Duplicate from armanakbari4/CircuitSense

Co-authored-by: Arman Akbari <armanakbari4@users.noreply.huggingface.co>

.gitattributes

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Analysis/curated/level0/zero_00001/q75_a.txt

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+4

Analysis/curated/level0/zero_00001/q75_category.txt

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+5

Analysis/curated/level0/zero_00001/q75_der.txt

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+The transmission matrix $([T])$ of a circuit is defined as follows:
+
+$$
+\left[\begin{array}{l}
+v_1(r) \\
+i_1(t)
+\end{array}\right]-\left[\begin{array}{ll}
+r_{11} & r_{12} \\
+r_{21} & r_{22}
+\end{array}\right]\left[\begin{array}{l}
+v_2(r) \\
+t_2(t)
+\end{array}\right]+[T]-\left[\begin{array}{ll}
+t_{11} & t_{12} \\
+t_{21} & t_{22}
+\end{array}\right]
+$$
+
+
+The transmission matrix of the circuit shown in Figure 10.13.2 can be determined as follows:
+Applying KVL in the loop:
+
+$$
+-v_1(r)+v_2(t)-0 \Rightarrow v_1(t)-v_2(t)
+$$
+
+
+Applying KVL in the right-side mesh:
+
+$$
+-v_2(r)+R\left(i_1(t)-i_2(r)\right)-0 \Rightarrow i_1(t)-\frac{1}{R} v_2(t)+i_2(t)
+$$
+
+
+Solving (1), (2), and (3):
+
+$$
+\left[\begin{array}{l}
+v_1(t) \\
+l_1(t)
+\end{array}\right]-\left[\begin{array}{ll}
+1 & 0 \\
+\frac{1}{R} & 1
+\end{array}\right]\left[\begin{array}{l}
+v_2(t) \\
+i_2(t)
+\end{array}\right]+\left[T_P\right]-\left[\begin{array}{cc}
+1 & 0 \\
+\frac{1}{R} & 1
+\end{array}\right]
+$$
+
+
+The transmission matrix of the circuit shown in Figure 10.13.3 can be determined as follows:
+Applying KVL in the loop:
+
+$$
+-v_1(t)+R i_2(t)+v_2(t)-0 \Rightarrow v_1(t)-R i_2(t)+v_2(t)
+$$
+
+
+From the circuit, it is clear that:
+
+$$
+i_1(t)-i_2(t)
+$$
+
+
+Solving (1), (5), and (6):
+
+$$
+\left[\begin{array}{l}
+v_1(r) \\
+v_1(t)
+\end{array}\right]=\left[\begin{array}{ll}
+1 & R \\
+0 & 1
+\end{array}\right]\left[\begin{array}{l}
+v_2(t) \\
+t_2(t)
+\end{array}\right] \Rightarrow\left|T_s\right|=\left[\begin{array}{ll}
+1 & R \\
+0 & 1
+\end{array}\right]
+$$
+
+
+The circuit of Figure 10.13.1 can be assumed like the series connection of seven small circuits. The total transmission matrix of sach a circuit can be determined as follows:
+
+$$
+\begin{aligned}
+{[T]-\left[T_{p 1}\right] \times\left[T_{s 1}\right] \times\left[T_{p 2}\right] \times\left[T_{s 1}\right] \times\left[T_{p 2}\right] \times\left[T_{s 1}\right] \times\left[T_{p 1}\right] } \\
+-\left[\begin{array}{ll}
+1 & 0 \\
+\frac{1}{1} & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 1 \\
+0 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 0 \\
+\frac{1}{0.5} & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 1 \\
+0 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 0 \\
+\frac{1}{0.5} & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 1 \\
+0 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 0 \\
+\frac{1}{1} & 1
+\end{array}\right] \\
+-\left[\begin{array}{ll}
+1 & 0 \\
+1 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 1 \\
+0 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 0 \\
+2 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 1 \\
+0 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 0 \\
+2 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 1 \\
+0 & 1
+\end{array}\right] \times\left[\begin{array}{ll}
+1 & 0 \\
+1 & 1
+\end{array}\right] \\
+\rightarrow[T]=\left[\begin{array}{ll}
+26 & 15 \\
+45 & 26
+\end{array}\right]
+\end{aligned}
+$$

