Chapter 11, Problem 35E

(a)

To determine

Find the average power delivered to each load, the apparent power supplied by the source, and the power factor of the combined loads for the circuit in Figure 11.43 in the textbook.

Answer to Problem 35E

The average power delivered to $Z_1$, $Z_2$, the apparent power supplied by the source, and the power factor of the combined loads are $\underset{\_}{139.8\text{W}}$, $\underset{\_}{259\text{W}}$, $\underset{\_}{408.35\text{VA}}$, and $\underset{\_}{0.977\text{lagging}}$, respectively.

Explanation of Solution

Given data:

Refer to Figure 11.43 in the textbook for the given circuit.

$\begin{array}{l}{V}_{\text{ff}}=119\angle 3°\text{V,}\text{rms}\\ Z_1=14\angle 32°\Omega \\ Z_2=22\Omega \end{array}$

Formula used:

Write the expression for average power delivered to load as follows:

$P=V_{\text{eff}}{I}_{\text{eff}}\cos \left(\theta -\varphi \right)$        (1)

Here,

$V_{\text{eff}}$ is the rms source voltage in the circuit,

$I_{\text{eff}}$ is the rms current in the circuit,

$\theta$ is the phase angle of rms source voltage,

$\varphi$ is the phase angle of the rms current, and

$\cos \left(\theta -\varphi \right)$ is the power factor.

Write the expression for rms current in the circuit as follows:

$I_{\text{eff}}=\frac{V_{\text{eff}}}{Z}$        (2)

Here,

$Z$ is the is the total impedance in the given circuit.

Write the expression for total impedance in the given circuit as follows:

$Z=Z_1+Z_2$        (3)

Write the expression for rms voltage across load as follows:

$V_{\text{eff}}=I_{\text{eff}}Z$        (4)

Write the expression for complex power supplied by the source as follows:

$S=V_{\text{eff}}{\left(I_{\text{eff}}\right)}^{*}$        (5)

Write the expression for power factor of the combined loads as follows:

$PF=\cos \left(\theta -\varphi \right)$        (6)

Calculation:

From Equation (3), substitute $\left(Z_1+Z_2\right)$ for $Z$ in Equation (2) as follows:

$I_{\text{eff}}=\frac{V_{\text{eff}}}{Z_1+Z_2}$        (7)

Substitute $119\angle 3°\text{V}$ for $V_{\text{ff}}$, $14\angle 32°\Omega$ for $Z_1$, and $22\Omega$ for $Z_2$ to obtain the value of rms current in the circuit.

$\begin{array}{c}{I}_{\text{eff}}=\frac{119\angle 3°\text{V}}{14\angle 32°\Omega +22\Omega }\\ =\frac{119\angle 3°}{11.8726+j7.4188+22}\text{A}\\ =\frac{119\angle 3°}{33.8726+j7.4188}\text{A}\\ =\frac{119\angle 3°}{34.6755\angle 12.35°}\text{A}\end{array}$

Simplify the expression as follows:

$I_{\text{eff}}=3.4315\angle -9.35°\text{A}$

Modify the expression in Equation (4) for the voltage across the load $Z_1$ as follows:

${\left(V_{\text{eff}}\right)}_1=I_{\text{eff}}{Z}_1$        (8)

Substitute $3.4315\angle -9.35°\text{A}$ for $I_{\text{eff}}$ and $14\angle 32°\Omega$ for $Z_1$ to obtain the voltage across the load $Z_1$.

$\begin{array}{c}{\left(V_{\text{eff}}\right)}_1=\left(3.4315\angle -9.35°\text{A}\right)\left(14\angle 32°\Omega \right)\\ =48.041\angle 22.65°\text{V}\end{array}$

Modify the expression in Equation (1) for the average power delivered to the load $Z_1$ as follows:

$P_1={\left(V_{\text{eff}}\right)}_1{I}_{\text{eff}}\cos \left({\theta }_1-\varphi \right)$        (9)

Substitute 48.041 V for ${\left(V_{\text{eff}}\right)}_1$, 3.4315 A for $I_{\text{eff}}$, $22.65°$ for ${\theta }_1$, and $-9.35°$ for $\varphi$ to obtain the average power delivered to the load $Z_1$.

