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V b = E Rx Rx+Ry V b = 15V 2.222kΩ 2.222kΩ+1 kΩ V b ≈10.34V The second sub-circuit will use the 6 volt source. Therefore, the 15 volt source will be replaced with its ideal internal resistance, a short. This new circuit is shown in Figure 6.13. The current directions are as follows: current exits the source and travels... | DCElectricalCircuitAnalysis_Page_179_Chunk1101 |
Computer Simulation Computer Simulation For further verification, the circuit of Example 6.4 is entered into a simulator as shown in Figure 6.14. A DC operating point analysis is performed, as shown in Figure 6.15. The voltage at node 2, which is Vb in the original circuit, agrees nicely with the values computed previo... | DCElectricalCircuitAnalysis_Page_180_Chunk1102 |
6.4 Thévenin's Theorem 6.4 Thévenin's Theorem Thévenin's theorem, named after Léon Charles Thévenin, is a powerful analysis tool. For DC, it states: Any single port linear network can be reduced to a simple voltage source, Eth, in series with an internal resistance, Rth. An example is shown in Figure 6.16. The phrase “... | DCElectricalCircuitAnalysis_Page_181_Chunk1103 |
As noted earlier, the original circuit could be cut in a number of different ways. We might, for example, want to determine the Thévenin equivalent that drives R2 in the circuit of Figure 6.17. This cut appears in Figure 6.20. Clearly, this will result in different values for both Eth and Rth. For example, Rth is now R... | DCElectricalCircuitAnalysis_Page_182_Chunk1104 |
Whether determined analytically or empirically, the Thévenin equivalent circuit can replace the original single port network regardless of what the original was connected to. The same voltages and currents will be seen in this other portion, and it won't matter if the other portion is comprised of a single resistor, mu... | DCElectricalCircuitAnalysis_Page_183_Chunk1105 |
It might seem that the Thévenin method is the “long way home” in this example, and it is, but it has the advantage of being more efficient if several different loads are being considered. For example, suppose we decided to determine the output voltage not just for the 1 kΩ, but for a group of a half dozen different res... | DCElectricalCircuitAnalysis_Page_184_Chunk1106 |
As a further check, the simulation is run again, but this time using an alternate load resistor value. The value chosen is the Thévenin resistance value. By matching the resistances, this should produce a 50/50 voltage divider and the load voltage should equal half of the Thévenin source voltage, or approximately 4.444... | DCElectricalCircuitAnalysis_Page_185_Chunk1107 |
Example 6.6 Determine the Thévenin equivalent of the circuit driving the resistor/voltage source combo in Figure 6.26. Verify that the equivalent produces the same voltage across this resistor as the original circuit. Determining Rth is not particularly difficult here. After shorting the 10 volt source and open the cur... | DCElectricalCircuitAnalysis_Page_186_Chunk1108 |
6.5 Norton's Theorem 6.5 Norton's Theorem Norton's theorem is credited to Edward Lawry Norton. In a nutshell, Norton's theorem is the current source version of Thévenin's theorem. That is, a single port DC network can be reduced to a single current source, IN, with parallel internal resistance, RN. Indeed, once you und... | DCElectricalCircuitAnalysis_Page_187_Chunk1109 |
6.6 Maximum Power Transfer Theorem 6.6 Maximum Power Transfer Theorem Given a simple voltage source with internal resistance, a useful question to ask is “What value of load resistance will yield the maximum amount of power in the load?” While it is not true that maximizing load power is a goal of all circuit designs, ... | DCElectricalCircuitAnalysis_Page_188_Chunk1110 |
then added together as illustrated in Figure 6.29. Of course, we must remember that we really want 1 / ( R2 + 2R + 1) and so we plot the reciprocal of the combo as shown in Figure 6.30 (red curve). We also include the numerator term, R. This is shown as a straight line (blue) with a slope of 1. Finally, these two curve... | DCElectricalCircuitAnalysis_Page_189_Chunk1111 |
A close examination of the power curve will show that the peak occurs at R = 1. This is easier to see if we plot the completed power curve using a logarithmic horizontal axis and also scale the vertical axis to 100%, as shown in Figure 6.31. The peak is more apparent and the curve is symmetrical in shape rather than lo... | DCElectricalCircuitAnalysis_Page_190_Chunk1112 |