Analysis/curated/level0/zero_00001/q75_mc.txt

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+1. $\left[\begin{array}{cc}26 & 15 \\ 7 & 6\end{array}\right]$
+2. $\left[\begin{array}{ll}26 & 15 \\ 40 & 20\end{array}\right]$
+3. $\left[\begin{array}{ll}26 & 15 \\ 21 & 15\end{array}\right]$
+4. $\left[\begin{array}{ll}26 & 15 \\ 45 & 26\end{array}\right]$
+5. Non of the above

Analysis/curated/level0/zero_00001/q75_question.txt

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+In the provided circuit image For the circuit, sdetermine the transmission matrix $([T])$.
+
+
+

Analysis/curated/level0/zero_00001/q75_ta.txt

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+$\left[\begin{array}{ll}26 & 15 \\ 45 & 26\end{array}\right]$

Analysis/curated/level0/zero_00002/q56_a.txt

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+2

Analysis/curated/level0/zero_00002/q56_category.txt

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+2

Analysis/curated/level0/zero_00002/q56_der.txt

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+Based on the information given in the problem, we have [1-2]:
+
+$$
+\begin{gathered}
+\text { Test I }: v_s=7 \mathrm{~V}, i_s=3 \mathrm{~A}, i_{\text {out }}=1 \mathrm{~A} \\
+\text { Test II }: v_s=9 \mathrm{~V}, i_s=1 \mathrm{~A}, i_{\text {out }}=3 \mathrm{~A} \\
+\text { Test III }: v_s=15 \mathrm{~V}, i_s=9 \mathrm{~A}, i_{\text {out }}=\text { Unknown }
+\end{gathered}
+$$
+
+
+Since the network only includes linear time-invariant (LTI) resistors and dependent sources, the following relation for the output current is held:
+
+$$
+i_{\text {out }}=\alpha i_s+\beta v_s
+$$
+
+
+Solving (1), (2), and (4):
+
+$$
+\left\{\begin{array}{c}
+1=3 \alpha+7 \beta \\
+3=\alpha+9 \beta
+\end{array} \Rightarrow \alpha=-0.6, \beta=0.4\right.
+$$
+
+
+Solving (3), (4), and (5):
+
+$$
+i_{o u t}=\alpha i_s+\beta v_s=-0.6 \times 9+0.4 \times 15=0.6 \mathrm{~A}
+$$

Analysis/curated/level0/zero_00002/q56_mc.txt

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+1. 0.3 A
+2. 0.6 A
+3. $3 A$
+4. $6 A$
+5. Non of the above

Analysis/curated/level0/zero_00002/q56_question.txt

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+In the provided circuit image, The three-port network, includes linear time-invariant (LTI) resistors and dependent sources [1-2]. The network is put under two tests as follows.
+Test I: $v_s=7 \mathrm{~V}, i_s=3 \mathrm{~A}, i_{\text {out }}=1 \mathrm{~A}$
+Test II: $v_s=9 \mathrm{~V}, i_s=1 \mathrm{~A}, i_{\text {out }}=3 \mathrm{~A}$
+Determine the value of $i_{\text {out }}$ for $v_s=15 \mathrm{~V}, i_s=9 \mathrm{~A}$
+