$\begin{array}{c}{P}_1=\left(48.041\text{V}\right)\left(3.4315\text{A}\right)\cos \left[22.65°-\left(-9.35°\right)\right]\\ =139.8030\text{W}\\ \cong 139.8\text{W}\end{array}$

Modify the expression in Equation (4) for the voltage across the load $Z_2$ as follows:

${\left(V_{\text{eff}}\right)}_2=I_{\text{eff}}{Z}_2$        (10)

Substitute $3.4315\angle -9.35°\text{A}$ for $I_{\text{eff}}$ and $22\Omega$ for $Z_2$ to obtain the voltage across the load $Z_2$.

$\begin{array}{c}{\left(V_{\text{eff}}\right)}_2=\left(3.4315\angle -9.35°\text{A}\right)\left(22\Omega \right)\\ =75.493\angle -9.35°\text{V}\end{array}$

Modify the expression in Equation (1) for the average power delivered to the load $Z_2$ as follows:

$P_2={\left(V_{\text{eff}}\right)}_2{I}_{\text{eff}}\cos \left({\theta }_2-\varphi \right)$        (11)

Substitute 75.493 V for ${\left(V_{\text{eff}}\right)}_2$, 3.4315 A for $I_{\text{eff}}$, $-9.35°$ for ${\theta }_2$, and $-9.35°$ for $\varphi$ to obtain the average power delivered to the load $Z_2$.

$\begin{array}{c}{P}_2=\left(75.493\text{V}\right)\left(3.4315\text{A}\right)\cos \left[-9.35°-\left(-9.35°\right)\right]\\ =259.0542\text{W}\\ \cong 259\text{W}\end{array}$

Substitute $119\angle 3°\text{V}$ for $V_{\text{ff}}$ and $3.4315\angle -9.35°\text{A}$ for $I_{\text{eff}}$ in Equation (5) to obtain the complex power supplied by the source.

$\begin{array}{c}S=\left(119\angle 3°\text{V}\right){\left(3.4315\angle -9.35°\text{A}\right)}^{*}\\ =\left(119\angle 3°\text{V}\right)\left(3.4315\angle 9.35°\text{A}\right)\\ =408.3485\angle 12.35°\text{VA}\\ =\left(398.8990+j87.3387\right)\text{VA}\end{array}$

Find the apparent power supplied by the source from the complex power as follows:

$\begin{array}{c}\left|S\right|=\left|408.3485\angle 12.35°\text{VA}\right|\\ =408.3485\text{VA}\\ \cong 408.35\text{VA}\end{array}$

Substitute $3°$ for $\theta$ and $-9.35°$ for $\varphi$ in Equation (6) to obtain the power factor of the combined loads.

$\begin{array}{c}PF=\cos \left[3°-\left(-9.35°\right)\right]\\ =0.9768\\ \cong 0.977\end{array}$

If the imaginary part of the complex power (reactive power) is positive value, then the load has lagging power factor. If the imaginary part is negative value, then the load has leading power factor.

As the imaginary part of the given complex power is positive value, the power factor is lagging power factor.

$\text{PF}=0.977\text{lagging}$

Conclusion:

Thus, the average power delivered to $Z_1$, $Z_2$, the apparent power supplied by the source, and the power factor of the combined loads are $\underset{\_}{139.8\text{W}}$, $\underset{\_}{259\text{W}}$, $\underset{\_}{408.35\text{VA}}$, and $\underset{\_}{0.977\text{lagging}}$, respectively.

(b)

To determine

Find the average power delivered to each load, the apparent power supplied by the source, and the power factor of the combined loads for the circuit in Figure 11.43 in the textbook.

Answer to Problem 35E

The average power delivered to $Z_1$, $Z_2$, the apparent power supplied by the source, and the power factor of the combined loads are $\underset{\_}{435.72\text{W}}$, $\underset{\_}{1307.2\text{W}}$, $\underset{\_}{1756.45\text{VA}}$, and $\underset{\_}{0.992\text{leading}}$, respectively.

Explanation of Solution

Given data:

$\begin{array}{l}{Z}_1=2\angle 0°\Omega \\ Z_2=\left(6-j\right)\Omega \end{array}$

Calculation:

Substitute $119\angle 3°\text{V}$ for $V_{\text{ff}}$, $2\angle 0°\Omega$ for $Z_1$, and $\left(6-j\right)\Omega$ for $Z_2$ in Equation (7) to obtain the value of rms current in the circuit.

