **not sufficient** for our total area of 0.9376 in².\n - The 2 inch IMC conduit has a fill capacity of 1.452 in², which **is sufficient** for our total area.\n - The 2-1/2 inch IMC conduit has a fill capacity of 2.054 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.452 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 32 #14 RHW-2 conductors, you should use at least a 2 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch IMC conduit will be 65% full with 32 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *20* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *20 * 0.0293 in² = 0.5860 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**1-1/2 inch IMC:** *0.89 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5860}{0.659} \approx 0.89$$**1-1/4 inch IMC conduit is 89% full with 20 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 20 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 20 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 20 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 20 \times 0.0293 = 0.5860 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which is **not sufficient** for our total area of 0.5860 in².\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is sufficient** for our total area.\n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.659 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 20 #14 RHW-2 conductors, you should use at least a 1-1/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch IMC conduit will be 89% full with 20 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *6* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *6 * 0.0293 in² = 0.1758 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1/2 inch IMC:** *0.137 in²* \n**3/4 inch IMC:** *0.235 in²* \n**1 inch IMC:** *0.384 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.1758}{0.235} \approx 0.75$$**3/4 inch IMC conduit is 75% full with 6 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 6 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 6 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 6 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 6 \times 0.0293 = 0.1758 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1/2 inch IMC conduit has a fill capacity of 0.137 in², which is **not sufficient** for our total area of 0.1758 in².\n - The 3/4 inch IMC conduit has a fill capacity of 0.235 in², which **is sufficient** for our total area.\n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 3/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.235 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 6 #14 RHW-2 conductors, you should use at least a 3/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 3/4 inch IMC conduit will be 75% full with 6 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *20* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *20 * 0.0293 in² = 0.5860 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**1-1/2 inch IMC:** *0.89 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5860}{0.659} \approx 0.89$$**1-1/4 inch IMC conduit is 89% full with 20 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 20 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 20 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 20 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 20 \times 0.0293 = 0.5860 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which is **not sufficient** for our total area of 0.5860 in².\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is sufficient** for our total area.\n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.659 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 20 #14 RHW-2 conductors, you should use at least a 1-1/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch IMC conduit will be 89% full with 20 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *40* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *40 * 0.0293 in² = 1.1720 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1-1/2 inch IMC:** *0.89 in²* \n**2 inch IMC:** *1.452 in²* \n**2-1/2 inch IMC:** *2.054 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{1.1720}{1.452} \approx 0.81$$**2 inch IMC conduit is 81% full with 40 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 40 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 40 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 40 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 40 \times 0.0293 = 1.1720 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which is **not sufficient** for our total area of 1.1720 in².\n - The 2 inch IMC conduit has a fill capacity of 1.452 in², which **is sufficient** for our total area.\n - The 2-1/2 inch IMC conduit has a fill capacity of 2.054 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.452 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 40 #14 RHW-2 conductors, you should use at least a 2 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch IMC conduit will be 81% full with 40 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *14* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *14 * 0.0293 in² = 0.4102 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1 inch IMC:** *0.384 in²* \n**1-1/4 inch IMC:** *0.659 in²* \n**1-1/2 inch IMC:** *0.89 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.4102}{0.659} \approx 0.62$$**1-1/4 inch IMC conduit is 62% full with 14 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 14 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 14 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 14 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 14 \times 0.0293 = 0.4102 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1 inch IMC conduit has a fill capacity of 0.384 in², which is **not sufficient** for our total area of 0.4102 in².\n - The 1-1/4 inch IMC conduit has a fill capacity of 0.659 in², which **is sufficient** for our total area.\n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.659 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 14 #14 RHW-2 conductors, you should use at least a 1-1/4 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch IMC conduit will be 62% full with 14 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *33* \n**Conduit Type:** *IMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *33 * 0.0293 in² = 0.9669 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)):</strong></u> \n**1-1/2 inch IMC:** *0.89 in²* \n**2 inch IMC:** *1.452 in²* \n**2-1/2 inch IMC:** *2.054 in²* \n**Link:** [Chapter 9 Table 4(IMC)](https://2023.antflix.net#table-4(IMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.9669}{1.452} \approx 0.67$$**2 inch IMC conduit is 67% full with 33 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size IMC conduit is needed to run 33 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 33 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 33 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 33 \times 0.0293 = 0.9669 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)), we can see the following: \n - The 1-1/2 inch IMC conduit has a fill capacity of 0.89 in², which is **not sufficient** for our total area of 0.9669 in².\n - The 2 inch IMC conduit has a fill capacity of 1.452 in², which **is sufficient** for our total area.\n - The 2-1/2 inch IMC conduit has a fill capacity of 2.054 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch IMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.452 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table-4(IMC)).\n\n**Conclusion** \nTo safely install 33 #14 RHW-2 conductors, you should use at least a 2 inch IMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch IMC conduit will be 67% full with 33 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *26* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *26 * 0.0293 in² = 0.7618 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.7618}{0.807} \approx 0.94$$**1-1/2 inch LFNC conduit is 94% full with 26 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 26 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 26 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 26 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 26 \times 0.0293 = 0.7618 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which is **not sufficient** for our total area of 0.7618 in².