Analysis/curated/level0/zero_00002/q56_ta.txt

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+0.6 A

Analysis/curated/level0/zero_00003/q57_a.txt

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+1

Analysis/curated/level0/zero_00003/q57_category.txt

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+2

Analysis/curated/level0/zero_00003/q57_der.txt

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+Based on Tellegen's theorem, we can write the following equation for the network:
+
+$$
+v_1(t)\left(-\widetilde{i}_1(t)\right)+v_2(t) \widetilde{i}_2(t)=\widetilde{v}_1(t)\left(-i_1(t)\right)+\widetilde{v}_2(t) i_2(t)
+$$
+
+
+In (1), a negative sign was applied for the polarity of the current of the left-side voltage source, as its current flows toward the network.
+
+Based on the information given in the problem, we have:
+
+$$
+\text { Test I : } v_1(t)=30 t V, v_2(t)=0, i_1(t)=5 t A, i_2(t)=2 t A
+$$
+
+
+Test II : $v_1(t)=30 t+60 V, v_2(t)=60 t+15 V, i_1(t)=$ Unknown
+Solving (1), (2), and (3):
+
+$$
+\begin{aligned}
+& 30 t \times\left(-\widetilde{i}_1(t)\right)+0 \times \widetilde{i}_2(t)=(30 t+60) \times(-5 t)+(60 t+15) \times 2 t \\
+& \Rightarrow \widetilde{i}_1(t)=\frac{-150 t^2-300 t+120 t^2+30 t}{-30 t}=\frac{-30 t^2-270 t}{-30 t}=(t+9) A
+\end{aligned}
+$$

Analysis/curated/level0/zero_00003/q57_mc.txt

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+1. $(t+9) A$
+2. $(9 t+11) A$
+3. $(5 t+10) A$
+4. $(-4 t-1) A$
+5. Non of the above

Analysis/curated/level0/zero_00003/q57_question.txt

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+In the provided circuit image, For the resistive linear time-invariant (LTI) two-port network, the information below is given.
+
+$$
+v_1(t)=30 t V, v_2(t)=0, i_1(t)=5 t A, i_2(t)=2 t A
+$$
+
+
+Determine the value of $i_1(t)$ in Ampere if $v_1(t)=30 t+60 V, v_2(t)=60 t+15 V$
+Difficulty level Easy
+

Analysis/curated/level0/zero_00003/q57_ta.txt

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+$(t+9) A$

Analysis/curated/level0/zero_00004/q58_a.txt

@@ -0,0 +1 @@
+4

Analysis/curated/level0/zero_00004/q58_category.txt

@@ -0,0 +1 @@
+2

Analysis/curated/level0/zero_00004/q58_der.txt

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+Based on the information given in the problem, we have:
+
+Test I : If $v_1=3 \mathrm{~V}, i_s=3 \mathrm{~A}, i_{\text {out }}=6 \mathrm{~A}$
+
+
+
+Test II : $v_1=0 A, i_s=-2 A, i_{\text {out }}=2 A$
+Test III : $v_1=-2 V, i_s=0 A, i_{\text {out }}=$ Unknown
+Since the network is linear time-invariant (LTI), the following relation for the output current is held:
+
+$$
+i_{o u t}=\alpha v_1+\beta i_s
+$$
+
+
+Solving (1), (2), and (4):
+
+$$
+\left\{\begin{array}{c}
+6=3 \alpha+3 \beta \\
+2=0 \times \alpha+(-2) \beta
+\end{array} \Rightarrow \alpha=3, \beta=-1\right.
+$$
+
+
+Solving (3), (4), and (5):
+
+$$
+i_{\text {out }}=\alpha v_1+\beta i_s=3 \times(-2)+(-1) \times 0=-6 A
+$$

Analysis/curated/level0/zero_00004/q58_mc.txt

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+1. $6 A$
+2. $-4 A$
+3. $4 A$
+4. $-6 A$
+5. Non of the above

Analysis/curated/level0/zero_00004/q58_question.txt

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+In the provided circuit image, For the resistive linear time-invariant (LTI) two-port network, the information below is given.
+
+Test I: If $v_1=3 \mathrm{~V}, i_s=3 \mathrm{~A}, i_{\text {out }}=6 \mathrm{~A}$
+Test II: $v_1=0 A, i_s=-2 A, i_{\text {out }}=2 A$
+Determine the value of $i_{\text {out }}(t)$ if $v_1(t)=-2 \mathrm{~V}, i_s(t)=0 \mathrm{~A}$
+
+