$\begin{array}{c}{I}_{\text{eff}}=\frac{119\angle 3°\text{V}}{2\angle 0°\Omega +\left(6-j\right)\Omega }\\ =\frac{119\angle 3°}{8-j}\text{A}\\ =\frac{119\angle 3°}{8.06225\angle -7.13°}\text{A}\\ =14.7601\angle 10.13°\text{A}\end{array}$

Substitute $14.7601\angle 10.13°\text{A}$ for $I_{\text{eff}}$ and $2\angle 0°\Omega$ for $Z_1$ in Equation (8) to obtain the voltage across the load $Z_1$.

$\begin{array}{c}{\left(V_{\text{eff}}\right)}_1=\left(14.7601\angle 10.13°\text{A}\right)\left(2\angle 0°\Omega \right)\\ =29.5202\angle 10.13°\text{V}\end{array}$

Substitute 29.5202 V for ${\left(V_{\text{eff}}\right)}_1$, 14.7601 A for $I_{\text{eff}}$, $10.13°$ for ${\theta }_1$, and $10.13°$ for $\varphi$ in Equation (9) to obtain the average power delivered to the load $Z_1$.

$\begin{array}{c}{P}_1=\left(29.5202\text{V}\right)\left(14.7601\text{A}\right)\cos \left(10.13°-10.13°\right)\\ =435.7211\text{W}\\ \cong 435.72\text{W}\end{array}$

Substitute $14.7601\angle 10.13°\text{A}$ for $I_{\text{eff}}$ and $\left(6-j\right)\Omega$ for $Z_2$ in Equation (10) to obtain the voltage across the load $Z_2$.

$\begin{array}{c}{\left(V_{\text{eff}}\right)}_2=\left(14.7601\angle 10.13°\text{A}\right)\left[\left(6-j\right)\Omega \right]\\ =\left(14.7601\angle 10.13°\text{A}\right)\left(6.0827\angle -9.46°\Omega \right)\\ =89.7812\angle 0.67°\text{V}\end{array}$

Substitute 89.7812 V for ${\left(V_{\text{eff}}\right)}_2$, 14.7601 A for $I_{\text{eff}}$, $0.67°$ for ${\theta }_2$, and $10.13°$ for $\varphi$ in Equation (11) to obtain the average power delivered to the load $Z_2$.

$\begin{array}{c}{P}_2=\left(89.7812\text{V}\right)\left(14.7601\text{A}\right)\cos \left(0.67°-10.13°\right)\\ =1307.1525\text{W}\\ \cong 1307.2\text{W}\end{array}$

Substitute $119\angle 3°\text{V}$ for $V_{\text{ff}}$ and $14.7601\angle 10.13°\text{A}$ for $I_{\text{eff}}$ in Equation (5) to obtain the complex power supplied by the source.

$\begin{array}{c}S=\left(119\angle 3°\text{V}\right){\left(14.7601\angle 10.13°\text{A}\right)}^{*}\\ =\left(119\angle 3°\text{V}\right)\left(14.7601\angle -10.13°\text{A}\right)\\ =1756.4519\angle -7.13°\text{VA}\\ =\left(1742.7515-j217.9756\right)\text{VA}\end{array}$

Find the apparent power supplied by the source from the complex power as follows:

$\begin{array}{c}\left|S\right|=\left|1756.4519\angle -7.13°\text{VA}\right|\\ =1756.4519\text{VA}\\ \cong 1756.45\text{VA}\end{array}$

Substitute $3°$ for $\theta$ and $10.13°$ for $\varphi$ in Equation (6) to obtain the power factor of the combined loads.

$\begin{array}{c}PF=\cos \left(3°-10.13°\right)\\ =0.992\end{array}$

As the imaginary part of the given complex power is negative value, the power factor is leading power factor.

$\text{PF}=0.992\text{leading}$

Conclusion:

Thus, the average power delivered to $Z_1$, $Z_2$, the apparent power supplied by the source, and the power factor of the combined loads are $\underset{\_}{435.72\text{W}}$, $\underset{\_}{1307.2\text{W}}$, $\underset{\_}{1756.45\text{VA}}$, and $\underset{\_}{0.992\text{leading}}$, respectively.

(c)

To determine

Find the average power delivered to each load, the apparent power supplied by the source, and the power factor of the combined loads for the circuit in Figure 11.43 in the textbook.

Answer to Problem 35E

The average power delivered to $Z_1$, $Z_2$, the apparent power supplied by the source, and the power factor of the combined loads are $\underset{\_}{16.3\text{W}}$, $\underset{\_}{0\text{W}}$, $\underset{\_}{82.11\text{VA}}$, and $\underset{\_}{0.199\text{lagging}}$, respectively.