\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is sufficient** for our total area.\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.807 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 26 #14 RHW-2 conductors, you should use at least a 1-1/2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch LFNC conduit will be 94% full with 26 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *20* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *20 * 0.0293 in² = 0.5860 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.5860}{0.601} \approx 0.98$$**1-1/4 inch LFNC conduit is 98% full with 20 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 20 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 20 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 20 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 20 \times 0.0293 = 0.5860 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.5860 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 20 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 98% full with 20 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *11* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *11 * 0.0293 in² = 0.3223 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**3/4 inch LFNC:** *0.214 in²* \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.3223}{0.342} \approx 0.94$$**1 inch LFNC conduit is 94% full with 11 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 11 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 11 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 11 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 11 \times 0.0293 = 0.3223 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 3/4 inch LFNC conduit has a fill capacity of 0.214 in², which is **not sufficient** for our total area of 0.3223 in².\n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which **is sufficient** for our total area.\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 11 #14 RHW-2 conductors, you should use at least a 1 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch LFNC conduit will be 94% full with 11 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *26* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *26 * 0.0293 in² = 0.7618 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.7618}{0.807} \approx 0.94$$**1-1/2 inch LFNC conduit is 94% full with 26 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 26 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 26 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 26 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 26 \times 0.0293 = 0.7618 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which is **not sufficient** for our total area of 0.7618 in².\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is sufficient** for our total area.\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.807 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 26 #14 RHW-2 conductors, you should use at least a 1-1/2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch LFNC conduit will be 94% full with 26 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *18* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *18 * 0.0293 in² = 0.5274 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.5274}{0.601} \approx 0.88$$**1-1/4 inch LFNC conduit is 88% full with 18 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 18 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 18 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 18 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 18 \times 0.0293 = 0.5274 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.5274 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 18 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 88% full with 18 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *32* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *32 * 0.0293 in² = 0.9376 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**6 inch LFNC:** *6.0 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.9376}{1.337} \approx 0.70$$**2 inch LFNC conduit is 70% full with 32 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 32 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 32 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 32 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 32 \times 0.0293 = 0.9376 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which is **not sufficient** for our total area of 0.9376 in².\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is sufficient** for our total area.\n - The 6 inch LFNC conduit has a fill capacity of 6.0 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.337 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 32 #14 RHW-2 conductors, you should use at least a 2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch LFNC conduit will be 70% full with 32 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *16* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *16 * 0.0293 in² = 0.4688 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.4688}{0.601} \approx 0.78$$**1-1/4 inch LFNC conduit is 78% full with 16 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 16 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 16 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 16 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 16 \times 0.0293 = 0.4688 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.4688 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 16 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 78% full with 16 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *20* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *20 * 0.0293 in² = 0.5860 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.5860}{0.601} \approx 0.98$$**1-1/4 inch LFNC conduit is 98% full with 20 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 20 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 20 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 20 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 20 \times 0.0293 = 0.5860 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.5860 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 20 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 98% full with 20 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *10* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *10 * 0.0293 in² = 0.2930 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**3/4 inch LFNC:** *0.214 in²* \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.2930}{0.342} \approx 0.86$$**1 inch LFNC conduit is 86% full with 10 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 10 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 10 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 10 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 10 \times 0.0293 = 0.2930 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 3/4 inch LFNC conduit has a fill capacity of 0.214 in², which is **not sufficient** for our total area of 0.2930 in².\n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which **is sufficient** for our total area.\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 10 #14 RHW-2 conductors, you should use at least a 1 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch LFNC conduit will be 86% full with 10 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *16* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *16 * 0.0293 in² = 0.4688 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.4688}{0.601} \approx 0.78$$**1-1/4 inch LFNC conduit is 78% full with 16 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 16 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 16 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 16 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 16 \times 0.0293 = 0.4688 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.4688 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 16 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 78% full with 16 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *9* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *9 * 0.0293 in² = 0.2637 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**3/4 inch LFNC:** *0.214 in²* \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.2637}{0.342} \approx 0.77$$**1 inch LFNC conduit is 77% full with 9 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 9 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 9 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 9 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 9 \times 0.0293 = 0.2637 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 3/4 inch LFNC conduit has a fill capacity of 0.214 in², which is **not sufficient** for our total area of 0.2637 in².