product, however, is at the maximum. Further, efficiency at maximum load power is only 50% (i.e., only half of all generated power goes to the load with the other half being wasted internally). Values of R greater than Ri will achieve higher efficiency but at reduced load power. Sometimes we favor efficiency over maxim... | DCElectricalCircuitAnalysis_Page_191_Chunk1113 |
It is possible to convert back and forth between delta and Y networks. That is, for every delta network, there exists a Y network such that the resistances seen between the X, Y and Z terminals are identical, and vice versa. Consequently, one configuration can replace another in order to simplify a larger circuit. Δ-Y ... | DCElectricalCircuitAnalysis_Page_192_Chunk1114 |
Y-Δ Conversion Y-Δ Conversion For the reverse process of converting Y to delta, start by noting the similarities of the expressions for Rd, Re and Rf. If two of these expressions are divided, a single equation for Ra, Rb or Rc will result. For example, using Equations 6.4 and 6.5: Rd Re = Ra Rb Ra+Rb+Rc Ra Rc Ra+Rb+Rc ... | DCElectricalCircuitAnalysis_Page_193_Chunk1115 |
Example 6.7 Determine the equivalent of the bridge circuit shown in Figure 6.34. This circuit uses a five resistor bridge which cannot be further simplified using basic series-parallel combinations. As a result, the techniques presented in Chapter 5 will not be sufficient to obtain a solution. A delta-Y conversion is p... | DCElectricalCircuitAnalysis_Page_194_Chunk1116 |
The equivalent network is placed back into the original circuit as shown in Figure 6.35 (red replaces blue). We now have a 333.333 Ω in series with 2.555556 kΩ || 5.666667 kΩ. To determine Vb and Vc the total equivalent resistance can be used to find the source current. From there, a current divider may be employed alo... | DCElectricalCircuitAnalysis_Page_195_Chunk1117 |
I 2k = I 1k −I5k I 2k ≈4.1032 mA −0.1548mA I 2k ≈3.9484 mA This result matches the current Ib calculated previously. For completeness sake, the process can be replicated for node c. Computer Simulation Computer Simulation For further verification, both the original circuit of Figure 6.34 and the converted circuit of Fi... | DCElectricalCircuitAnalysis_Page_196_Chunk1118 |
The voltages match each other perfectly, along with the manual calculation. In closing, we see that it is possible to swap delta networks with Y networks, and achieve identical results. 6.8 Summary 6.8 Summary In this chapter we have examined several new techniques and theorems to assist with the analysis of DC electri... | DCElectricalCircuitAnalysis_Page_197_Chunk1119 |
Thévenin's and Norton's theorems allow the simplification of complex linear single port (i.e., two connecting points) networks. The Thévenin equivalent consists of a voltage source with series a resistance while the Norton equivalent consists of a current source with a parallel resistance. These equivalents, when repla... | DCElectricalCircuitAnalysis_Page_198_Chunk1120 |
6.9 Exercises 6.9 Exercises Analysis Analysis 1. For the circuit shown in Figure 6.38, determine the equivalent current source. 2. Given the circuit shown in Figure 6.39, determine the equivalent current source. 3. Determine the equivalent current source for the circuit shown in Figure 6.40. 4. For the circuit shown in... | DCElectricalCircuitAnalysis_Page_199_Chunk1121 |
5. For the circuit shown in Figure 6.42, determine the equivalent voltage source. 6. Given the circuit shown in Figure 6.43, determine the equivalent voltage source. 7. Determine the equivalent voltage source for the circuit shown in Figure 6.44. 8. For the circuit shown in Figure 6.45, determine the equivalent voltage... | DCElectricalCircuitAnalysis_Page_200_Chunk1122 |
9. Given the circuit shown in Figure 6.46, determine the equivalent voltage source. 10. Using source conversion, find Vb for the circuit shown in Figure 6.47. 11. Using source conversion, find the current through the 3 kΩ resistor in the circuit of Figure 6.48. 12. Using source conversion, find Vb for the circuit shown... | DCElectricalCircuitAnalysis_Page_201_Chunk1123 |
13. Using source conversion, find Va for the circuit shown in Figure 6.50. 14. Using source conversion, find Vb for the circuit shown in Figure 6.51. 15. Using superposition, determine Vb for the circuit shown in Figure 6.47. 16. Using superposition, find the current through the 3 kΩ resistor for the circuit of Figure ... | DCElectricalCircuitAnalysis_Page_202_Chunk1124 |
