Chapter 11, Problem 53E

■ FIGURE 11.49

Instead of including a capacitor as indicated in Fig. 11.49, the circuit is erroneously constructed using two identical inductors, each having an impedance of j30 W at the operating frequency of 50 Hz. (a) Compute the complex power delivered to each passive component. (b) Verify your solution by calculating the complex power supplied by the source. (c) At what power factor is the source operating?

To determine

Find the complex power delivered to each passive element when $-j25\Omega$ capacitor is replace with $j30\Omega$ inductor in the circuit in Figure 11.49 in the textbook.

Answer to Problem 53E

The complex power delivered to $j30\Omega$ inductive reactance of the left-loop inductor, $10\Omega$ resistor, $j30\Omega$ inductive reactance of the right-loop inductor and $15\Omega$ resistor are $\underset{\_}{68.7\angle 90°\text{VA}}$, $\underset{\_}{16.89\angle 0°\text{VA}}$, $\underset{\_}{\text{4}\text{.50}\angle 90°\text{VA}}$, and $\underset{\_}{2.26\angle 0°\text{VA}}$, respectively.

Explanation of Solution

Given data:

Refer to Figure 11.49 in the textbook for the given circuit.

$V_s=50\angle -17°\text{V}\text{rms}$

$-j25\Omega$ capacitor is replace with $j30\Omega$ inductor in the given circuit.

Operating frequency is 50 Hz.

Formula used:

Write the expression for complex power delivered to the element as follows:

$S=V{I}^{*}$        (1)

Here,

$V$ is the voltage across the element and

$I$ is the current through the element.

Write the expression for current in terms of voltage and impedance as follows:

$I=\frac{V}{Z}$        (2)

Calculation:

Consider $-j25\Omega$ capacitor is replace with $j30\Omega$ inductor in the given circuit and find the equivalent impedance for the circuit to the right of $j30\Omega$ inductive reactance of the inductor as follows:

$\begin{array}{c}{Z}_{\text{eq}}=\left[\frac{\left(15+j30\right)\left(10\right)}{\left(15+j30\right)+\left(10\right)}\right]\Omega \\ =\left(8.3606+j1.9672\right)\Omega \\ =8.5889\angle 13.24°\Omega \end{array}$

Use the expression in Equation (2) and find the source current as follows:

$I_s=\frac{V_s}{j30\Omega +Z_{\text{eq}}}$

Substitute $50\angle -17°\text{V}$ for $V_s$ and $\left(8.3606+j1.9672\right)\Omega$ for $Z_{\text{eq}}$ to obtain the source current as follows:

$\begin{array}{c}{I}_s=\frac{50\angle -17°\text{V}}{j30\Omega +\left(8.3606+j1.9672\right)\Omega }\\ =\frac{50\angle -17°\text{V}}{\left(8.3606+j31.9672\right)\Omega }\\ =\frac{50\angle -17°}{33.0424\angle 75.34°}\text{A}\\ =1.5132\angle -92.34°\text{A}\end{array}$

Use voltage division rule and find the voltage across $j30\Omega$ left-loop inductor as follows:

$V_{j30\Omega -\text{left}}=\left(\frac{j30\Omega }{j30\Omega +Z_{\text{eq}}}\right)V_s$

Substitute $50\angle -17°\text{V}$ for $V_s$ and $\left(8.3606+j1.9672\right)\Omega$ for $Z_{\text{eq}}$ to obtain the value of $V_{j30\Omega }$.

$\begin{array}{c}{V}_{j30\Omega -\text{left}}=\left[\frac{j30\Omega }{j30\Omega + \left(8.3606+j1.9672\right)\Omega }\right]\left(50\angle -17°\text{V}\right)\\ =\left[\frac{30\angle 90°}{33.0424\angle 75.34°}\right]\left(50\angle -17°\text{V}\right)\\ =45.3962\angle -2.34°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to $j30\Omega$ inductive reactance of the left-loop inductor as follows:

$S_{j30\Omega -\text{left}}=V_{j30\Omega -\text{left}}{\left(I_s\right)}^{*}$

Substitute $45.3962\angle -2.34°\text{V}$ for $V_{j30\Omega -\text{left}}$ and $1.5132\angle -92.34°\text{A}$ for $I_s$ to obtain the complex power delivered to $j30\Omega$ inductive reactance of the inductor.