\n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which **is sufficient** for our total area.\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 9 #14 RHW-2 conductors, you should use at least a 1 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch LFNC conduit will be 77% full with 9 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *23* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *23 * 0.0293 in² = 0.6739 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.6739}{0.807} \approx 0.84$$**1-1/2 inch LFNC conduit is 84% full with 23 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 23 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 23 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 23 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 23 \times 0.0293 = 0.6739 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which is **not sufficient** for our total area of 0.6739 in².\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is sufficient** for our total area.\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.807 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 23 #14 RHW-2 conductors, you should use at least a 1-1/2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch LFNC conduit will be 84% full with 23 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *40* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *40 * 0.0293 in² = 1.1720 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**6 inch LFNC:** *6.0 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{1.1720}{1.337} \approx 0.88$$**2 inch LFNC conduit is 88% full with 40 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 40 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 40 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 40 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 40 \times 0.0293 = 1.1720 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which is **not sufficient** for our total area of 1.1720 in².\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is sufficient** for our total area.\n - The 6 inch LFNC conduit has a fill capacity of 6.0 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.337 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 40 #14 RHW-2 conductors, you should use at least a 2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch LFNC conduit will be 88% full with 40 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *25* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *25 * 0.0293 in² = 0.7325 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.7325}{0.807} \approx 0.91$$**1-1/2 inch LFNC conduit is 91% full with 25 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 25 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 25 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 25 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 25 \times 0.0293 = 0.7325 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which is **not sufficient** for our total area of 0.7325 in².\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is sufficient** for our total area.\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.807 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 25 #14 RHW-2 conductors, you should use at least a 1-1/2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch LFNC conduit will be 91% full with 25 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *40* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *40 * 0.0293 in² = 1.1720 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**6 inch LFNC:** *6.0 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{1.1720}{1.337} \approx 0.88$$**2 inch LFNC conduit is 88% full with 40 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 40 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 40 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 40 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 40 \times 0.0293 = 1.1720 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which is **not sufficient** for our total area of 1.1720 in².\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is sufficient** for our total area.\n - The 6 inch LFNC conduit has a fill capacity of 6.0 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.337 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 40 #14 RHW-2 conductors, you should use at least a 2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch LFNC conduit will be 88% full with 40 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *10* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *10 * 0.0293 in² = 0.2930 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**3/4 inch LFNC:** *0.214 in²* \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.2930}{0.342} \approx 0.86$$**1 inch LFNC conduit is 86% full with 10 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 10 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 10 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 10 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 10 \times 0.0293 = 0.2930 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 3/4 inch LFNC conduit has a fill capacity of 0.214 in², which is **not sufficient** for our total area of 0.2930 in².\n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which **is sufficient** for our total area.\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 10 #14 RHW-2 conductors, you should use at least a 1 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch LFNC conduit will be 86% full with 10 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *38* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *38 * 0.0293 in² = 1.1134 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**6 inch LFNC:** *6.0 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{1.1134}{1.337} \approx 0.83$$**2 inch LFNC conduit is 83% full with 38 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 38 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 38 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 38 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 38 \times 0.0293 = 1.1134 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which is **not sufficient** for our total area of 1.1134 in².\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is sufficient** for our total area.\n - The 6 inch LFNC conduit has a fill capacity of 6.0 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.337 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 38 #14 RHW-2 conductors, you should use at least a 2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch LFNC conduit will be 83% full with 38 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *12* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *12 * 0.0293 in² = 0.3516 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.3516}{0.601} \approx 0.59$$**1-1/4 inch LFNC conduit is 59% full with 12 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 12 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 12 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 12 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 12 \times 0.0293 = 0.3516 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.3516 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 12 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 59% full with 12 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *31* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *31 * 0.0293 in² = 0.9083 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**6 inch LFNC:** *6.0 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.9083}{1.337} \approx 0.68$$**2 inch LFNC conduit is 68% full with 31 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 31 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 31 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 31 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 31 \times 0.0293 = 0.9083 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which is **not sufficient** for our total area of 0.9083 in².