25. Using superposition, find the current through the 1.5 kΩ resistor for the circuit of Figure 6.52. 26. Using superposition, determine Vab for the circuit shown in Figure 6.52. 27. Using superposition, determine Vb for the circuit shown in Figure 6.53. 28. Using superposition, find the current through the 200 Ω resis... | DCElectricalCircuitAnalysis_Page_203_Chunk1125 |
32. Using superposition, determine Vbc for the circuit shown in Figure 6.55. 33. Using superposition, find the current through the 10 kΩ resistor for the circuit of Figure 6.55. 34. Using superposition, find the currents through the 100 Ω and 700 Ω resistors for the circuit shown in Figure 6.56. 35. Using superposition... | DCElectricalCircuitAnalysis_Page_204_Chunk1126 |
38. Using superposition, find the current through the 20 kΩ resistor for the circuit shown in Figure 6.57. 39. Using superposition, find the current through the 12 kΩ resistor for the circuit shown in Figure 6.58. 40. Using superposition, determine Vb for the circuit shown in Figure 6.58. 41. Using superposition, deter... | DCElectricalCircuitAnalysis_Page_205_Chunk1127 |
45. Given the circuit shown in Figure 6.61, determine the Thévenin equivalent circuit that is driving the 4 kΩ resistor. 46. Given the circuit shown in Figure 6.61, determine the Norton equivalent circuit driving the 4 kΩ resistor. 47. Given the circuit shown in Figure 6.61, determine the Norton equivalent circuit driv... | DCElectricalCircuitAnalysis_Page_206_Chunk1128 |
51. Given the circuit shown in Figure 6.64, determine the Norton equivalent circuit driving the 40 Ω resistor. 52. Determine the Thévenin equivalent circuit driving the 10 Ω resistor for the circuit shown in Figure 6.64. 53. Given the circuit shown in Figure 6.65, determine the Norton equivalent circuit driving the 12 ... | DCElectricalCircuitAnalysis_Page_207_Chunk1129 |
Design Design 58. Consider the 4 kΩ resistor to be the load in Figure 6.61. Determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 59. Consider the 12 kΩ resistor to be the load in Figure 6.62. Determine a new value for the load in order to achieve maximum load... | DCElectricalCircuitAnalysis_Page_208_Chunk1130 |
67. Convert the pi network of Figure 6.67 into a T network. 68. Convert the Y network of Figure 6.68 into a delta network. 69. Convert the T network of Figure 6.69 into a pi network. 209 Figure 6.67 Figure 6.68 Figure 6.69 | DCElectricalCircuitAnalysis_Page_209_Chunk1131 |
Challenge Challenge 70. Redesign the circuit of Figure 6.70 so that it uses only current sources and produces the same node voltages as the original circuit. 71. Using any combination of techniques, find the current through the 9 kΩ resistor for the circuit of Figure 6.70. 72. Using any combination of techniques, deter... | DCElectricalCircuitAnalysis_Page_210_Chunk1132 |
75. Using superposition, find Vab in the circuit of Figure 6.71. 76. Redesign the circuit of Figure 6.71 using only voltage sources so that it achieves the same node voltages as the original. 77. Is it possible to determine Vbc in Figure 6.72 by using just source conversions or just superposition? Why/why not? 78. Usin... | DCElectricalCircuitAnalysis_Page_211_Chunk1133 |
Simulation Simulation 84. Using DC bias simulations, compare the original circuit of problem 1 to its converted equivalent. Do this by connecting a resistor to the output terminals, trying several different resistance values and checking to see if the two circuits always produce the same voltage across this resistor. 8... | DCElectricalCircuitAnalysis_Page_212_Chunk1134 |
96. Using either Monte Carlo simulation or multiple DC bias simulations, verify that the resistance calculated in problem 56 achieves maximum power. Do this by trying several resistor values near the calculated value, and determining the power for each based on the squared load voltages. This XKCD comic is too large to... | DCElectricalCircuitAnalysis_Page_213_Chunk1135 |
7 7 Nodal & Mesh Analysis Nodal & Mesh Analysis, Dependent Sources , Dependent Sources 7.0 Chapter Learning Objectives 7.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize nodal analysis techniques to solve for voltages in multi-source series-parallel networks. • Utilize mesh ... | DCElectricalCircuitAnalysis_Page_214_Chunk1136 |
7.2 Nodal Analysis 7.2 Nodal Analysis Nodal analysis is a technique that can be applied to virtually any circuit. In general use, it might be considered a universal solution technique as there are no practical circuit configurations that it cannot handle. Nodal analysis relies on the application of Kirchhoff's current ... | DCElectricalCircuitAnalysis_Page_215_Chunk1137 |