Analysis/curated/level0/zero_00004/q58_ta.txt

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+$-6 A$

Analysis/curated/level0/zero_00005/q59_a.txt

@@ -0,0 +1 @@
+1

Analysis/curated/level0/zero_00005/q59_category.txt

@@ -0,0 +1 @@
+5

Analysis/curated/level0/zero_00005/q59_der.txt

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+Based on Tellegen's theorem, we can write:
+
+$$
+v_1(t) \bar{\mu}_1(t)+v_2(r) \bar{\mu}_2(t)+v_3(r) \bar{\mu}_3(t)-\bar{v}_1(r) \mu_1(t)+\bar{v}_2(t) \mu_2(t)+\bar{v}_3(t) \mu_3(t)
+$$
+
+
+From the circuits of Figs 8.4.1-2, we have:
+\begin{tabular}{|l|l|l|l|}
+\hline \multirow{6}{*}{Fig. 1 :} & $v_1(t)=$ Unknown & & $\bar{v}_1(t)-0$ \\
+\hline & $v_2(\mathrm{r})=\left(e^{-\mathrm{t}}-e^{-2 \mathrm{r}}\right) \alpha(t) V$ & & $\bar{v}_2(t)=$ Unknown \\
+\hline & $v_3(t)=0$ & & $\bar{v}_1(t)=e^{-t} u(t) V$ \\
+\hline & $i_1(t)=-e^{-2} u(t) A$ & & $\bar{T}_1(t)=0$ \\
+\hline & $i_2(t)=0$ & & $\bar{T}_2(t)=-e^{-t} u(t) A$ \\
+\hline & $i_3(t)=\left(2 e^{-2 t}-e^{-t}\right) u(t) A$ & & $\bar{B}_3(t)=$ Unknown \\
+\hline
+\end{tabular}
+
+In (2), based on Tellegen's theorem, a negative sign was applied for the polarity of the currents flowing toward the network. Now, for this problem, we should transfer to Laplace domain. Therefore, the quantities are as follows:
+Linear. . .
+Fig. $1:\left\{\begin{array}{r}V_1(s)=\text { Unknown } \\ V_2(s)=\frac{1}{s+1}-\frac{1}{s+2} \\ V_3(s)=0 \\ I_1(s)=-\frac{1}{s+2} \\ I_2(s)=0 \\ I_3(s)=\frac{2}{s+2}-\frac{1}{s+1}\end{array}\right.$.
+Fig.2: $\left\{\begin{array}{r}\nabla_1(x)=0 \\ \nabla_2(x)=\text { Unknown } \\ \bar{V}_3(x)=\frac{1}{x+1} \\ \bar{I}_1(x)=0 \\ \bar{I}_2(x)=-\frac{1}{x+1} \\ \bar{I}_3(x)=\text { Unknown }\end{array}\right.$
+
+For this problem, we should use Tellegen's theorem in Laplace domain, as follows:
+
+$$
+V_1(s) \bar{M}_1(s)+V_2(s) \bar{U}_2(s)+V_3(s) \bar{M}_3(s)-\bar{V}_1(s) A_1(s)+\bar{V}_2(s) H_2(s)+\bar{V}_3(s) A_3(s)
+$$
+
+
+Solving (3) and (4):
+
+$$
+\begin{gathered}
+V_1(s) \times 0+\left(\frac{1}{s+1}-\frac{1}{s+2}\right)\left(-\frac{1}{s+1}\right)+0 \times \bar{I}_3(s)-\bar{V}_1(s)\left(-\frac{1}{s+2}\right)+\bar{V}_2(s) \times 0+\left(\frac{1}{s+1}\right)\left(\frac{2}{s+2}-\frac{1}{s+1}\right) \\
+\Rightarrow \bar{V}_1(s)\left(\frac{1}{s+2}\right)-\left(\frac{1}{x+1}\right)\left(\frac{1}{x+2}\right) \Rightarrow \bar{V}_1(s)-\frac{1}{s+1} \Rightarrow V_1(s)-e^{-1} w(s) V
+\end{gathered}
+$$

Analysis/curated/level0/zero_00005/q59_mc.txt

@@ -0,0 +1,5 @@
+1. $e^{-t} u(t) V$
+2. $(t-1) e^{-t} u(t) V$
+3. $(t-2) e^{-t} u(t) V$
+4. $(t+1) e^{-t} u(t) V$
+5. Non of the above

Analysis/curated/level0/zero_00005/q59_question.txt

@@ -0,0 +1,2 @@
+In the provided circuit image, shows a resistive three-port network that a test is done on it. The same network is put under another test which is illustrated. Determine the value of $\widetilde{v}_1(t)$.
+

