Chapter 5, Problem 35E

(a) Obtain a value for the Thévenin equivalent resistance seen looking into the open terminals of the circuit in Fig. 5.76 by first finding V_oc and I_sc. (b) Connect a 1 A test source to the open terminals of the original circuit after shorting the voltage source, and use this to obtain R_TH. (c) Connect a 1 V test source to the open terminals of the original circuit after again zeroing the 2 V source, and use this now to obtain R_TH.

FIGURE 5.76

To determine

Find the value for the Thevenin’s equivalent resistance seen looking into the open terminals of the circuit by first finding $V_{OC}$ and $I_{SC}$.

Answer to Problem 35E

The Thevenin’s equivalent resistance is $\underset{\_}{35.43\Omega }$.

Explanation of Solution

Calculation:

The redrawn circuit diagram is given in Figure 1.

Apply KCL ay node 1,

$\frac{v_1-v}{R_1}+\frac{v_1}{R_2}+\frac{v_1-v_2}{R_3}=0\text{A}$ (1)

Here,

$v_1$ is the voltage at node 1,

$v_2$ is the voltage at node 2,

$v$ is the voltage of $2\text{V}$ independent source,

$R_1$ is the resistance of $10\Omega$ resistor,

$R_2$ is the resistance of $7\Omega$ resistor,

$R_3$ is the resistance of $20\Omega$ resistor,

Substitute $2\text{V}$ for $v$, $10\Omega$ for $R_1$, $7\Omega$ for $R_2$ and $20\Omega$ for $R_3$ in equation (1),

$\begin{array}{l}\frac{v_1-2\text{V}}{10\Omega }+\frac{v_1}{7\Omega }+\frac{v_1-v_2}{20\Omega }=0\text{A}\\ \frac{14v_1+28\text{V}+\text{20}{v}_1+7v_1-7v_2}{140\Omega }=0\text{A}\\ \frac{41v_1+28\text{V}-7v_2}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$ and $v_2$,

$41v_1+28\text{V}-7v_2=0\text{V}$

$41v_1-7v_2=-28\text{V}$ (2)

Apply KCL at node 2,

$\frac{v_2-v_1}{R_3}+\frac{v_2}{R_4}=0\text{A}$ (3)

Here,

$R_4$ is the resistance of $7\Omega$ resistor.

Substitute $20\Omega$ for $R_3$ and $7\Omega$ for $R_4$ in equation (3),

$\begin{array}{l}\frac{v_2-v_1}{20\Omega }+\frac{v_2}{7\Omega }=0\text{A}\\ \frac{7v_2-7v_1+20v_2}{140\Omega }=0\text{A}\\ \frac{27v_2-7v_1}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_2$,

$\begin{array}{l}27v_2-7v_1=0\text{V}\\ 27v_2=7v_1\end{array}$

$v_2=\frac{7}{27}{v}_1$ (4)

Substitute $\frac{7}{27}{v}_1$ for $v_2$ in equation (2),

$\begin{array}{l}41v_1-7\left(\frac{7}{27}{v}_1\right)=-28\text{V}\\ 41v_1-\frac{49}{27}{v}_1=-28\text{V}\\ \frac{1107v_1-49v_1}{27}=-28\text{V}\\ 1107v_1-49v_1=-\left(28\times 27\right)\text{V}\end{array}$

$1058v_1=-756\text{V}$

Rearrange for $v_1$,

$\begin{array}{c}{v}_1=\frac{-756}{1058}\text{V}\\ =-\text{0}\text{.71456}\text{V}\end{array}$

Substitute $-\text{0}\text{.71456}\text{V}$ for $v_1$ in equation (4),

$\begin{array}{c}{v}_2=\frac{7\times \left(-\text{0}\text{.71456}\text{V}\right)}{27}\\ =-0.1853\text{V}\end{array}$

So, the voltage $V_{OC}$ is $-0.1853\text{V}$.