Next, we describe these currents in terms of the node voltages and associated components via Ohm's law. For example, I3 is the node b voltage divided by R3 while I1 is the voltage across R1 divided by R1. This voltage is Va − Vb. Therefore, V a −V b R1 +V c −V b R2 = V b R3 Noting that Va = E1 and Vc = E2, with a littl... | DCElectricalCircuitAnalysis_Page_216_Chunk1138 |
The currents are then described by their Ohm's law equivalents: Node a: I 1 = V a −V b R3 +V a R1 Node b: −I 2 =−V a −V b R3 +V b R2 Expanding and collecting terms yields: Node a: I 1 =( 1 R1 + 1 R3)V a −( 1 R3)V b Node b: −I 2 =−( 1 R3)V a +( 1 R3 + 1 R2)V b As the resistor values and currents are known, simultaneous ... | DCElectricalCircuitAnalysis_Page_217_Chunk1139 |
and solving for Vb, V b = 16.5 mA 1mS +0.25mS +0.2mS V b = 11.379volts As Vb is higher than the 6 volt source, our assumed current direction for the 4 kΩ resistor was incorrect; we assumed right to left but it is in fact left to right, flowing from 11.379 volts to 6 volts. As you can see, this did not present a problem... | DCElectricalCircuitAnalysis_Page_218_Chunk1140 |
4. Sum the current sources feeding the node of interest. Entering is deemed positive while exiting is deemed negative. The sum is placed on one side of the equals sign. 5. Next, find all of the resistors connected to the node of interest and write them as a sum of conductances on the other side of the equals sign, the ... | DCElectricalCircuitAnalysis_Page_219_Chunk1141 |
so far. Repeat for all remaining nodes except the ground reference. In this example there is only one other node, node b, and thus only one iteration. 800 mA −2 A =( 1 10Ω + 1 50Ω + 1 100Ω)V a −( 1 50Ω + 1 100 Ω)V b Finally, simplify the constants and coefficients, and the first expression is complete: −1.2 A = 130mSV ... | DCElectricalCircuitAnalysis_Page_220_Chunk1142 |
Let's verify that these values are correct. We can perform a KCL summation at node a and see if it balances. We already know that 2 amps and 75.6 milliamps exit while 800 milliamps enters. We only need to find the currents through the 10 Ω and 50 Ω resistors. First, notice that the 50 Ω sees the same voltage as the 100... | DCElectricalCircuitAnalysis_Page_221_Chunk1143 |
1.5A = (0.2S +0.25S)V a −(0.25S)V b −(0)V c Finally, simplify the constants and coefficients, and the first expression is complete: 1.5A = 0.45SV a −0.25SV b −0V c We leave the zero term in just for alignment. Now we repeat the entire process for the next equation. The node of interest is now node b. Here is the result... | DCElectricalCircuitAnalysis_Page_222_Chunk1144 |
Here is the trap: in the converted circuit, although R1 still connects to node a, the other end no longer connects to the voltage source. Rather, the right side now connects to node c. Therefore, the voltage drop across R1 in the converted circuit is not likely to equal the voltage drop seen across R1 in the original c... | DCElectricalCircuitAnalysis_Page_223_Chunk1145 |
In this version we have replaced the voltage source with its ideal internal resistance, a short. We have also labeled the two nodes of interest, a and b, and labeled the currents, drawn with convenient directions. Due to the shorted voltage source, nodes a and b are now the same node. Consider the currents entering and... | DCElectricalCircuitAnalysis_Page_224_Chunk1146 |
Example 7.4 Find Va and Vb for the circuit of Figure 7.13. As shown in Figure 7.14, we short the 60 volt source and write a current summation at the a b supernode: Σ Iin = Σ I out 1A+I 1+I 2 = 2.5 A+I 1+I 2+I 3+I 4 This can be simplified to: −1.5 A = I 3 +I 4 Writing this in terms of Ohm's law we have: −1.5A = 1 4ΩV a ... | DCElectricalCircuitAnalysis_Page_225_Chunk1147 |
Doing likewise for node b, and assuming I1 enters as drawn: I 1 = 2.5 A+ V b 10Ω −V a−V b 20Ω I 1 = 2.5 A+−47.143 V 10Ω −12.857V−(−47.143V) 20Ω I 1 =−5.2143A (negative enter means it's exiting) These currents match meaning that the current through the voltage source is verified to be the same at both terminals, as it m... | DCElectricalCircuitAnalysis_Page_226_Chunk1148 |
On to node b: 1A = I 3+I 4 1A = 0.5S(V b −V a)+0.25S(V b −V c) 1A = 0.5S(V b −V a)+0.25S(V b −(V a −8V)) 1A = 0.5S(V b −V a)+0.25S(V b −V a+8 V) −1 A =−0.75V a+0.75V b And finally node c: I 4 = I1+I5 I 1 = I 4 −I 5 I 1 = 0.25S(V b −V c)−0.1SV c I 1 = 0.25S(V b −(V a−8V))−0.1S(V a −8 V) I 1 = 0.25S(V b −V a+8 V)−0.1S(V ... | DCElectricalCircuitAnalysis_Page_227_Chunk1149 |
Computer Simulation Computer Simulation In order to verify the result of Example 7.5, the circuit is entered into a simulator as shown in Figure 7.17. A DC operating point simulation is run. The results are shown in Figure 7.18 and match the calculated values perfectly. Node 1 corresponds to Vb, node 2 corresponds to V... | DCElectricalCircuitAnalysis_Page_228_Chunk1150 |
