Chapter 4, Problem 17P

Use superposition to obtain v_x in the circuit of Fig. 4.85. Check your result using PSpice or MultiSim.

Figure 4.85

To determine

Find the value of the voltage $v_x$ in Figure 4.85 by using source transformation and also verify using PSpice.

Answer to Problem 17P

The value of the voltage $v_x$ in the given circuit is $-8.57\text{1}\text{V}$ and verified using PSpice.

Explanation of Solution

Given data:

Refer to Figure 4.85 in the textbook.

Calculation:

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

$v_x=v_1+v_2+v_3$

Where $v_1$, $v_2$, and $v_3$ are contributions due to the $90\text{V}$, $6\text{A}$, and $40\text{V}$ sources.

When $90\text{V}$ voltage source is active:

The given circuit is modified as shown in Figure 1.

In Figure 1, the voltage source with series resistance is converted into current source with parallel resistance by source transformation method.

That is,

$\begin{array}{c}I=\frac{90\text{V}}{30\Omega }\\ =3\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

The source transformation is shown in Figure 2.

In Figure 2, $30\Omega$ and $60\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}30\Omega \left|\right|60\Omega =\frac{30\Omega \times 60\Omega }{30\Omega +60\Omega }\\ =20\Omega \end{array}$

Similarly, $30\Omega$ and $20\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}30\Omega \left|\right|60\Omega =\frac{30\Omega \times 20\Omega }{30\Omega +20\Omega }\\ =12\Omega \end{array}$

The modified Figure is shown in Figure 3.

In Figure 3, the current $i_1$ through the 10 ohms resistor is calculated by using current division principle.

$\begin{array}{c}{i}_1=\left(3\text{A}\right)\left(\frac{20\Omega }{20\Omega +22\Omega }\right)\\ =\left(3\text{A}\right)\left(\frac{20\Omega }{42\Omega }\right)\\ =1.4286\text{A}\end{array}$

The voltage $v_1$ across the 10 ohms resistor is calculated by using Ohm’s law as follows.

$v_1=10i_1$

Substitute $1.4286$ for $i_1$ to find the voltage $v_1$ in volts.

$\begin{array}{c}{v}_1=10\left(1.4286\right)\\ =14.286\text{V}\end{array}$

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

The given circuit is modified as shown in Figure 4.

In Figure 4, $30\Omega$ and $60\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}30\Omega \left|\right|60\Omega =\frac{30\Omega \times 60\Omega }{30\Omega +60\Omega }\\ =20\Omega \end{array}$

Similarly, $30\Omega$ and $20\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}30\Omega \left|\right|60\Omega =\frac{30\Omega \times 20\Omega }{30\Omega +20\Omega }\\ =12\Omega \end{array}$

The modified Figure is shown in Figure 5.

In Figure 5, the current $i_2$ through the 10 ohms resistor is calculated by using current division principle.

$\begin{array}{c}{i}_2=\left(6\text{A}\right)\left(\frac{12\Omega }{12\Omega +30\Omega }\right)\\ =\left(6\text{A}\right)\left(\frac{12\Omega }{42\Omega }\right)\\ =1.7143\text{A}\end{array}$

The voltage $v_2$ across the 10 ohms resistor is calculated by using Ohm’s law as follows.

$v_2=-10i_2$

Substitute $1.7143$ for $i_2$ to find the voltage $v_2$ in volts.

$\begin{array}{c}{v}_2=-10\left(1.7143\right)\\ =-17.143\text{V}\end{array}$

When $40\text{V}$ voltage source is active:

The given circuit is modified as shown in Figure 6.

In Figure 6, the voltage source with series resistance is converted into current source with parallel resistance by source transformation method.

That is,

$\begin{array}{c}{I}_1=\frac{40\text{V}}{20\Omega }\\ =2\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

The source transformation is shown in Figure 7.