The redrawn circuit diagram is given in Figure 2,

Refer to the redrawn Figure 2,

Apply KCL at node 1,

$\frac{v_1-v}{R_1}+\frac{v_1}{R_2}+\frac{v_1-v_2}{R_3}=0\text{A}$ (5)

Substitute $2\text{V}$ for $v$, $10\Omega$ for $R_1$, $7\Omega$ for $R_2$ and $20\Omega$ for $R_3$ in equation (1),

$\begin{array}{l}\frac{v_1-2\text{V}}{10\Omega }+\frac{v_1}{7\Omega }+\frac{v_1-v_2}{20\Omega }=0\text{A}\\ \frac{14v_1+28\text{V}+\text{20}{v}_1+7v_1-7v_2}{140\Omega }=0\text{A}\\ \frac{41v_1+28\text{V}-7v_2}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$ and $v_2$,

$41v_1+28\text{V}-7v_2=0\text{V}$

$41v_1-7v_2=-28\text{V}$ (6)

Apply KCL at node 2,

$\frac{v_2-v_1}{R_3}+\frac{v_2}{R_4}+\frac{v_2}{R_5}=0\text{A}$ (7)

Here,

$R_4$ is the resistance of $7\Omega$ resistor and

$R_5$ is the resistance of $30\Omega$ resistor.

Substitute $20\Omega$ for $R_3$, $7\Omega$ for $R_4$ and $30\Omega$ for $R_5$ in equation (7),

$\begin{array}{l}\frac{v_2-v_1}{20\Omega }+\frac{v_2}{7\Omega }+\frac{v_2}{30\Omega }=0\text{A}\\ \frac{21v_2-21v_1+60v_2+14v_2}{420\Omega }=0\text{A}\\ \frac{95v_2-21v_1}{420\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$,

$\begin{array}{l}95v_2-21v_1=0\text{V}\\ 95v_2=21v_1\end{array}$

$v_1=\frac{95}{21}{v}_2$ (8)

Substitute $\frac{95}{21}{v}_2$ for $v_1$ in equation (6),

$\begin{array}{l}\left(41\right)\times \left(\frac{95}{21}{v}_2\right)-7v_2=-28\text{V}\\ \text{185}{\text{.476v}}_2-7v_2=-28\text{V}\\ 178.476v_2=-28\text{V}\\ v_2=\frac{-28}{178.476}\text{}\text{V}\end{array}$

$v_2=-0.1569\text{}\text{V}$

The expression for the current flowing through $30\Omega$ resistor is as follows,

$I_N=\frac{v_2}{R_5}$ (9)

Here,

$I_N$ is the current flowing through $30\Omega$ resistor.

Substitute $-0.1569\text{}\text{V}$ for $v_2$ and $30\Omega$ for $R_5$ in equation (9),

$\begin{array}{c}{I}_N=\frac{-0.1569\text{}\text{V}}{30\Omega }\\ =-5.229\times {10}^{-3}\text{A}\end{array}$

So, the current $I_{SC}$ is $5.229\times {10}^{-3}\text{A}$.

The expression for the Thevenin’s equivalent resistance is as follows,

$R_{TH}=\frac{V_{OC}}{I_{SC}}$ (10)

Here,

$R_{TH}$ is the Thevenin’s equivalent resistance,

$V_{OC}$ is the Thevenin’s voltage and

$I_{SC}$ is the Norton current.

Substitute $-0.1853\text{V}$ for $V_{OC}$ and $-5.229\times {10}^{-3}\text{A}$ for $I_{SC}$ in equation (10),

$\begin{array}{c}{R}_{TH}=\frac{-0.1853\text{V}}{-5.229\times {10}^{-3}\text{A}}\\ =35.43\Omega \end{array}$

Conclusion:

Thus, the Thevenin’s equivalent resistance is $\underset{\_}{35.43\Omega }$.

To determine

Connect a $1\text{ A}$test source to the open terminals of the original circuit after shorting the voltage source, and use this to obtain $R_{TH}$.

Answer to Problem 35E

The Thevenin’s equivalent resistance is $\underset{\_}{35.43\Omega }$.