\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is sufficient** for our total area.\n - The 6 inch LFNC conduit has a fill capacity of 6.0 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.337 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 31 #14 RHW-2 conductors, you should use at least a 2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch LFNC conduit will be 68% full with 31 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *12* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *12 * 0.0293 in² = 0.3516 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.3516}{0.601} \approx 0.59$$**1-1/4 inch LFNC conduit is 59% full with 12 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 12 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 12 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 12 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 12 \times 0.0293 = 0.3516 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.3516 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 12 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 59% full with 12 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *15* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *15 * 0.0293 in² = 0.4395 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.4395}{0.601} \approx 0.73$$**1-1/4 inch LFNC conduit is 73% full with 15 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 15 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 15 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 15 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 15 \times 0.0293 = 0.4395 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.4395 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 15 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 73% full with 15 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *15* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *15 * 0.0293 in² = 0.4395 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1 inch LFNC:** *0.342 in²* \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.4395}{0.601} \approx 0.73$$**1-1/4 inch LFNC conduit is 73% full with 15 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 15 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 15 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 15 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 15 \times 0.0293 = 0.4395 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1 inch LFNC conduit has a fill capacity of 0.342 in², which is **not sufficient** for our total area of 0.4395 in².\n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which **is sufficient** for our total area.\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.601 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 15 #14 RHW-2 conductors, you should use at least a 1-1/4 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch LFNC conduit will be 73% full with 15 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *38* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *38 * 0.0293 in² = 1.1134 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**6 inch LFNC:** *6.0 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{1.1134}{1.337} \approx 0.83$$**2 inch LFNC conduit is 83% full with 38 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 38 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 38 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 38 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 38 \times 0.0293 = 1.1134 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which is **not sufficient** for our total area of 1.1134 in².\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is sufficient** for our total area.\n - The 6 inch LFNC conduit has a fill capacity of 6.0 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.337 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 38 #14 RHW-2 conductors, you should use at least a 2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch LFNC conduit will be 83% full with 38 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *27* \n**Conduit Type:** *LFNC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *27 * 0.0293 in² = 0.7911 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)):</strong></u> \n**1-1/4 inch LFNC:** *0.601 in²* \n**1-1/2 inch LFNC:** *0.807 in²* \n**2 inch LFNC:** *1.337 in²* \n**Link:** [Chapter 9 Table 4(LFNC)](https://2023.antflix.net#table_4(LFNC-A)) \n \n$$\text{Space occupied by conductors} = \frac{0.7911}{0.807} \approx 0.98$$**1-1/2 inch LFNC conduit is 98% full with 27 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size LFNC conduit is needed to run 27 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 27 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 27 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 27 \times 0.0293 = 0.7911 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)), we can see the following: \n - The 1-1/4 inch LFNC conduit has a fill capacity of 0.601 in², which is **not sufficient** for our total area of 0.7911 in².\n - The 1-1/2 inch LFNC conduit has a fill capacity of 0.807 in², which **is sufficient** for our total area.\n - The 2 inch LFNC conduit has a fill capacity of 1.337 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch LFNC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.807 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(LFNC-A)).\n\n**Conclusion** \nTo safely install 27 #14 RHW-2 conductors, you should use at least a 1-1/2 inch LFNC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch LFNC conduit will be 98% full with 27 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *6* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *6 * 0.0293 in² = 0.1758 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1/2 inch RMC:** *0.125 in²* \n**3/4 inch RMC:** *0.22 in²* \n**1 inch RMC:** *0.355 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.1758}{0.22} \approx 0.80$$**3/4 inch RMC conduit is 80% full with 6 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 6 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 6 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 6 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 6 \times 0.0293 = 0.1758 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1/2 inch RMC conduit has a fill capacity of 0.125 in², which is **not sufficient** for our total area of 0.1758 in².\n - The 3/4 inch RMC conduit has a fill capacity of 0.22 in², which **is sufficient** for our total area.\n - The 1 inch RMC conduit has a fill capacity of 0.355 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 3/4 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.22 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 6 #14 RHW-2 conductors, you should use at least a 3/4 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 3/4 inch RMC conduit will be 80% full with 6 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *35* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *35 * 0.0293 in² = 1.0255 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**2-1/2 inch RMC:** *1.946 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{1.0255}{1.363} \approx 0.75$$**2 inch RMC conduit is 75% full with 35 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 35 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 35 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 35 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 35 \times 0.0293 = 1.0255 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which is **not sufficient** for our total area of 1.0255 in².\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is sufficient** for our total area.