$\begin{array}{c}{S}_{j30\Omega -\text{left}}=\left(45.3962\angle -2.34°\text{V}\right){\left(1.5132\angle -92.34°\text{A}\right)}^{*}\\ =\left(45.3962\angle -2.34°\text{V}\right)\left(1.5132\angle 92.34°\text{A}\right)\\ =68.6935\angle 90°\text{VA}\\ \cong 68.7\angle 90°\text{VA}\end{array}$

Consider the node voltage across the shunt branches as $V_1$ and find the value of $V_1$ by voltage division rule as follows:

$V_1=\left(\frac{Z_{\text{eq}}}{j30\Omega +Z_{\text{eq}}}\right)V_s$

Substitute $50\angle -17°\text{V}$ for $V_s$ and $\left(8.3606+j1.9672\right)\Omega$ for $Z_{\text{eq}}$ to obtain the value of $V_1$.

$\begin{array}{c}{V}_1=\left[\frac{ \left(8.3606+j1.9672\right)\Omega }{j30\Omega + \left(8.3606+j1.9672\right)\Omega }\right]\left(50\angle -17°\text{V}\right)\\ =\left[\frac{8.5885\angle 13.24°\Omega }{33.0424\angle 75.34°}\right]\left(50\angle -17°\text{V}\right)\\ =12.9961\angle -79.1°\text{V}\end{array}$

Use current division rule and find the current through $10\Omega$ resistor as follows:

$I_{10\Omega }=\left[\frac{\left(15+j30\right)\Omega }{10\Omega +\left(15+j30\right)\Omega }\right]I_s$

Substitute $1.5132\angle -92.34°\text{A}$ for $I_s$ to obtain the value of current through $10\Omega$ resistor.

$\begin{array}{c}{I}_{10\Omega }=\left[\frac{\left(15+j30\right)\Omega }{10\Omega +\left(15+j30\right)\Omega }\right]\left(1.5132\angle -92.34°\text{A}\right)\\ =\left(\frac{33.5410\angle 63.43°\Omega }{39.0512\angle 50.19°\Omega }\right)\left(1.5132\angle -92.34°\text{A}\right)\\ =1.2996\angle -79.1°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the $10\Omega$ resistor as follows:

$S_{10\Omega }=V_1{\left(I_{10\Omega }\right)}^{*}$

Substitute $12.9961\angle -79.1°\text{V}$ for $V_1$ and $1.2996\angle -79.1°\text{A}$ for $I_{10\Omega }$ to obtain the complex power delivered to the $10\Omega$ resistor.

$\begin{array}{c}{S}_{10\Omega }=\left(12.9961\angle -79.1°\text{V}\right){\left(1.2996\angle -79.1°\text{A}\right)}^{*}\\ =\left(12.9961\angle -79.1°\text{V}\right)\left(1.2996\angle 79.1°\text{A}\right)\\ =16.8896\angle 0°\text{VA}\\ =16.89\angle 0°\text{VA}\end{array}$

Use voltage division rule and find the voltage across $j30\Omega$ right-loop inductor as follows:

$V_{j30\Omega -\text{right}}=\left(\frac{j30\Omega }{15\Omega +j30\Omega }\right)V_1$

Substitute $12.9961\angle -79.1°\text{V}$ for $V_1$ to obtain the voltage across $j30\Omega$ right-loop inductor.