Explanation of Solution

Given data:

$\begin{array}{l}{Z}_1=100\angle 70°\Omega \\ Z_2=75\angle 90°\Omega \end{array}$

Calculation:

Substitute $119\angle 3°\text{V}$ for $V_{\text{ff}}$, $100\angle 70°\Omega$ for $Z_1$, and $75\angle 90°\Omega$ for $Z_2$ in Equation (7) to obtain the value of rms current in the circuit.

$\begin{array}{c}{I}_{\text{eff}}=\frac{119\angle 3°\text{V}}{100\angle 70°\Omega +75\angle 90°\Omega }\\ =\frac{119\angle 3°}{\left(34.202+j93.969\right)+\left(0+j75\right)}\text{A}\\ =\frac{119\angle 3°}{172.3957\angle 78.56°}\text{A}\\ =0.69\angle -75.56°\text{A}\end{array}$

Substitute $0.69\angle -75.56°\text{A}$ for $I_{\text{eff}}$ and $100\angle 70°\Omega$ for $Z_1$ in Equation (8) to obtain the voltage across the load $Z_1$.

$\begin{array}{c}{\left(V_{\text{eff}}\right)}_1=\left(0.69\angle -75.56°\text{A}\right)\left(100\angle 70°\Omega \right)\\ =69\angle -5.56°\text{V}\end{array}$

Substitute 69 V for ${\left(V_{\text{eff}}\right)}_1$, 0.69 A for $I_{\text{eff}}$, $-5.56°$ for ${\theta }_1$, and $-75.56°$ for $\varphi$ in Equation (9) to obtain the average power delivered to the load $Z_1$.

$\begin{array}{c}{P}_1=\left(69\text{V}\right)\left(0.69\text{A}\right)\cos \left[-5.56°-\left(-75.56°\right)\right]\\ =16.2835\text{W}\\ \cong 16.3\text{W}\end{array}$

Substitute $0.69\angle -75.56°\text{A}$ for $I_{\text{eff}}$ and $75\angle 90°\Omega$ for $Z_2$ in Equation (10) to obtain the voltage across the load $Z_2$.

$\begin{array}{c}{\left(V_{\text{eff}}\right)}_2=\left(0.69\angle -75.56°\text{A}\right)\left(75\angle 90°\Omega \right)\\ =51.75\angle 14.44°\text{V}\end{array}$

Substitute 51.75 V for ${\left(V_{\text{eff}}\right)}_2$, 0.69 A for $I_{\text{eff}}$, $14.44°$ for ${\theta }_2$, and $-75.56°$ for $\varphi$ in Equation (11) to obtain the average power delivered to the load $Z_2$.

$\begin{array}{c}{P}_2=\left(51.75\text{V}\right)\left(0.69\text{A}\right)\cos \left[14.44°-\left(-75.56°\right)\right]\\ =0\text{W}\end{array}$

Substitute $119\angle 3°\text{V}$ for $V_{\text{ff}}$ and $0.69\angle -75.56°\text{A}$ for $I_{\text{eff}}$ in Equation (5) to obtain the complex power supplied by the source.

$\begin{array}{c}S=\left(119\angle 3°\text{V}\right){\left(0.69\angle -75.56°\text{A}\right)}^{*}\\ =\left(119\angle 3°\text{V}\right)\left(0.69\angle 75.56°\text{A}\right)\\ =82.11\angle 78.56°\text{VA}\\ =\left(16.2858+j80.4787\right)\text{VA}\end{array}$

Find the apparent power supplied by the source from the complex power as follows:

$\begin{array}{c}\left|S\right|=\left|82.11\angle 78.56°\text{VA}\right|\\ =82.11\text{VA}\end{array}$

Substitute $3°$ for $\theta$ and $-75.56°$ for $\varphi$ in Equation (6) to obtain the power factor of the combined loads.

$\begin{array}{c}PF=\cos \left[3°-\left(-75.56°\right)\right]\\ \cong 0.199\end{array}$

As the imaginary part of the given complex power is positive value, the power factor is lagging power factor.

$\text{PF}=0.199\text{lagging}$

Conclusion:

Thus, the average power delivered to $Z_1$, $Z_2$, the apparent power supplied by the source, and the power factor of the combined loads are $\underset{\_}{16.3\text{W}}$, $\underset{\_}{0\text{W}}$, $\underset{\_}{82.11\text{VA}}$, and $\underset{\_}{0.199\text{lagging}}$, respectively.