Explanation of Solution

Calculation:

The redrawn circuit diagram is given in Figure 3,

Refer to the redrawn Figure 3,

Apply KCL ay node 1,

$\frac{v_1}{R_1}+\frac{v_1}{R_2}+\frac{v_1-v_2}{R_3}=0\text{A}$ (11)

Here,

$v_1$ is the voltage at node 1,

$v_2$ is the voltage at node 2,

$R_1$ is the resistance of $10\Omega$ resistor,

$R_2$ is the resistance of $7\Omega$ resistor,

$R_3$ is the resistance of $20\Omega$ resistor,

Substitute $10\Omega$ for $R_1$, $7\Omega$ for $R_2$ and $20\Omega$ for $R_3$ in equation (11),

$\begin{array}{l}\frac{v_1}{10\Omega }+\frac{v_1}{7\Omega }+\frac{v_1-v_2}{20\Omega }=0\text{A}\\ \frac{14v_1+\text{20}{v}_1+7v_1-7v_2}{140\Omega }=0\text{A}\\ \frac{41v_1-7v_2}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$,

$\begin{array}{l}41v_1-7v_2=0\text{V}\\ 41v_1=7v_2\end{array}$

$v_1=\frac{7}{41}{v}_2$ (12)

Apply KCL at node 2,

$\frac{v_2-v_1}{R_3}+\frac{v_2}{R_4}+\frac{v_2-v_3}{R_5}=0\text{A}$ (13)

Here,

$v_3$ is the voltage at node 3,

$R_4$ is the resistance of $7\Omega$ resistor and

$R_5$ is the resistance of $30\Omega$ resistor.

Substitute $20\Omega$ for $R_3$, $7\Omega$ for $R_4$ and $30\Omega$ for $R_5$ in equation (3),

$\begin{array}{l}\frac{v_2-v_1}{20\Omega }+\frac{v_2}{7\Omega }+\frac{v_2-v_3}{30\Omega }=0\text{A}\\ \frac{21v_2-21v_1+60v_2+14v_2-14v_3}{420\Omega }=0\text{A}\\ \frac{95v_2-21v_1-14v_3}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$, $v_2$ and $v_3$,

$95v_2-21v_1-14v_3=0\text{V}$ (14)

The expression for the current flowing through $30\Omega$ resistor is as follows,

$i_0=\frac{v_2-v_3}{R_5}$ (15)

Here,

$i_0$ is the 1 A test source.

Substitute 1 A for $i_0$ and $30\Omega$ for $R_5$ in equation (15),

$1\text{A}=\frac{v_2-v_3}{30\Omega }$

Rearrange for $v_3$,

$v_2-v_3=30\text{V}$

$v_3=v_2-30\text{V}$ (16)

Substitute $v_2-30\text{V}$ for $v_3$ in equation (14),

$\begin{array}{l}95v_2-21v_1-14\left(v_2-30\text{V}\right)=0\text{V}\\ 95v_2-21v_1-14v_2+420\text{V}=0\text{V}\end{array}$

$81v_2-21v_1+420\text{V}\text{=}\text{0V}$ (17)

Substitute $\frac{7}{41}{v}_2$ for $v_1$ in equation (17),

$\begin{array}{l}81v_2-21\left(\frac{7}{41}{v}_2\right)+420\text{V}\text{=}\text{0V}\\ 81v_2-3.585v_2+420\text{V}\text{=}\text{0V}\end{array}$

Rearrange for $v_2$,

$\begin{array}{l}\text{77}\text{.414}6v_2=-420\text{V}\\ v_2=\frac{-420}{77.4146}\text{V}\\ v_2=-5.4253\text{V}\end{array}$

Substitute $-5.4253\text{V}$ for $v_2$ in equation (16),

$\begin{array}{l}{v}_3=-5.4253\text{V}-30\text{V}\\ v_3=-35.43\text{V}\end{array}$

The expression for the Thevenin’s equivalent resistance is as follows,

$R_{TH}=\frac{v_3}{i_0}$ (18)

Here,

$R_{TH}$ is the Thevenin’s equivalent resistance.

