Chapter 4, Problem 11P

Use the superposition principle to find i_o and v_o in the circuit of Fig. 4.79.

Figure 4.79

To determine

Find the value of the voltage $v_o$ and the current $i_o$ for the given circuit in PSPICE by using superposition principle.

Answer to Problem 11P

The value of the voltage $v_o$ is $18\text{V}$ and the current $i_o$ is $1.8\text{A}$ for the given circuit using PSPICE.

Explanation of Solution

Given data:

Refer to Figure 4.79 in the textbook.

Calculation:

The circuit in Figure 4.79 involves a current-controlled current source, which must be left intact. Let

$v_o=v_1+v_2$        (1)

Where $v_1$ and $v_2$ are contributions due to the current source $6\text{A}$ and the voltage source $80\text{V}$ respectively.

To obtain $v_1$, set the voltage source to $0\text{V}$ (or a short circuit). The modified circuit is shown in Figure 1.

Apply nodal analysis at node $v_a$ in Figure 1.

$\begin{array}{l}6=\frac{v_a}{40}+\frac{v_a-v_b}{10}\\ 6=\frac{v_a+4\left(v_a-v_b\right)}{40}\\ 240=v_a+4v_a-4v_b\end{array}$

$240=5v_a-4v_b$        (2)

Apply nodal analysis at node $v_b$ in Figure 1.

$\begin{array}{l}-i_1-4i_1+\frac{\left(v_b-0\right)}{20}=0\\ -5i_1+\frac{v_b}{20}=0\end{array}$

$-100i_1+v_b=0$        (3)

Rearranging the equation (3),

$-v_b=-100i_1$

$v_b=100i_1$        (4)

In Figure 1, the current $i_1$ through the 10 ohm resistor is,

$i_1=\frac{v_a-v_b}{10}$        (5)

Substitute $\frac{v_a-v_b}{10}$ for $i_1$ in equation (4) to find $v_b$.

$\begin{array}{l}{v}_b=100\left(\frac{v_a-v_b}{10}\right)\\ v_b=10\left(v_a-v_b\right)\end{array}$

$v_b=10v_a-10v_b$        (6)

Rearranging the equation (6),

$\begin{array}{l}{v}_b+10v_b=10v_a\\ 11v_b=10v_a\\ v_b=\frac{10}{11}{v}_a\end{array}$

$v_b=0.9091v_a$        (7)

Substitute $0.9091v_a$ for $v_b$ in equation (2) to find $v_a$.

$\begin{array}{l}240=5v_a-4\left(0.9091v_a\right)\\ 240=5v_a-3.636v_a\end{array}$

$240=1.364v_a$        (8)

Rearrange the equation (8) to find the node voltage $v_a$ in volts.

$\begin{array}{c}{v}_a=\frac{240}{1.364}\\ =175.95\text{V}\end{array}$

Substitute $175.95\text{V}$ for $v_a$ in equation (7) to find the node voltage $v_b$ in volts.

$\begin{array}{c}{v}_b=0.9091\times 175.95\text{V}\\ =159.96\text{V}\end{array}$

The voltage $v_1$ across the 10 ohm resistor is,

$v_1=v_a-v_b$        (9)

Substitute $175.95\text{V}$ for $v_a$ and $159.96\text{V}$ for $v_b$ in equation (9) to find the voltage $v_1$ in volts.

$\begin{array}{c}{v}_1=175.95\text{V}-159.96\text{V}\\ =15.99\text{V}\end{array}$

To obtain $v_2$, set the current source to $0\text{A}$ (or an open circuit). The modified circuit is shown in Figure 2.

Apply nodal analysis at node $v_c$ in Figure 2.

$\frac{0-v_c}{50}+4i_2+\frac{\left(-30-v_c\right)}{20}=0$

$-\frac{v_c}{50}+4i_2-\frac{\left(30+v_c\right)}{20}=0$        (10)

In Figure 2, the resistors 10 ohms and 40 ohms are connected in series. Therefore, the equivalent resistance for series connection is,

$10\Omega +40\Omega =50\Omega$

The modified circuit is shown in Figure 3.

In Figure 3, the current $i_2$ flowing through the 10 ohms resistor is calculated by using Ohms law in amperes.

$i_2=\frac{-v_c}{50}$        (11)

Substitute $\frac{-v_c}{50}$ for $i_2$ in equation (10) to find the node voltage $v_c$ in volts.