7.3 Mesh Analysis 7.3 Mesh Analysis In some respects mesh analysis is a mirror of nodal analysis. While nodal analysis leverages KCL to create a series of node equations that are used to solve for node voltages, mesh analysis uses KVL to create a series of loop equations that can be solved for mesh currents. A mesh cur... | DCElectricalCircuitAnalysis_Page_229_Chunk1151 |
Expand the voltage terms using Ohm's law. Loop 1: E1 = I1 ∙ R1 + (I1 − I2) R3 Loop 2: −E2 = I2 ∙ R2 + (I2 − I1) R3 Expanding and collecting terms yields: Loop 1: E1 = (R1 + R3) I1 − R3 ∙ I2 Loop 2: −E2 = R3 ∙ I1 + (R2 + R3) I2 Assuming that the resistor values and source voltages are known, we have two equations with t... | DCElectricalCircuitAnalysis_Page_230_Chunk1152 |
Loop 1: 12 V = 1k(I 1 −I2)+2 k(I1 −I 2) Loop 2: 0V = 3 k(I2)+5k(I 2 −I 3)+1k(I 2 −I 1) Loop 3: 0V = 5 k(I 3 −I 2)+4k(I 3)+2k(I 3 −I1) The terms are expanded: Loop 1: 12 V = 1k I 1 −1 k I 2+2 k I 1 −2 k I 3 Loop 2: 0V = 3 k I 2+5k I 2 −5k I3+1k I 2 −1 k I 1 Loop 3: 0V = 5 k I 3 −5k I 2+4k I 3+2k I 3 −2k I1 Like terms ar... | DCElectricalCircuitAnalysis_Page_231_Chunk1153 |
simulation will be a randomly generated value between 950 Ω and 1050 Ω. We then state how many trials we'd like, each one using their own unique randomized values for each component. The results of a Monte Carlo analysis using the DC operating point simulation is shown in Figure 7.23. Ten randomized trials were generat... | DCElectricalCircuitAnalysis_Page_232_Chunk1154 |
Inspection Method Inspection Method Like nodal analysis, there is a method to generate the equations by inspection. Simply focus on one loop, which we'll call the loop under inspection, and ask the following questions: what is the total source voltage in this loop? This yields the voltage constant for the equation. The... | DCElectricalCircuitAnalysis_Page_233_Chunk1155 |
Example 7.7 Find Vc for the circuit of Figure 7.24. First, we'll label the loops, as shown in Figure 7.25. Now, beginning with loop 1, sum all of the voltage sources in this loop. That's +50 volts. Secondly, sum all of the resistors in this loop. That's 100 Ω + 200 Ω, or 300 Ω. This is the coefficient for the I1 term. ... | DCElectricalCircuitAnalysis_Page_234_Chunk1156 |
Supermesh Supermesh Sometimes you may run across a current source which has no associated internal resistance, such as found in the circuit of Figure 7.26. This is similar to the situation seen previously under nodal analysis where a voltage source does not have a specified internal resistance. There are two ways of so... | DCElectricalCircuitAnalysis_Page_235_Chunk1157 |
We now perform a KVL summation around the supermesh loop, similar to what we have done previously. The difference this time is that we need to recognize that components each see one of the original mesh currents; namely I1 or I2 in this case. We do not solve for a supermesh current, we simply use the supermesh to defin... | DCElectricalCircuitAnalysis_Page_236_Chunk1158 |
By inspection we know 3A = I2 −I 1 or I 2 = I 1+3A We can substitute this expression into the prior supermesh expression and solve for I1: 8 V = 5ΩI 1+8ΩI 2 8 V = 5ΩI 1+8Ω(I1+3A) 8 V = 5ΩI 1+8ΩI1+24 V −16 V = 13Ω I 1 I 1 ≈−1.231A Thus, I2 = −1.231 A + 3 A, or 1.769 A. To determine Vb we simply subtract the drop across ... | DCElectricalCircuitAnalysis_Page_237_Chunk1159 |
7.4 Dependent Sources 7.4 Dependent Sources A dependent source is a current or voltage source whose value is not fixed (i.e., independent) but rather which depends on some other circuit current or voltage. The general form for the value of a dependent source is Y = kX where X and Y are currents and/or voltages and k is... | DCElectricalCircuitAnalysis_Page_238_Chunk1160 |
In general, there are two possible configurations: isolated and coupled. A example of the isolated form is shown in Figure 7.31. In this configuration, the dependent source (center) does not interact with the sub- circuit on the left driven by the independent source, thus it can be analyzed as two separate circuits. So... | DCElectricalCircuitAnalysis_Page_239_Chunk1161 |
We begin by defining current directions. Assume that the currents through R1 and R3 are flowing into node a, the current through R2 is flowing out of node a, and the current through R4 is flowing out of node b. We shall number the branch currents to reflect the associated resistor. The resulting KCL equations are: Σ I ... | DCElectricalCircuitAnalysis_Page_240_Chunk1162 |
