Chapter 3, Problem 1P

(a)

To determine

Show that the given circuit satisfies Kirchhoff’s current law at junction terminals x-y.

Answer to Problem 1P

Yes, the given circuit satisfies Kirchhoff’s current law at junction terminals x-y.

Explanation of Solution

Given data:

Refer to Figure given in the textbook.

The voltage delivered by the source is $120\text{V}$.

PSPICE Simulation:

Draw the circuit diagram in PSpice as shown in Figure 1.

Save the circuit and provide the Simulation Settings as shown in Figure 2.

Now run the simulation and the results will be displayed as shown in Figure 3 by enabling the “Enable Bias Current Display” icon.

From Figure 3, source current $i_s=12\text{A}$ and the node currents $i_1=4\text{A}$ and $i_2=8\text{A}$.

Kirchhoff’s current law states that the current entering the node is equal to the current leaving the node.

In Figure 3, apply Kirchhoff current law at node b. Therefore,

$-i_s+i_1+i_2=0$ (1)

Rearrange the equation (1) as follows,

$i_s=i_1+i_2$ (2)

Substitute $12\text{A}$ for $i_s$, $4\text{A}$ for $i_1$ and $8\text{A}$ for $i_2$ in equation (2).

$\begin{array}{l}12\text{A}=4\text{A}+8\text{A}\\ 12\text{A}=12\text{A}\\ 12\text{A}-12\text{A}=\text{0}\end{array}$

Hence, the given circuit satisfies Kirchhoff’s current law at junction terminals x-y.

Conclusion:

Thus, yes, the given circuit satisfies Kirchhoff’s current law at junction terminals x-y.

(b)

To determine

Show that the given circuit satisfies Kirchhoff’s voltage law.

Answer to Problem 1P

Yes, the given circuit satisfies Kirchhoff’s voltage law.

Explanation of Solution

Given data:

Refer to Figure given in the textbook.

Voltage delivered by the source is $120\text{V}$.

PSPICE Simulation:

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 3 by enabling the “Enable Bias Current Display” icon and “Enable Bias Voltage Display” icon.

From Figure 6, the voltage $v_1$ across the $4\Omega$ resistor is,

$\begin{array}{c}{v}_1=4\Omega \times 12\text{A}\\ =\text{48}\text{V}\left\{\because 1\text{V}=1\Omega \times 1\text{A}\right\}\end{array}$

The voltage $v_2$ across the $18\Omega$ resistor is,

$\begin{array}{c}{v}_2=18\Omega \times 4\text{A}\\ =7\text{2}\text{V}\left\{\because 1\text{V}=1\Omega \times 1\text{A}\right\}\end{array}$

The voltage $v_3$ across the $3\Omega$ resistor is,

$\begin{array}{c}{v}_3=3\Omega \times 8\text{A}\\ =24\text{V}\left\{\because 1\text{V}=1\Omega \times 1\text{A}\right\}\end{array}$

The voltage $v_4$ across the $6\Omega$ resistor is,

$\begin{array}{c}{v}_4=6\Omega \times 8\text{A}\\ =48\text{V}\left\{\because 1\text{V}=1\Omega \times 1\text{A}\right\}\end{array}$

Kirchhoff’s voltage law states that the sum of the voltage rise around any closed loop must be equal to the sum of voltage drops around that loop.

In Figure 6, apply Kirchhoff’s voltage law to the loop abda.

$v_s=v_1+v_2$ (3)

From Figure 6, the source voltage $v_s$ is $120\text{V}$.

Substitute $120\text{V}$ for $v_s$, $48\text{V}$ for $v_1$, and $72\text{V}$ for $v_2$ in equation (3). Therefore,

$\begin{array}{l}120\text{V}=48\text{V}+72\text{V}\\ 120\text{V}=120\text{V}\\ \text{120}\text{V}-\text{120}\text{V}=\text{0}\end{array}$

In Figure 6, apply Kirchhoff’s voltage law to the loop bcdb.

$v_2=v_3+v_4$ (4)

Substitute $72\text{V}$ for $v_2$, $24\text{V}$ for $v_3$, and $48\text{V}$ for $v_4$ in equation (4). Therefore,

$\begin{array}{l}72\text{V}=24\text{V}+48\text{V}\\ 72\text{V}=72\text{V}\\ 72\text{V}-72\text{V}=\text{0}\end{array}$

In Figure 6, apply Kirchhoff’s voltage law to the loop abcda.

$v_s=v_1+v_3+v_4$ (5)

Substitute $120\text{V}$ for $v_s$, $48\text{V}$ for $v_1$, $24\text{V}$ for $v_3$, and $48\text{V}$ for $v_4$ in equation (5). Therefore,

$\begin{array}{l}120\text{V}=\text{48}\text{V}+24\text{V}+48\text{V}\\ 120\text{V}=120\text{V}\\ 120\text{V}-120\text{V}=\text{0}\end{array}$

Hence, the given circuit satisfies Kirchhoff’s voltage law around every closed loop.

Conclusion:

Thus, yes, the given circuit satisfies Kirchhoff’s voltage law.