In Figure 7, $30\Omega$ and $60\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}30\Omega \left|\right|60\Omega =\frac{30\Omega \times 60\Omega }{30\Omega +60\Omega }\\ =20\Omega \end{array}$

Similarly, $30\Omega$ and $20\Omega$ resistance are connected in parallel. Therefore, the equivalent resistance for the parallel connection is,

$\begin{array}{l}30\Omega \left|\right|20\Omega =\frac{30\Omega \times 20\Omega }{30\Omega +20\Omega }\\ =12\Omega \end{array}$

The modified Figure is shown in Figure 8.

In Figure 8, the current $i_3$ through the 10 ohms resistor is calculated by using current division principle.

$\begin{array}{c}{i}_3=\left(2\text{A}\right)\left(\frac{12\Omega }{12\Omega +30\Omega }\right)\\ =\left(2\text{A}\right)\left(\frac{12\Omega }{42\Omega }\right)\\ =0.5714\text{A}\end{array}$

The voltage $v_3$ across the 10 ohms resistor is calculated by using ohms law as follows.

$v_3=-10i_3$

Substitute $0.5714$ for $i_3$ to find the voltage $v_3$ in volts.

$\begin{array}{c}{v}_3=-10\left(0.5714\right)\\ =-5.714\text{V}\end{array}$

The total voltage $v_x$ across the 10 ohms resistor is,

$v_x=v_1+v_2+v_3$ (1)

Substitute $14.286\text{V}$ for $v_1$, $-17.143\text{V}$ for $v_2$, and $-5.714\text{V}$ for $v_3$ in equation (1) to find the total voltage $v_x$ across the 10 ohms resistor in volts.

$\begin{array}{c}{v}_x=14.286\text{V}-17.143\text{V}-5.714\text{V}\\ =-8.57\text{1}\text{V}\end{array}$

PSPICE Simulation:

Refer to Figure 3, when $3\text{A}$ current source is active,

Draw the circuit diagram in PSPICE as shown in Figure 9.

Save the circuit and provide the Simulation Settings as shown in Figure 10.

Now run the simulation and the results will be displayed as shown in Figure 11 by enabling “Enable Bias Voltage Display” icon.

Refer to Figure 5, when $6\text{A}$ current source is active,

Draw the circuit diagram in PSPICE as shown in Figure 12.

Save the circuit and provide the Simulation Settings as shown in Figure 13.

Now run the simulation and the results will be displayed as shown in Figure 14 by enabling “Enable Bias Voltage Display” icon.

Refer to Figure 8, when $2\text{A}$ current source is active,

Draw the circuit diagram in PSPICE as shown in Figure 15.

Save the circuit and provide the Simulation Settings as shown in Figure 16.

Now run the simulation and the results will be displayed as shown in Figure 17 by enabling “Enable Bias Voltage Display” icon.

From Figure 11, the voltage $v_1$ across the 10 ohms resistor is,

$\begin{array}{c}{v}_1=31.43\text{V}-\text{17}\text{.14}\text{V}\\ =\text{14}\text{.29}\text{V}\end{array}$

From Figure 14, the voltage $v_2$ across the 10 ohms resistor is,

$\begin{array}{c}{v}_2=34.29\text{V}-51.43\text{V}\\ =-17.14\text{V}\end{array}$

From Figure 17, the voltage $v_3$ across the 10 ohms resistor is,

$\begin{array}{c}{v}_3=11.43\text{V}-17.14\text{V}\\ =-5.71\text{V}\end{array}$

Substitute $14.29\text{V}$ for $v_1$, $-17.14\text{V}$ for $v_2$, and $-5.71\text{V}$ for $v_3$ in equation (1) to find the total voltage $v_x$ across the 10 ohms resistor in volts.

$\begin{array}{c}{v}_x=14.29\text{V}-17.14\text{V}-5.71\text{V}\\ =-8.56\text{V}\end{array}$

Conclusion:

Thus, the value of the voltage $v_x$ in the given circuit is $-8.57\text{1}\text{V}$ and verified using PSpice.