Example 7.9 Find Vb and Vc for the circuit of Figure 7.33. This CCCS would be typical of a simple model of a bipolar junction transistor (nodes a, b, and c). Ideally, the output voltage, Vc, would be equal to the input voltage (1 V) times the resistor ratio of 15 kΩ / 2 kΩ and inverted, or approximately −7.5 volts. In ... | DCElectricalCircuitAnalysis_Page_241_Chunk1163 |
Computer Simulation Computer Simulation For verification, the dependent source circuit of Example 7.9 is entered into a simulator as shown in Figure 7.34. A DC operating point analysis is run, the results of which are shown in Figure 7.35. The output voltage shows roughly −7.08 volts and a Vb of just under 1 volt, thus... | DCElectricalCircuitAnalysis_Page_242_Chunk1164 |
7.5 Summary 7.5 Summary Nodal analysis can be used to solve virtually any complex multi-source DC electrical circuit. It is based on KCL, writing expressions involving each node in the circuit. A system of equations results, there being as many equations as there are nodes in the circuit, minus the reference node (whic... | DCElectricalCircuitAnalysis_Page_243_Chunk1165 |
Review Questions Review Questions 1. Describe the practical differences between nodal analysis and mesh analysis. 2. What is diagonal symmetry? Of what use is it? 3. What are the differences between the general method and the inspection method of nodal analysis? 4. What are the differences between the general method an... | DCElectricalCircuitAnalysis_Page_244_Chunk1166 |
5. Using mesh analysis, determine the value of Vb for the circuit shown in Figure 7.37. 6. For the circuit shown in Figure 7.37, use mesh analysis to determine the current through the 500 Ω resistor. 7. Given the circuit in Figure 7.38, write the mesh loop equations. 8. Using mesh analysis, determine the value of Vb fo... | DCElectricalCircuitAnalysis_Page_245_Chunk1167 |
13. Given the circuit in Figure 7.40, write the mesh loop equations. 14. Using mesh analysis, determine the value of Vac for the circuit shown in Figure 7.40. 15. For the circuit shown in Figure 7.40, use mesh analysis to determine the current through the 4 kΩ resistor. 16. Given the circuit in Figure 7.41, write the m... | DCElectricalCircuitAnalysis_Page_246_Chunk1168 |
20. Using mesh analysis, determine the value of Vbd for the circuit shown in Figure 7.42. 21. For the circuit shown in Figure 7.42, use mesh analysis to determine the current through the 500 Ω resistor. 22. Given the circuit in Figure 7.43, write the mesh loop equations. 23. Using mesh analysis, determine the value of ... | DCElectricalCircuitAnalysis_Page_247_Chunk1169 |
26. Using mesh analysis, determine the value of Ve for the circuit shown in Figure 7.44. 27. For the circuit shown in Figure 7.44, use mesh analysis to determine the current through the 9 kΩ resistor. 28. Given the circuit in Figure 7.45, write the mesh loop equations and the associated determinants. 29. Using mesh ana... | DCElectricalCircuitAnalysis_Page_248_Chunk1170 |
34. For the circuit in Figure 7.47, write the mesh loop equations. 35. Using mesh analysis, determine the value of Va for the circuit shown in Figure 7.47. 36. For the circuit shown in Figure 7.47, use mesh analysis to determine the current through the 3 kΩ resistor. 37. Given the circuit in Figure 7.48, write the mesh... | DCElectricalCircuitAnalysis_Page_249_Chunk1171 |
41. Using mesh analysis, determine the value of Va for the circuit shown in Figure 7.49. 42. For the circuit shown in Figure 7.49, use mesh analysis to determine the current passing through the 30 kΩ resistor. 43. Given the circuit in Figure 7.50, write the mesh loop equations (consider using source conversion). 44. Us... | DCElectricalCircuitAnalysis_Page_250_Chunk1172 |
49. Given the circuit in Figure 7.52, write the node equations. 50. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.52. 51. For the circuit shown in Figure 7.52, use nodal analysis to determine the current passing through the 12 kΩ resistor. 52. Given the circuit in Figure 7.53, write ... | DCElectricalCircuitAnalysis_Page_251_Chunk1173 |
56. Using nodal analysis, determine the value of Vac for the circuit shown in Figure 7.54. 57. For the circuit shown in Figure 7.54, use nodal analysis to determine the current passing through the 20 kΩ resistor. 58. Given the circuit in Figure 7.55, write the node equations. 59. Using nodal analysis, determine the val... | DCElectricalCircuitAnalysis_Page_252_Chunk1174 |