$\begin{array}{c}{V}_{j30\Omega -\text{right}}=\left(\frac{j30\Omega }{15\Omega +j30\Omega }\right)\left(12.9961\angle -79.1°\text{V}\right)\\ =\left(\frac{30\angle 90°\Omega }{33.5410\angle 63.43°\Omega }\right)\left(12.9961\angle -79.1°\text{V}\right)\\ =11.6240\angle -52.53°\text{V}\end{array}$

Use current division rule and find the current through $\left(15+j30\right)\Omega$ branch as follows:

$I_{\left(15+j30\right)\Omega }=\left[\frac{10\Omega }{10\Omega +\left(15+j30\right)\Omega }\right]I_s$

Substitute $1.5132\angle -92.34°\text{A}$ for $I_s$ to obtain the value of current through $\left(15+j30\right)\Omega$ branch.

$\begin{array}{c}{I}_{\left(15+j30\right)\Omega }=\left[\frac{10\Omega }{10\Omega +\left(15+j30\right)\Omega }\right]\left(1.5132\angle -92.34°\text{A}\right)\\ =\left(\frac{10\angle 0°\Omega }{39.0512\angle 50.19°\Omega }\right)\left(1.5132\angle -92.34°\text{A}\right)\\ =0.3875\angle -142.53°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the $j30\Omega$ right-loop inductor as follows:

$S_{j30\Omega -\text{right}}=V_{j30\Omega -\text{right}}{\left(I_{\left(15+j30\right)\Omega }\right)}^{*}$

Substitute $11.6240\angle -52.53°\text{V}$ for $V_{j30\Omega -\text{right}}$ and $0.3875\angle -142.53°\text{A}$ for $I_{\left(15+j30\right)\Omega }$ to obtain the complex power delivered to the $j30\Omega$ right-loop inductor.

$\begin{array}{c}{S}_{j30\Omega -\text{right}}=\left(11.6240\angle -52.53°\text{V}\right){\left(0.3875\angle -142.53°\text{A}\right)}^{*}\\ =\left(11.6240\angle -52.53°\text{V}\right)\left(0.3875\angle 142.53°\text{A}\right)\\ =4.5043\angle 90°\text{VA}\\ \cong \text{4}\text{.50}\angle 90°\text{VA}\end{array}$

Use voltage division rule and find the voltage across $15\Omega$ resistor as follows:

$V_{15\Omega }=\left(\frac{15\Omega }{15\Omega +j30\Omega }\right)V_1$

Substitute $12.9961\angle -79.1°\text{V}$ for $V_1$ to obtain the voltage across $15\Omega$ resistor.

$\begin{array}{c}{V}_{15\Omega }=\left(\frac{15\Omega }{15\Omega +j30\Omega }\right)\left(12.9961\angle -79.1°\text{V}\right)\\ =\left(\frac{15\Omega }{33.5410\angle 63.43°\Omega }\right)\left(12.9961\angle -79.1°\text{V}\right)\\ =5.8120\angle -142.53°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the $15\Omega$ resistor as follows:

$S_{15\Omega }=V_{15\Omega }{\left(I_{\left(15+j30\right)\Omega }\right)}^{*}$

Substitute $5.8120\angle -142.53°\text{V}$ for $V_{15\Omega }$ and $0.3875\angle -142.53°\text{A}$ for $I_{\left(15+j30\right)\Omega }$ to obtain the complex power delivered to the $15\Omega$ resistor.

$\begin{array}{c}{S}_{15\Omega }=\left(5.8120\angle -142.53°\text{V}\right){\left(0.3875\angle -142.53°\text{A}\right)}^{*}\\ =\left(5.8120\angle -142.53°\text{V}\right)\left(0.3875\angle 142.53°\text{A}\right)\\ =2.2521\angle 0°\text{VA}\\ \cong 2.26\angle 0°\text{VA}\end{array}$

Conclusion:

Thus, the complex power delivered to $j30\Omega$ inductive reactance of the left-loop inductor, $10\Omega$ resistor, $j30\Omega$ inductive reactance of the right-loop inductor and $15\Omega$ resistor are $\underset{\_}{68.7\angle 90°\text{VA}}$, $\underset{\_}{16.89\angle 0°\text{VA}}$, $\underset{\_}{\text{4}\text{.50}\angle 90°\text{VA}}$, and $\underset{\_}{2.26\angle 0°\text{VA}}$, respectively.