Substitute 35.43 V for $v_3$ and 1 A for $i_0$ in equation (18),

$\begin{array}{c}{R}_{TH}=\frac{35.43\text{V}}{1\text{ A}}\\ =35.43\Omega \end{array}$

Conclusion:

Thus, the Thevenin’s equivalent resistance is $\underset{\_}{35.43\Omega }$.

To determine

Connect a 1 V test source to the open terminals of the original circuit after again

zeroing the 2 V source, and use this now to obtain $R_{TH}$.

Answer to Problem 35E

The Thevenin’s equivalent resistance is $\underset{\_}{35.43\Omega }$.

Explanation of Solution

Calculation:

The redrawn circuit diagram is given in Figure 4,

Refer to the redrawn Figure 4,

Apply KCL ay node 1,

$\frac{v_1}{R_1}+\frac{v_1}{R_2}+\frac{v_1-v_2}{R_3}=0\text{A}$ (19)

Here,

$v_1$ is the voltage at node 1,

$v_2$ is the voltage at node 2,

$R_1$ is the resistance of $10\Omega$ resistor,

$R_2$ is the resistance of $7\Omega$ resistor,

$R_3$ is the resistance of $20\Omega$ resistor,

Substitute $10\Omega$  for $R_1$, $7\Omega$ for $R_2$ and $20\Omega$ for $R_3$ in equation (19),

$\begin{array}{l}\frac{v_1}{10\Omega }+\frac{v_1}{7\Omega }+\frac{v_1-v_2}{20\Omega }=0\text{A}\\ \frac{14v_1+\text{20}{v}_1+7v_1-7v_2}{140\Omega }=0\text{A}\\ \frac{41v_1-7v_2}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$,

$\begin{array}{l}41v_1-7v_2=0\text{V}\\ 41v_1=7v_2\end{array}$

$v_1=\frac{7}{41}{v}_2$ (20)

Apply KCL at node 2,

$\frac{v_2-v_1}{R_3}+\frac{v_2}{R_4}+\frac{v_2-v_0}{R_5}=0\text{A}$ (21)

Here,

$v_3$ is the voltage at node 3,

$v_0$ is the $1\text{V}$ test source,

$R_4$ is the resistance of $7\Omega$ resistor and

$R_5$ is the resistance of $30\Omega$ resistor.

Substitute $20\Omega$ for $R_3$, $7\Omega$ for $R_4$, $30\Omega$ for $R_5$ and 1 V for $v_0$ in equation (21),

$\begin{array}{l}\frac{v_2-v_1}{20\Omega }+\frac{v_2}{7\Omega }+\frac{v_2-1\text{V}}{30\Omega }=0\text{A}\\ \frac{21v_2-21v_1+60v_2+14v_2-14\text{V}}{420\Omega }=0\text{A}\\ \frac{95v_2-21v_1-14\text{V}}{140\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$ and $v_2$,

$95v_2-21v_1-14\text{V}=0\text{V}$

$95v_2-21v_1=14\text{V}$ (22)

Substitute $\frac{7}{41}{v}_2$ for $v_1$ in equation (22),

$\begin{array}{l}95v_2-21\left(\frac{7}{41}{v}_2\right)=14\text{V}\\ 95v_2-3.585v_2=14\text{V}\\ \text{91}\text{.4146}{v}_2=14\text{V}\\ v_2=\frac{14}{91.4146}\text{V}\end{array}$

$v_2=0.1532\text{V}$

The expression for the Thevenin’s equivalent resistance is as follows,

$R_{TH}=\frac{v_0}{\left(\frac{v_0-v_2}{R_5}\right)}$ (23)

Here,

$R_{TH}$ is the Thevenin’s equivalent resistance.

Substitute 0.1532 V  for $v_2$, 1 V for $v_0$ and $30\Omega$ for $R_5$ in equation (23),

$\begin{array}{c}{R}_{TH}=\frac{1\text{V}}{\left(\frac{1\text{V}-0.1532\text{V}}{30\Omega }\right)}\\ =\frac{1\text{V}}{0.02823\text{A}}\\ =35.43\Omega \end{array}$

Conclusion:

Thus, the Thevenin’s equivalent resistance is $\underset{\_}{35.43\Omega }$.