64. Given the circuit in Figure 7.48, write the node equations. 65. Using nodal analysis, determine the value of Vd for the circuit shown in Figure 7.48. 66. For the circuit shown in Figure 7.48, use nodal analysis to determine the current passing through the 20 kΩ resistor. 67. Given the circuit in Figure 7.49, write ... | DCElectricalCircuitAnalysis_Page_253_Chunk1175 |
75. Given the circuit of Figure 7.59, determine Vb. 76. Find the current through the 10 kΩ resistor given the circuit of Figure 7.60. 77. Given the circuit of Figure 7.61, determine Vc. 78. In the circuit of Figure 7.62, determine Va. 254 Figure 7.59 Figure 7.60 Figure 7.61 Figure 7.62 | DCElectricalCircuitAnalysis_Page_254_Chunk1176 |
79. For the circuit of Figure 7.63, determine Va. 80. For the circuit of Figure 7.64, determine Va. Challenge Challenge 81. Given the circuit in Figure 7.43, write the node equations. 82. Using nodal analysis, determine the value of Vb for the circuit shown in Figure 7.43. 83. For the circuit shown in Figure 7.43, use ... | DCElectricalCircuitAnalysis_Page_255_Chunk1177 |
89. For the circuit shown in Figure 7.45, use nodal analysis to determine the current through the 2 kΩ resistor. 90. Given the circuit of Figure 7.65, determine Vc. 91. Find the current through the 10 kΩ resistor in the circuit of Figure 7.66. 92. Given the circuit of Figure 7.67, determine Vc. 256 Figure 7.65 Figure 7... | DCElectricalCircuitAnalysis_Page_256_Chunk1178 |
93. Given the circuit of Figure 7.68, determine the current through the 5 kΩ resistor. 94. For the circuit of Figure 7.69, determine Vb. 95. For the circuit of Figure 7.70, determine Vc. 257 Figure 7.68 Figure 7.69 Figure 7.70 | DCElectricalCircuitAnalysis_Page_257_Chunk1179 |
96. For the circuit of Figure 7.71, determine Va. 97. For the circuit of Figure 7.72, determine Vb. Simulation Simulation 98. Perform a DC bias simulation on the circuit depicted in Figure 7.42 to verify the component currents. 99. Perform a DC bias simulation on the circuit depicted in Figure 7.44 to verify the compon... | DCElectricalCircuitAnalysis_Page_258_Chunk1180 |
Notes Notes ♫♫ ♫♫ 259 | DCElectricalCircuitAnalysis_Page_259_Chunk1181 |
8 8 Capacitors Capacitors 8.0 Chapter Learning Objectives 8.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe the theoretical and practical aspects of capacitor construction. • Describe the current-voltage characteristic behavior of capacitors. • Utilize component data sheet... | DCElectricalCircuitAnalysis_Page_260_Chunk1182 |
8.2 Capacitance and Capacitors 8.2 Capacitance and Capacitors A capacitor is a device that stores energy. Capacitors store energy in the form of an electric field. At its most simple, a capacitor can be little more than a pair of metal plates separated by air. As this constitutes an open circuit, DC current will not fl... | DCElectricalCircuitAnalysis_Page_261_Chunk1183 |
Unsurprisingly, the energy stored in capacitor is proportional to the capacitance. It is also proportional to the square of the voltage across the capacitor. W =1 2 CV 2 (8.3) Where W is the energy in joules, C is the capacitance in farads, V is the voltage in volts. The basic capacitor consists of two conducting plate... | DCElectricalCircuitAnalysis_Page_262_Chunk1184 |
From Equation 8.4 it is obvious that the permittivity of the dielectric plays a major role in determining the volumetric efficiency of the capacitor, in other words, the amount of capacitance that can be packed into a given sized component. Some dielectrics are notably more efficient than others. To make comparisons ea... | DCElectricalCircuitAnalysis_Page_263_Chunk1185 |
set an upper limit on how large of a voltage may be placed across a capacitor before it is damaged. Breakdown strength is measured in volts per unit distance, thus, the closer the plates, the less voltage the capacitor can withstand. For example, halving the plate distance doubles the capacitance but also halves its vo... | DCElectricalCircuitAnalysis_Page_264_Chunk1186 |
consequence of Equation 8.4. Modest surface mount capacitors can be quite small while the power supply filter capacitors commonly used in consumer electronics devices such as an audio amplifier can be considerably larger than a D cell battery. A sampling of capacitors is shown in Figure 8.6. Toward the front and left s... | DCElectricalCircuitAnalysis_Page_265_Chunk1187 |
The first two digits are the precision portion and the third digit is the power of ten multiplier. The result is in picofarads. Thus, 152 is 1500 pf. The schematic symbols for capacitors are shown in Figure 8.8. There are three symbols in wide use. The first symbol, using two parallel lines to echo the two plates, is f... | DCElectricalCircuitAnalysis_Page_266_Chunk1188 |