Analysis/curated/level0/zero_00005/q59_ta.txt

@@ -0,0 +1 @@
+$e^{-t} u(t) V$

Analysis/curated/level0/zero_00006/q60_a.txt

@@ -0,0 +1 @@
+3

Analysis/curated/level0/zero_00006/q60_category.txt

@@ -0,0 +1 @@
+1

Analysis/curated/level0/zero_00006/q60_der.txt

@@ -0,0 +1,60 @@
+Based on Tellegen's theorem, we can write the following equation for the network:
+
+$$
+v_1(t) \tilde{i}_1(t)+v_2(t) \tilde{i}_2(t)=\tilde{v}_1(t) i_1(t)+\tilde{v}_2(t) i_2(t)
+$$
+
+
+Based on the information given in the problem, we have:
+
+$$
+\text { Test } 1:\left\{\begin{array}{r}
+v_1(t)=4 \sin \left(\omega t+45^{\circ}\right) V \\
+v_2(t)=0 \\
+i_1(t)=2 \sin \left(\omega t+30^{\circ}\right) A \\
+i_2(t)=\sin \left(\omega t+60^{\circ}\right) A
+\end{array}, \text { Test } 2:\left\{\begin{array}{r}
+\tilde{v}_1(t)=\text { Requested } \\
+\tilde{v}_2(t)=\sin (\omega t) V \\
+\tilde{i}_1(t)=0.5 \sin \left(\omega t+15^{\circ}\right) A \\
+\tilde{i}_2(t)=\text { Unknown }
+\end{array}\right.\right.
+$$
+
+
+We need to solve the problem in phasor domain. Therefore, the quantities will be as follows:
+
+$$
+\text { Test 1: }\left\{\begin{array} { r } 
+{ \mathbf { V } _ { 1 } = 4 e ^ { j 4 5 ^ { \circ } } } \\
+{ \mathbf { V } _ { 2 } = 0 } \\
+{ \mathbf { I } _ { 1 } = 2 e ^ { j 3 0 ^ { \circ } } } \\
+{ \mathbf { I } _ { 2 } = e ^ { j 6 0 ^ { \circ } } }
+\end{array} , \text { Test 2: } \left\{\begin{array}{r}
+\tilde{\mathbf{V}}_1=\text { Requested } \\
+\tilde{\mathbf{V}}_2=1 \\
+\tilde{\mathbf{I}}_1=0.5 e^{j 15^{\circ}} \\
+\tilde{\mathbf{I}}_2=\text { Unknown }
+\end{array}\right.\right.
+$$
+
+
+Tellegen's theorem in phasor domain for this problem is as follows:
+
+$$
+\mathbf{V}_1 \widetilde{\mathbf{I}}_1+\mathbf{V}_2 \widetilde{\mathbf{I}}_2=\widetilde{\mathbf{V}}_1 \mathbf{I}_1+\widetilde{\mathbf{V}}_2 \mathbf{I}_2
+$$
+
+
+Solving (3) and (4):
+
+$$
+4 e^{j 45^{\circ}} \times 0.5 e^{j 15^{\circ}}+0 \times \widetilde{\mathbf{I}}_2=\widetilde{\mathbf{V}}_1 \times 2 e^{j 30^{\circ}}+1 \times e^{j 60^{\circ}} \Rightarrow \widetilde{\mathbf{V}}_1 \times 2 e^{\beta 30^{\circ}}=e^{j 60^{\circ}} \Rightarrow \widetilde{\mathbf{V}}_1=0.5 e^{j 30^{\circ}}
+$$
+
+
+By transferring to time domain, we have:
+
+$$
+\widetilde{v}_1(t)=0.5 \sin \left(\omega t+30^{\circ}\right) V
+$$

Analysis/curated/level0/zero_00006/q60_mc.txt

@@ -0,0 +1,5 @@
+1. $\sin (\omega t) V$
+2. $\sin \left(\omega t-60^{\circ}\right) V$
+3. $0.5 \sin \left(\omega t+30^{\circ}\right) V$
+4. $4 \sin \left(\omega t-15^{\circ}\right) V$
+5. Non of the above