To determine

Verify the solution obtained in Part (a) by calculating the complex power supplied by the source.

Explanation of Solution

Calculation:

Modify the expression in Equation (1) for the complex power supplied by the source as follows:

$S_s=V_s{\left(I_s\right)}^{*}$

Substitute $50\angle -17°\text{V}$ for $V_s$ and $1.5132\angle -92.34°\text{A}$ for $I_s$ to obtain the complex power supplied by the source.

$\begin{array}{c}{S}_s=\left(50\angle -17°\text{V}\right){\left(1.5132\angle -92.34°\text{A}\right)}^{*}\\ =\left(50\angle -17°\text{V}\right)\left(1.5132\angle 92.34°\text{A}\right)\\ =75.66\angle 75.34°\text{VA}\end{array}$

Write the expression for sum of complex delivered to (or absorbed by) each passive element as follows:

${\displaystyle \sum S_{\text{abs}}}=S_{j30\Omega -\text{left}}+S_{10\Omega }+S_{j30\Omega -\text{right}}+S_{15\Omega }$

From Part (a), substitute $68.7\angle 90°\text{VA}$ for $S_{j30\Omega -\text{left}}$, $16.89\angle 0°\text{VA}$ for $S_{10\Omega }$, $\text{4}\text{.50}\angle 90°\text{VA}$ for $S_{j30\Omega -\text{right}}$, and $2.26\angle 0°\text{VA}$ for $S_{15\Omega }$ to obtain the complex power absorbed by each passive element.

$\begin{array}{c}{\displaystyle \sum S_{\text{abs}}}=68.7\angle 90°\text{VA}+16.89\angle 0°\text{VA}+\text{4}\text{.50}\angle 90°\text{VA}+2.26\angle 0°\text{VA}\\ =\left[\left(0+j68.7\right)+\left(16.89+0j\right)+\left(0+j4.50\right)+\left(2.26+0j\right)\right]\text{VA}\\ =\left(19.15+j73.2\right)\text{VA}\\ =75.66\angle 75.34°\text{VA}\end{array}$

From the calculation, it is clear that, the complex power supplied by the source is equal to the complex power delivered to each passive element.

Conclusion:

Thus, the solution obtained in Part (a) is verified.

To determine

Find the power factor of the source.

Answer to Problem 53E

The power factor of the source is 0.253 lagging.

Explanation of Solution

Formula used:

Write the expression for complex power in the rectangular form as follows:

$S=P+jQ$        (3)

Here,

$P$ is the average power and

$Q$ is the reactive power.

Write the expression for power factor as follows:

$\text{PF}=\cos \left[{\tan }^{-1}\left(\frac{Q}{P}\right)\right]$        (4)

Calculation:

Rewrite the expression for complex power supplied by the source in rectangular form as follows:

$S_s=\left(19.1482+j73.1968\right)\text{VA}$

Compare the complex power supplied by the source with the expression in Equation (3) and write the average and reactive power supplied by the source as follows:

$\begin{array}{l}P=19.1482\text{W}\\ Q=73.1968\text{VAR}\end{array}$

Substitute 19.1482 W for $P$ and 73.1968 VAR for $Q$ in Equation (4) to obtain the value of source power factor.

$\begin{array}{c}\text{PF}=\cos \left[{\tan }^{-1}\left(\frac{73.1968\text{VAR}}{19.1482\text{W}}\right)\right]\\ =\cos \left(75.34°\right)\\ =0.2530\end{array}$

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.

$\text{PF}=0.253\text{lagging}$

Conclusion:

Thus, the power factor of the source is 0.253 lagging.