Example 8.1 Find the equivalent capacitance of the network shown in Figure 8.11. These capacitors are all in parallel, and thus, the equivalent value is the sum of the three capacitances: CTotal = C1+C 2+C 3 CTotal = 1μ F+100nF+560nF CTotal = 1.66μ F 267 Figure 8.11 Circuit for Example 8.1. Figure 8.10 Capacitor data s... | DCElectricalCircuitAnalysis_Page_267_Chunk1189 |
Example 8.2 Find the equivalent capacitance of the network shown in Figure 8.12. In this circuit, we find that the left and middle capacitors are in parallel. This combination is in series with the capacitor to the right: Cleft = C1+C 2 Cleft = 3.3μF+4.7μF Cleft = 8μ F CTotal = Cleft C3 Cleft+C3 CTotal = 8μ F16μ F 8μ F... | DCElectricalCircuitAnalysis_Page_268_Chunk1190 |
From here we determine the total charge: Q = V C Q = 12 V1.143μ F Q = 13.71μC Charge is constant across all of the series capacitors, therefore: V 2uF = Q C V 2uF = 13.71μC 2μ F V 2uF = 6.855V V 4uF = Q C V 4uF = 13.71μC 4μ F V 4uF = 3.427V V 8uF = Q C V 8uF = 13.71μC 8μ F V 8uF = 1.714V The sum of the three voltages i... | DCElectricalCircuitAnalysis_Page_269_Chunk1191 |
Current-Voltage Relationship Current-Voltage Relationship The fundamental current-voltage relationship of a capacitor is not the same as that of resistors. Capacitors do not so much resist current; it is more productive to think in terms of them reacting to it. The current through a capacitor is equal to the capacitanc... | DCElectricalCircuitAnalysis_Page_270_Chunk1192 |
As time progresses, the voltage across the capacitor increases with a positive polarity from top to bottom. With a theoretically perfect capacitor and source, this would continue forever, or until the current source was turned off. In reality, this line would either begin to deflect horizontally as the source reached i... | DCElectricalCircuitAnalysis_Page_271_Chunk1193 |
8.3 Initial and Steady-State Analysis of RC Circuits 8.3 Initial and Steady-State Analysis of RC Circuits When analyzing resistor-capacitor circuits, always remember that capacitor voltage cannot change instantaneously. If we assume that a capacitor in a circuit is not initially charged, then its voltage must be zero. ... | DCElectricalCircuitAnalysis_Page_272_Chunk1194 |
Continuing with the example, at steady-state both capacitors behave as opens. This is shown in Figure 8.20. This leaves E to drop across R1 and R2. This will create a simple voltage divider. The steady-state voltage across C1 will equal that of R2. As C2 is also open, the voltage across R3 will be zero while the voltag... | DCElectricalCircuitAnalysis_Page_273_Chunk1195 |
drawn below in Figure 8.23. The 3 kΩ resistor is now out of the picture, leaving us with the 6 kΩ in series with the 1 kΩ resistor. Once again, a voltage divider may be used to determine the voltage across the 6 kΩ. V 6k = E Rx Rx+Ry V 6k = 24 V 6k Ω 6k Ω+1 kΩ V 6k = 20.57V 8.4 Transient Response of RC Circuits 8.4 Tra... | DCElectricalCircuitAnalysis_Page_274_Chunk1196 |
The dashed red line represents the initial rate of change of capacitor voltage. This trajectory is what would be expected if an ideal current source drove the capacitor, as in Example 8.4. As noted previously, the rate of voltage change versus times is equal to i/C, and therefore in this case, E/RC. If the initial rate... | DCElectricalCircuitAnalysis_Page_275_Chunk1197 |
This sort of recursively dependent operation is characteristic of exponential functions. The equation for the capacitor's voltage charging curve is: V C (t) = E(1 −ϵ −t τ) (8.12) Where VC(t) is the capacitor voltage at time t, E is the source voltage, t is the time of interest, τ is the time constant, ε (also written e... | DCElectricalCircuitAnalysis_Page_276_Chunk1198 |
Steady-state will be reached in five time constants, or 500 milliseconds. Therefore we know that VC(0) = 0 volts and VC(1) = 100 volts. To find VC(50 ms) we simply solve Equation 8.12. V C(t) = E(1 −ϵ −t τ) V C (50 ms) = 100V(1 −ϵ −50 ms 100 ms) V C (50 ms) ≈39.35V This value can also be determined graphically from Fig... | DCElectricalCircuitAnalysis_Page_277_Chunk1199 |
steady-state. Third, at steady-state the capacitor voltage has virtually reached the maximum value set be the source, or 100 volts. Finally, at 50 milliseconds, we see that the capacitor voltage has reached roughly 40 volts, just as predicted. As mentioned previously, it is possible for a circuit to have different char... | DCElectricalCircuitAnalysis_Page_278_Chunk1200 |
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