Analysis/curated/level0/zero_00006/q60_question.txt

@@ -0,0 +1,9 @@
+In the provided circuit image shows a linear time-invariant (LTI) two-port network that includes RLC components.
+
+$$
+i_1(t)=2 \sin \left(\omega t+30^{\circ}\right) A, v_1(t)=4 \sin \left(\omega t+45^{\circ}\right) V, v_2(t)=0, i_2(t)=\sin \left(\omega t+60^{\circ}\right) A
+$$
+
+
+Determine the value of $v_1(t)$ in Ampere if $i_1(t)=0.5 \sin \left(\omega t+15^{\circ}\right) A, v_2(t)=\sin (\omega t) V$
+

Analysis/curated/level0/zero_00006/q60_ta.txt

@@ -0,0 +1 @@
+$0.5 \sin \left(\omega t+30^{\circ}\right) V$

Analysis/curated/level0/zero_00007/q61_a.txt

@@ -0,0 +1 @@
+4

Analysis/curated/level0/zero_00007/q61_category.txt

@@ -0,0 +1 @@
+5

Analysis/curated/level0/zero_00007/q61_der.txt

@@ -0,0 +1,40 @@
+Based on the information given in the problem, we have:
+
+Test I : If $R_L=\infty,\left|\mathbf{V}_{\text {out }}\right|=13 \mathrm{~V}$
+Test II : If $R_L=3 \Omega,\left|\mathbf{V}_{\text {out }}\right|=3 \mathrm{~V}$
+Test III : If $R_L=14 \Omega,\left|\mathbf{V}_{\text {out }}\right|=9.1 \mathrm{~V}$
+
+Since the network is linear time-invariant (LTI), it can be replaced by its Thevenin equivalent circuit, as can be seen in Figure 8.6.2.
+
+By using Figure 8.6.2 and (1), we can conclude:
+
+$$
+\left|\mathbf{V}_{\mathbf{T h}}\right|=\left|\mathbf{V}_{\text {out }}\right|=13 \mathrm{~V}
+$$
+
+
+Using Figure 8.6.2 and applying voltage division and (2) and (4):
+
+$$
+\begin{gathered}
+\left|\mathbf{V}_{\text {out }}\right|=\left|\frac{R_L}{R_L+a+j b} \times \mathbf{V}_{\text {Th }}\right| \Rightarrow 3=\left|\frac{3}{3+a+j b} \times 13\right| \Rightarrow 1=\left|\frac{13}{3+a+j b}\right| \Rightarrow|3+a+j b|=13 \\
+\Rightarrow(3+a)^2+b^2=169
+\end{gathered}
+$$
+
+
+Using Figure 8.6.2 and applying voltage division and (3) and (4):
+
+$$
+\begin{gathered}
+\left|\mathbf{V}_{\text {out }}\right|=\left|\frac{R_L}{R_L+a+j b} \times \mathbf{V}_{\text {Th }}\right| \Rightarrow 9.1=\left|\frac{14}{14+a+j b} \times 13\right| \Rightarrow 9.1=\left|\frac{182}{14+a+j b}\right| \Rightarrow|14+a+j b|=20 \\
+\Rightarrow(14+a)^2+b^2=400
+\end{gathered}
+$$
+
+
+Solving (5) and (6):
+
+$$
+a=2, b=12 \Rightarrow Z_{T h}=(2+j 12) \Omega
+$$

Analysis/curated/level0/zero_00007/q61_mc.txt

@@ -0,0 +1,5 @@
+1. $2 \Omega$
+2. $12 \Omega$
+3. $(12+j 2) \Omega$
+4. $(2+j 12) \Omega$
+5. Non of the above

Analysis/curated/level0/zero_00007/q61_question.txt

@@ -0,0 +1,9 @@
+In the provided circuit image, The two-port network, is linear time-invariant (LTI). The tests below have been carried out on the resistive load. Calculate the impedance of the network.
+
+Test I: If $R_L=\infty,\left|\mathbf{V}_{\text {out }}\right|=13 \mathrm{~V}$
+Test II: If $R_L=3 \Omega$, | $\mathbf{V}_{\text {out }} \mid=3 \mathrm{~V}$
+Test III: If $R_L=14 \Omega,\left|\mathbf{V}_{\text {out }}\right|=9.1 \mathrm{~V}$
+
+
+
+