\n - The 2-1/2 inch RMC conduit has a fill capacity of 1.946 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.363 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 35 #14 RHW-2 conductors, you should use at least a 2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch RMC conduit will be 75% full with 35 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *25* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *25 * 0.0293 in² = 0.7325 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/4 inch RMC:** *0.61 in²* \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.7325}{0.829} \approx 0.88$$**1-1/2 inch RMC conduit is 88% full with 25 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 25 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 25 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 25 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 25 \times 0.0293 = 0.7325 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/4 inch RMC conduit has a fill capacity of 0.61 in², which is **not sufficient** for our total area of 0.7325 in².\n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which **is sufficient** for our total area.\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.829 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 25 #14 RHW-2 conductors, you should use at least a 1-1/2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch RMC conduit will be 88% full with 25 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *34* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *34 * 0.0293 in² = 0.9962 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**2-1/2 inch RMC:** *1.946 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.9962}{1.363} \approx 0.73$$**2 inch RMC conduit is 73% full with 34 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 34 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 34 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 34 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 34 \times 0.0293 = 0.9962 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which is **not sufficient** for our total area of 0.9962 in².\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is sufficient** for our total area.\n - The 2-1/2 inch RMC conduit has a fill capacity of 1.946 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.363 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 34 #14 RHW-2 conductors, you should use at least a 2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch RMC conduit will be 73% full with 34 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *30* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *30 * 0.0293 in² = 0.8790 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**2-1/2 inch RMC:** *1.946 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.8790}{1.363} \approx 0.64$$**2 inch RMC conduit is 64% full with 30 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 30 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 30 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 30 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 30 \times 0.0293 = 0.8790 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which is **not sufficient** for our total area of 0.8790 in².\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is sufficient** for our total area.\n - The 2-1/2 inch RMC conduit has a fill capacity of 1.946 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.363 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 30 #14 RHW-2 conductors, you should use at least a 2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch RMC conduit will be 64% full with 30 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *27* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *27 * 0.0293 in² = 0.7911 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/4 inch RMC:** *0.61 in²* \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.7911}{0.829} \approx 0.95$$**1-1/2 inch RMC conduit is 95% full with 27 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 27 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 27 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 27 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 27 \times 0.0293 = 0.7911 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/4 inch RMC conduit has a fill capacity of 0.61 in², which is **not sufficient** for our total area of 0.7911 in².\n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which **is sufficient** for our total area.\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.829 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 27 #14 RHW-2 conductors, you should use at least a 1-1/2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch RMC conduit will be 95% full with 27 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *20* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *20 * 0.0293 in² = 0.5860 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1 inch RMC:** *0.355 in²* \n**1-1/4 inch RMC:** *0.61 in²* \n**1-1/2 inch RMC:** *0.829 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5860}{0.61} \approx 0.96$$**1-1/4 inch RMC conduit is 96% full with 20 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 20 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 20 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 20 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 20 \times 0.0293 = 0.5860 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1 inch RMC conduit has a fill capacity of 0.355 in², which is **not sufficient** for our total area of 0.5860 in².\n - The 1-1/4 inch RMC conduit has a fill capacity of 0.61 in², which **is sufficient** for our total area.\n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.61 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 20 #14 RHW-2 conductors, you should use at least a 1-1/4 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch RMC conduit will be 96% full with 20 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *33* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *33 * 0.0293 in² = 0.9669 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**2-1/2 inch RMC:** *1.946 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.9669}{1.363} \approx 0.71$$**2 inch RMC conduit is 71% full with 33 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 33 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 33 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 33 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 33 \times 0.0293 = 0.9669 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which is **not sufficient** for our total area of 0.9669 in².\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is sufficient** for our total area.\n - The 2-1/2 inch RMC conduit has a fill capacity of 1.946 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.363 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 33 #14 RHW-2 conductors, you should use at least a 2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch RMC conduit will be 71% full with 33 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *17* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *17 * 0.0293 in² = 0.4981 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1 inch RMC:** *0.355 in²* \n**1-1/4 inch RMC:** *0.61 in²* \n**1-1/2 inch RMC:** *0.829 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.4981}{0.61} \approx 0.82$$**1-1/4 inch RMC conduit is 82% full with 17 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 17 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 17 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 17 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 17 \times 0.0293 = 0.4981 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1 inch RMC conduit has a fill capacity of 0.355 in², which is **not sufficient** for our total area of 0.4981 in².\n - The 1-1/4 inch RMC conduit has a fill capacity of 0.61 in², which **is sufficient** for our total area.\n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.61 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 17 #14 RHW-2 conductors, you should use at least a 1-1/4 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch RMC conduit will be 82% full with 17 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *18* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *18 * 0.0293 in² = 0.5274 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1 inch RMC:** *0.355 in²* \n**1-1/4 inch RMC:** *0.61 in²* \n**1-1/2 inch RMC:** *0.829 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.5274}{0.61} \approx 0.86$$**1-1/4 inch RMC conduit is 86% full with 18 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 18 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 18 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 18 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 18 \times 0.0293 = 0.5274 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1 inch RMC conduit has a fill capacity of 0.355 in², which is **not sufficient** for our total area of 0.5274 in².