Chapter 11, Problem 62P

For the circuit in Fig. 11.81, find V_s.

To determine

Find the voltage $V_s$ of the circuit shown in Figure 11.81.

Answer to Problem 62P

The voltage $V_s$ of the circuit is $120.06\angle 0.03°\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 11.81 in the textbook.

The voltage $V_2$ is $120\text{V}$.

For load A,

The real power $P$ is $10\text{W}$.

The power factor $\text{pf}$ is $0.9\left(\text{lagging}\right)$.

For load B,

The real power $P$ is $15\text{W}$.

The power factor $\text{pf}$ is $0.8\left(\text{leading}\right)$.

Formula used:

Write the expression to find the complex power.

$S=P+jQ$ (1)

Here,

$P$ is the real power, and

$Q$ is the reactive power.

Write the expression to find the power factor $\text{pf}$.

$\text{pf}=\cos \left(\theta \right)$ (2)

Here,

$\theta$ is the phase angle.

Write the expression to find the real power.

$P=S\cos \left(\theta \right)$ (3)

Write the expression to find the reactive power.

$Q=S\sin \theta$ (4)

Write the expression to find the output voltage.

$I^{*}=\frac{S}{V}$ (5)

Calculation:

The given Figure 11.81 is redrawn as shown in Figure 1.

For load A:

Substitute $0.9$ for $\text{pf}$ in equation (2) to find the phase angle.

$\begin{array}{l}0.9=\cos \left(\theta \right)\\ \theta ={\cos }^{-1}\left(0.9\right)\\ \theta =25.84°\end{array}$

Substitute $25.84°$ for $\theta$ and $\text{10}\text{W}$ for P in equation (3) to find the apparent power S in VA.

$\text{10}\text{W}=S\cos \left(25.84°\right)$

Rearrange the equation as follows,

$\begin{array}{c}S=\frac{10\text{W}}{\cos \left(25.84°\right)}\left\{\because 1\text{W}=1\text{V}\cdot 1\text{A}\right\}\\ =11.11\text{VA}\end{array}$

Substitute $\text{11}\text{.11}\text{VA}$ for S and $25.84°$ for $\theta$ in equation (4) to find the reactive power Q in VAR.

$\begin{array}{c}Q=\text{11}\text{.11}\text{VA}\sin \left(25.84°\right)\\ =4.843\text{VAR}\end{array}$

Substitute $\text{10}\text{W}$ for P and $\text{4}\text{.843}\text{VAR}$ for $Q$ in equation (1) to find the complex power in VA.

$S_A=\left(10+j4.843\right)\text{VA}$

For load B:

Substitute $0.8$ for $\text{pf}$ in equation (2) to find the phase angle.

$\begin{array}{l}0.8=\cos \left(\theta \right)\\ \theta ={\cos }^{-1}\left(0.8\right)\\ \theta =36.87°\end{array}$

Substitute $36.87°$ for $\theta$ and $\text{15}\text{W}$ for P in equation (3) to find the apparent power S in VA.

$\text{15}\text{W}=S\cos \left(36.87°\right)$

Rearrange the equation as follows,

$\begin{array}{c}S=\frac{15\text{VA}}{\cos \left(36.87°\right)}\left\{\because 1\text{W}=1\text{V}\cdot 1\text{A}\right\}\\ =18.75\text{VA}\end{array}$

Substitute $\text{18}\text{.75}\text{VA}$ for S and $36.87°$ for $\theta$ in equation (4) to find the reactive power Q in VAR.

$\begin{array}{c}Q=\text{18}\text{.75}\text{VA}\sin \left(36.87°\right)\\ =11.25\text{VAR}\end{array}$

Substitute $\text{15}\text{W}$ for P and $\text{11}\text{.25}\text{VAR}$ for $Q$ in equation (1) to find the complex power $S_B$ in VA.

$S_B=\left(15+j11.25\right)\text{VA}$

As the power factor is leading, the load is capacitive. Therefore, the equation becomes,

$S_B=\left(15-j11.25\right)\text{VA}$

The modified Figure is shown in Figure 2.

Substitute $\left(15-j11.25\right)\text{VA}$ for $S_B$ and $120\text{V}$ for $V_2$ in equation (5) to find the current $I_2$ in amperes.

$\begin{array}{l}{I}_2^{*}=\frac{\left(15-j11.25\right)\text{VA}}{120\text{V}}\\ I_2^{*}=\left(0.125-j0.09375\right)\text{A}\\ I_2=\left(0.125+j0.09375\right)\text{A}\end{array}$

The voltage $V_1$ is,

$V_1=V_2+I_2\left(0.3+j0.15\right)\Omega$

Substitute $120\text{V}$ for $V_2$ and $\left(0.125+j0.09375\right)\text{A}$ for $I_2$ in the equation to find the voltage $V_1$ in volts.

$\begin{array}{c}{V}_1=\left(120\text{V}\right)+\left(0.125+j0.09375\right)\left(0.3+j0.15\right)\text{A}\cdot \Omega \\ =\left(120\text{V}\right)+\left(0.02+j0.0469\right)\text{V}\left\{\because 1\text{V}=\text{A}\cdot \Omega \right\}\\ =\left(120.02+j0.0469\right)\text{V}\end{array}$

Substitute $\left(10+j4.843\right)\text{VA}$ for $S_A$ and $\left(120.02+j0.0469\right)\text{V}$ for $V_1$ in equation (5) to find the current $I_1$ in amperes.

$\begin{array}{l}{I}_1^{*}=\frac{\left(10+j4.843\right)\text{VA}}{\left(120.02+j0.0469\right)\text{V}}\left\{\because 1\text{k}={10}^3\right\}\\ I_1^{*}=\frac{11.111\angle 25.84°\text{A}}{120.02\angle 0.02°}\\ I_1^{*}=0.093\angle 25.82°\text{A}\\ I_1=0.093\angle -25.82°\text{A}\end{array}$

Convert the equation from polar to rectangular form.

$I_1=\left(0.0837-j0.0405\right)\text{A}$

The current $I$ is,

$I=I_1+I_2$

Substitute $\left(0.0837-j0.0405\right)\text{A}$ for $I_1$ and $\left(0.125+j0.09375\right)\text{A}$ for $I_2$ in the equation to find the current $I$ in amperes.

$\begin{array}{c}I=\left(0.0837-j0.0405\right)\text{A}+\left(0.125+j0.09375\right)\text{A}\\ =\left(0.2087+j0.053\right)\text{A}\end{array}$

The voltage supplied by the source is,

$V_s=V_1+I\left(0.2+j0.04\right)\Omega$

Substitute $\left(120.02+j0.0469\right)\text{V}$ for $V_1$ and $\left(0.2087+j0.053\right)\text{A}$ for $I$ in the equation to find the voltage supplied by the source in volts.

$\begin{array}{c}{V}_s=\left(120.02+j0.0469\right)\text{V}+\left(0.2087+j0.053\right)\left(0.2+j0.04\right)\text{A}\cdot \Omega \\ =\left(120.02+j0.0469\right)\text{V}+\left(0.04+j0.0189\right)\text{V}\left\{\because 1\text{V}=\text{A}\cdot \Omega \right\}\\ =\left(120.06+j0.0658\right)\text{V}\end{array}$

Convert the equation from rectangular to polar form.

$V_s=120.06\angle 0.03°\text{V}$

Conclusion:

Thus, the voltage $V_s$ of the circuit is $120.06\angle 0.03°\text{V}$.