$\begin{array}{l}-\frac{v_{}}{50}+4\left(\frac{-v_c}{50}\right)-\frac{\left(30+v_c\right)}{20}=0\\ -\frac{v_c}{50}-\frac{4v_c}{50}-\frac{\left(30+v_c\right)}{20}=0\\ -\frac{5v_c}{50}-\frac{\left(30+v_c\right)}{20}=0\end{array}$

$\frac{-2\left(5v_c\right)-5\left(30+v_c\right)}{100}=0$        (12)

The equation (12) becomes,

$-10v_c-150-5v_c=0$

$-15v_c-150=0$        (13)

Rearrange the equation (13) to find the node voltage $v_c$ in volts.

$\begin{array}{l}-15v_c=150\\ v_c=-\frac{150}{15}\\ v_c=-10\text{V}\end{array}$

Substitute $-10\text{V}$ for $v_c$ in equation (11) to find the current $i_2$ in amperes.

$\begin{array}{c}{i}_2=\frac{-\left(-10\text{V}\right)}{50}\\ =\frac{10\text{V}}{50}\\ =0.2\text{V}\end{array}$

Therefore, in Figure 2 the voltage $v_2$ across the 10 ohm resistor is calculated by using Ohms law.

$v_2=10i_2$        (14)

Substitute $0.2\text{V}$ for $i_2$ in equation (14) to find the voltage $v_2$ in volts.

$\begin{array}{c}{v}_2=10\times 0.2\text{V}\\ =2\text{V}\end{array}$

Substitute $15.99\text{V}$ for $v_1$ and $2\text{V}$ for $v_2$ in equation (1) to find the output voltage $v_o$ in volts.

$\begin{array}{c}{v}_o=15.99\text{V}+2\text{V}\\ =\text{17}\text{.99}\text{V}\\ \cong \text{18}\text{V}\end{array}$

In Figure 4.79, the current $i_o$ through the 10 ohm resistor is calculated by using Ohms law in amperes.

$\begin{array}{c}{i}_o=\frac{17.99\text{V}}{10\Omega }\\ =1.799\text{A}\left\{1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\\ \cong 1.8\text{A}\end{array}$

PSPICE Simulation:

In the given circuit, since there are two sources, let

$v_o=v_1+v_2$

Where $v_1$, and $v_2$ are contributions due to the $6\text{A}$ and $30\text{V}$ sources.

When $6\text{A}$ current source is active:

Draw the circuit diagram in PSPICE as shown in Figure 4.

Save the circuit and provide the Simulation Settings as shown in Figure 5.

Now run the simulation and the results will be displayed as shown in Figure 6 by enabling “Enable Bias Voltage Display” icon and “Enable Bias Current Display” icon.

From Figure 6, the voltage $v_1$ is calculated by using ohms law as follows,

$\begin{array}{c}{v}_1=\left(1.6\text{A}\right)\left(10\Omega \right)\\ =16\text{V}\left\{\because 1\text{V}=1\text{A}\times 1\Omega \right\}\end{array}$

When $30\text{V}$ voltage source is active:

Draw the circuit diagram in PSPICE as shown in Figure 7.

Save the circuit and provide the Simulation Settings as shown in Figure 8.

Now run the simulation and the results will be displayed as shown in Figure 9 by enabling “Enable Bias Voltage Display” icon and “Enable Bias Current Display” icon.

From Figure 9, the voltage $v_2$ is calculated by using ohms law as follows,

$\begin{array}{c}{v}_2=\left(200\times {10}^{-3}\text{A}\right)\left(10\Omega \right)\left\{\because 1\text{m}={10}^{-3}\right\}\\ =2\text{V}\left\{\because 1\text{V}=1\text{A}\times 1\Omega \right\}\end{array}$

Therefore, the total voltage $v_o$ in the circuit is,

$v_o=v_1+v_2$

Substitute $16\text{V}$ for $v_1$ and $2\text{V}$ for $v_2$ to find the total voltage $v_o$ in volts.

$\begin{array}{c}{v}_o=16\text{V}+\text{2}\text{V}\\ =18\text{V}\end{array}$

The current $i_o$ is calculated by using ohms law as follows.

$i_o=\frac{v_o}{10}$

Substitute 18 for $v_o$ to find the current $i_o$ in amperes.

$\begin{array}{c}{i}_o=\frac{18}{10}\\ =1.8\text{A}\end{array}$

Conclusion:

Thus, the value of the voltage $v_o$ is $18\text{V}$ and the current $i_o$ is $1.8\text{A}$ for the given circuit using PSPICE.