\n - The 1-1/4 inch RMC conduit has a fill capacity of 0.61 in², which **is sufficient** for our total area.\n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.61 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 18 #14 RHW-2 conductors, you should use at least a 1-1/4 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch RMC conduit will be 86% full with 18 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *30* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *30 * 0.0293 in² = 0.8790 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**2-1/2 inch RMC:** *1.946 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.8790}{1.363} \approx 0.64$$**2 inch RMC conduit is 64% full with 30 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 30 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 30 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 30 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 30 \times 0.0293 = 0.8790 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which is **not sufficient** for our total area of 0.8790 in².\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is sufficient** for our total area.\n - The 2-1/2 inch RMC conduit has a fill capacity of 1.946 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.363 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 30 #14 RHW-2 conductors, you should use at least a 2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch RMC conduit will be 64% full with 30 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *17* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *17 * 0.0293 in² = 0.4981 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1 inch RMC:** *0.355 in²* \n**1-1/4 inch RMC:** *0.61 in²* \n**1-1/2 inch RMC:** *0.829 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{0.4981}{0.61} \approx 0.82$$**1-1/4 inch RMC conduit is 82% full with 17 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 17 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 17 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 17 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 17 \times 0.0293 = 0.4981 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1 inch RMC conduit has a fill capacity of 0.355 in², which is **not sufficient** for our total area of 0.4981 in².\n - The 1-1/4 inch RMC conduit has a fill capacity of 0.61 in², which **is sufficient** for our total area.\n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 0.61 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 17 #14 RHW-2 conductors, you should use at least a 1-1/4 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch RMC conduit will be 82% full with 17 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#14* \n**Number of Conductors:** *35* \n**Conduit Type:** *RMC* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0293 in²* \n**Total Area occupied by conductors:** *35 * 0.0293 in² = 1.0255 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)):</strong></u> \n**1-1/2 inch RMC:** *0.829 in²* \n**2 inch RMC:** *1.363 in²* \n**2-1/2 inch RMC:** *1.946 in²* \n**Link:** [Chapter 9 Table 4(RMC)](https://2023.antflix.net#table_4(RMC)) \n \n$$\text{Space occupied by conductors} = \frac{1.0255}{1.363} \approx 0.75$$**2 inch RMC conduit is 75% full with 35 #14 RHH, RHW, RHW-2 wires**\nQuestion: What size RMC conduit is needed to run 35 #14 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 35 #14 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #14 RHW-2 conductor is 0.0293 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 35 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 35 \times 0.0293 = 1.0255 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)), we can see the following: \n - The 1-1/2 inch RMC conduit has a fill capacity of 0.829 in², which is **not sufficient** for our total area of 1.0255 in².\n - The 2 inch RMC conduit has a fill capacity of 1.363 in², which **is sufficient** for our total area.\n - The 2-1/2 inch RMC conduit has a fill capacity of 1.946 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch RMC conduit.\n* The maximum allowable fill capacity for this conduit size is 1.363 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(RMC)).\n\n**Conclusion** \nTo safely install 35 #14 RHW-2 conductors, you should use at least a 2 inch RMC conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch RMC conduit will be 75% full with 35 #14 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *24* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *24 * 0.0353 in² = 0.8472 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.8472}{1.342} \approx 0.63$$**2 inch EMT conduit is 63% full with 24 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 24 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 24 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 24 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 24 \times 0.0353 = 0.8472 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 0.8472 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 24 #12 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 63% full with 24 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *8* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *8 * 0.0353 in² = 0.2824 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**3/4 inch EMT:** *0.213 in²* \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.2824}{0.346} \approx 0.82$$**1 inch EMT conduit is 82% full with 8 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 8 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 8 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 8 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 8 \times 0.0353 = 0.2824 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 3/4 inch EMT conduit has a fill capacity of 0.213 in², which is **not sufficient** for our total area of 0.2824 in².\n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which **is sufficient** for our total area.\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.346 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 8 #12 RHW-2 conductors, you should use at least a 1 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch EMT conduit will be 82% full with 8 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *35* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *35 * 0.0353 in² = 1.2355 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{1.2355}{1.342} \approx 0.92$$**2 inch EMT conduit is 92% full with 35 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 35 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 35 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 35 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 35 \times 0.0353 = 1.2355 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 1.2355 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 35 #12 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 92% full with 35 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *6* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *6 * 0.0353 in² = 0.2118 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1/2 inch EMT:** *0.122 in²* \n**3/4 inch EMT:** *0.213 in²* \n**1 inch EMT:** *0.346 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.2118}{0.213} \approx 0.99$$**3/4 inch EMT conduit is 99% full with 6 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 6 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 6 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 6 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 6 \times 0.0353 = 0.2118 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1/2 inch EMT conduit has a fill capacity of 0.122 in², which is **not sufficient** for our total area of 0.2118 in².\n - The 3/4 inch EMT conduit has a fill capacity of 0.213 in², which **is sufficient** for our total area.\n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 3/4 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.213 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 6 #12 RHW-2 conductors, you should use at least a 3/4 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 3/4 inch EMT conduit will be 99% full with 6 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *9* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *9 * 0.0353 in² = 0.3177 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**3/4 inch EMT:** *0.213 in²* \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.3177}{0.346} \approx 0.92$$**1 inch EMT conduit is 92% full with 9 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 9 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 9 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 9 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 9 \times 0.0353 = 0.3177 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 3/4 inch EMT conduit has a fill capacity of 0.213 in², which is **not sufficient** for our total area of 0.3177 in².\n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which **is sufficient** for our total area.\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.346 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 9 #12 RHW-2 conductors, you should use at least a 1 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch EMT conduit will be 92% full with 9 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *11* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *11 * 0.0353 in² = 0.3883 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.3883}{0.598} \approx 0.65$$**1-1/4 inch EMT conduit is 65% full with 11 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 11 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 11 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 11 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 11 \times 0.0353 = 0.3883 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which is **not sufficient** for our total area of 0.3883 in².\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is sufficient** for our total area.\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.598 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 11 #12 RHW-2 conductors, you should use at least a 1-1/4 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch EMT conduit will be 65% full with 11 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *9* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *9 * 0.0353 in² = 0.3177 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**3/4 inch EMT:** *0.213 in²* \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.3177}{0.346} \approx 0.92$$**1 inch EMT conduit is 92% full with 9 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 9 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 9 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 9 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 9 \times 0.0353 = 0.3177 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 3/4 inch EMT conduit has a fill capacity of 0.213 in², which is **not sufficient** for our total area of 0.3177 in².\n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which **is sufficient** for our total area.\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.346 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 9 #12 RHW-2 conductors, you should use at least a 1 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch EMT conduit will be 92% full with 9 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *9* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *9 * 0.0353 in² = 0.3177 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**3/4 inch EMT:** *0.213 in²* \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.3177}{0.346} \approx 0.92$$**1 inch EMT conduit is 92% full with 9 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 9 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 9 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 9 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 9 \times 0.0353 = 0.3177 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 3/4 inch EMT conduit has a fill capacity of 0.213 in², which is **not sufficient** for our total area of 0.3177 in².\n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which **is sufficient** for our total area.\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.346 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 9 #12 RHW-2 conductors, you should use at least a 1 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1 inch EMT conduit will be 92% full with 9 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *15* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *15 * 0.0353 in² = 0.5295 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1 inch EMT:** *0.346 in²* \n**1-1/4 inch EMT:** *0.598 in²* \n**1-1/2 inch EMT:** *0.814 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.5295}{0.598} \approx 0.89$$**1-1/4 inch EMT conduit is 89% full with 15 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 15 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 15 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 15 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 15 \times 0.0353 = 0.5295 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1 inch EMT conduit has a fill capacity of 0.346 in², which is **not sufficient** for our total area of 0.5295 in².\n - The 1-1/4 inch EMT conduit has a fill capacity of 0.598 in², which **is sufficient** for our total area.\n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.598 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 15 #12 RHW-2 conductors, you should use at least a 1-1/4 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch EMT conduit will be 89% full with 15 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *28* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *28 * 0.0353 in² = 0.9884 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{0.9884}{1.342} \approx 0.74$$**2 inch EMT conduit is 74% full with 28 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 28 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 28 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 28 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 28 \times 0.0353 = 0.9884 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 0.9884 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 28 #12 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 74% full with 28 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *34* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *34 * 0.0353 in² = 1.2002 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**1-1/2 inch EMT:** *0.814 in²* \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{1.2002}{1.342} \approx 0.89$$**2 inch EMT conduit is 89% full with 34 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 34 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 34 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 34 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 34 \times 0.0353 = 1.2002 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 1-1/2 inch EMT conduit has a fill capacity of 0.814 in², which is **not sufficient** for our total area of 1.2002 in².\n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which **is sufficient** for our total area.\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.342 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 34 #12 RHW-2 conductors, you should use at least a 2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch EMT conduit will be 89% full with 34 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *39* \n**Conduit Type:** *EMT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *39 * 0.0353 in² = 1.3767 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)):</strong></u> \n**2 inch EMT:** *1.342 in²* \n**2-1/2 inch EMT:** *2.343 in²* \n**3 inch EMT:** *3.538 in²* \n**Link:** [Chapter 9 Table 4(EMT)](https://2023.antflix.net#table_4(EMT)) \n \n$$\text{Space occupied by conductors} = \frac{1.3767}{2.343} \approx 0.59$$**2-1/2 inch EMT conduit is 59% full with 39 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size EMT conduit is needed to run 39 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 39 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 39 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 39 \times 0.0353 = 1.3767 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)), we can see the following: \n - The 2 inch EMT conduit has a fill capacity of 1.342 in², which is **not sufficient** for our total area of 1.3767 in².\n - The 2-1/2 inch EMT conduit has a fill capacity of 2.343 in², which **is sufficient** for our total area.\n - The 3 inch EMT conduit has a fill capacity of 3.538 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2-1/2 inch EMT conduit.\n* The maximum allowable fill capacity for this conduit size is 2.343 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(EMT)).\n\n**Conclusion** \nTo safely install 39 #12 RHW-2 conductors, you should use at least a 2-1/2 inch EMT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2-1/2 inch EMT conduit will be 59% full with 39 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *18* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *18 * 0.0353 in² = 0.6354 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.6354}{0.794} \approx 0.80$$**1-1/2 inch ENT conduit is 80% full with 18 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 18 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 18 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 18 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 18 \times 0.0353 = 0.6354 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which is **not sufficient** for our total area of 0.6354 in².\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is sufficient** for our total area.\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.794 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 18 #12 RHW-2 conductors, you should use at least a 1-1/2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch ENT conduit will be 80% full with 18 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *31* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *31 * 0.0353 in² = 1.0943 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{1.0943}{1.316} \approx 0.83$$**2 inch ENT conduit is 83% full with 31 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 31 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 31 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 31 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 31 \times 0.0353 = 1.0943 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which is **not sufficient** for our total area of 1.0943 in².\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.316 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 31 #12 RHW-2 conductors, you should use at least a 2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch ENT conduit will be 83% full with 31 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *17* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *17 * 0.0353 in² = 0.6001 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.6001}{0.794} \approx 0.76$$**1-1/2 inch ENT conduit is 76% full with 17 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 17 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 17 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 17 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 17 \times 0.0353 = 0.6001 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which is **not sufficient** for our total area of 0.6001 in².\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is sufficient** for our total area.\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.794 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 17 #12 RHW-2 conductors, you should use at least a 1-1/2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch ENT conduit will be 76% full with 17 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *33* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *33 * 0.0353 in² = 1.1649 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{1.1649}{1.316} \approx 0.89$$**2 inch ENT conduit is 89% full with 33 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 33 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 33 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 33 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 33 \times 0.0353 = 1.1649 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which is **not sufficient** for our total area of 1.1649 in².\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 1.316 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 33 #12 RHW-2 conductors, you should use at least a 2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 2 inch ENT conduit will be 89% full with 33 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *18* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *18 * 0.0353 in² = 0.6354 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**2 inch ENT:** *1.316 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.6354}{0.794} \approx 0.80$$**1-1/2 inch ENT conduit is 80% full with 18 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 18 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 18 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 18 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 18 \times 0.0353 = 0.6354 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which is **not sufficient** for our total area of 0.6354 in².\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is sufficient** for our total area.\n - The 2 inch ENT conduit has a fill capacity of 1.316 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/2 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.794 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 18 #12 RHW-2 conductors, you should use at least a 1-1/2 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/2 inch ENT conduit will be 80% full with 18 #12 RHH, RHW, RHW-2 wires.**
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Context: \n<u><strong>User Provided Information:</strong></u> \n**Conductor Type:** *RHH, RHW, RHW-2* \n**Conductor Size:** *#12* \n**Number of Conductors:** *14* \n**Conduit Type:** *ENT* \n \n<u><strong>Conductor Information from [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)):</strong></u> \n**Conductor Area:** *0.0353 in²* \n**Total Area occupied by conductors:** *14 * 0.0353 in² = 0.4942 in²* \n**Link:** [Chapter 9 Table 5(RHH, RHW, RHW-2)](https://2023.antflix.net#table_5(RHH)) \n \n<u><strong>Conduit Information from [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)):</strong></u> \n**1 inch ENT:** *0.333 in²* \n**1-1/4 inch ENT:** *0.581 in²* \n**1-1/2 inch ENT:** *0.794 in²* \n**Link:** [Chapter 9 Table 4(ENT)](https://2023.antflix.net#table_4(ENT)) \n \n$$\text{Space occupied by conductors} = \frac{0.4942}{0.581} \approx 0.85$$**1-1/4 inch ENT conduit is 85% full with 14 #12 RHH, RHW, RHW-2 wires**\nQuestion: What size ENT conduit is needed to run 14 #12 RHH, RHW, RHW-2 wires?
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To determine the appropriate conduit size for 14 #12 RHW-2 conductors, follow these steps:\n\n**Step 1: Identify the Conductor Area** \n* The area for one #12 RHW-2 conductor is 0.0353 in². \n* This value can be found in [Chapter 9, Table 5](https://2023.antflix.net#table_5(RHH)) of the NEC codebook.\n\n**Step 2: Calculate the Total Area Occupied by Conductors** \n* The total area occupied by 14 conductors is calculated as: \n $$ \text{Total Area} = \text{Number of Conductors} \times \text{Conductor Area} $$\n $$ \text{Total Area} = 14 \times 0.0353 = 0.4942 \text{ in²} $$\n\n**Step 3: Determine the Required Conduit Size** \n* Looking at [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)), we can see the following: \n - The 1 inch ENT conduit has a fill capacity of 0.333 in², which is **not sufficient** for our total area of 0.4942 in².\n - The 1-1/4 inch ENT conduit has a fill capacity of 0.581 in², which **is sufficient** for our total area.\n - The 1-1/2 inch ENT conduit has a fill capacity of 0.794 in², which **is also sufficient** for our total area.\n \n* The minimum required conduit size that can accommodate this total area, without exceeding the NEC's 40% fill capacity, is the 1-1/4 inch ENT conduit.\n* The maximum allowable fill capacity for this conduit size is 0.581 in², which can be found in [Chapter 9, Table 4](https://2023.antflix.net#table_4(ENT)).\n\n**Conclusion** \nTo safely install 14 #12 RHW-2 conductors, you should use at least a 1-1/4 inch ENT conduit. This will ensure that the fill capacity does not exceed the NEC's 40% limit.\n\n**The 1-1/4 inch ENT conduit will be 85% full with 14 #12 RHH, RHW, RHW-2 wires.**
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