Chapter 11, Problem 48E

To determine

Find the complex power absorbed by each passive element in the circuit in Figure 11.47 in the textbook and power factor of the source.

Answer to Problem 48E

The complex power absorbed by $18\Omega$ series resistor, $18\Omega$ shunt resistor, $j10\Omega$ inductive reactance of the inductor, $-j5\Omega$ capacitive reactance of the capacitor, $1000\Omega$ resistor, and the source power factor are $\underset{\_}{2.78\angle 0°\text{kVA}}$, $\underset{\_}{126.77\angle 0°\text{VA}}$, $\underset{\_}{70.43\angle 90°\text{VA}}$, $\underset{\_}{597.23\angle -90°\text{VA}}$, $\underset{\_}{2.98\angle 0°\text{VA}}$, and 0.975 leading, respectively.

Explanation of Solution

Given data:

Refer to Figure 11.47 in the textbook for the given circuit.

$V_s=240\angle 45°\text{V}\text{rms}$

Formula used:

Write the expression for complex power absorbed by 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)

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:

Find the equivalent impedance of the shunt components in the circuit as follows:

$\begin{array}{c}{Z}_{\text{eq-shunt}}=\frac{1}{\frac{1}{18+j10}+\frac{1}{-j5}+\frac{1}{1000}}\Omega \\ =\frac{1}{0.0424+j0.0235+j0.2+0.001}\Omega \\ =\left(0.8372-j4.3116\right)\Omega \\ =4.4\angle -79°\Omega \end{array}$

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

$I_s=\frac{V_s}{18\Omega +Z_{\text{eq-shunt}}}$

Substitute $240\angle 45°\text{V}$ for $V_s$ and $\left(0.8372-j4.3116\right)\Omega$ for $Z_{\text{eq-shunt}}$ to obtain the source current as follows:

$\begin{array}{c}{I}_s=\frac{240\angle 45°\text{V}}{18\Omega +\left(0.8372-j4.3116\right)\Omega }\\ =\frac{240\angle 45°\text{V}}{\left(18.8372-j4.3116\right)\Omega }\\ =\frac{240\angle 45°}{19.3243\angle -12.89°}\text{A}\\ =12.4195\angle 57.89°\text{A}\end{array}$

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  $240\angle 45°\text{V}$ for $V_s$ and $12.4195\angle 57.89°\text{A}$ for $I_s$ to obtain the complex power supplied by the source.

$\begin{array}{c}{S}_s=\left(240\angle 45°\text{V}\right){\left(12.4195\angle 57.89°\text{A}\right)}^{*}\\ =\left(240\angle 45°\text{V}\right)\left(12.4195\angle -57.89°\text{A}\right)\\ =2980.68\angle -12.89°\text{VA}\\ \cong 2.98\angle -12.89°\text{kVA}\end{array}$

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

$S_s=\left(2.9049-j0.6647\right)\text{kVA}$

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:

$\begin{array}{l}P=2.9049\text{kW}\\ Q=0.6647\text{kVAR}\end{array}$

Substitute $2.9049\text{kW}$ for $P$ and $0.6647\text{kVAR}$ for $Q$ in Equation (4) to obtain the value of source power factor.

$\begin{array}{c}\text{PF}=\cos \left[{\tan }^{-1}\left(\frac{0.6647\text{VAR}}{2.9049\text{W}}\right)\right]\\ =\cos \left(12.89°\right)\\ \cong 0.975\end{array}$

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 negative value, the power factor is leading power factor.

$\text{PF}=0.975\text{leading}$

Use voltage division rule and find the voltage across $18\Omega$ series resistor as follows:

$V_{18\Omega -\text{series}}=\left(\frac{18\Omega }{18\Omega +Z_{\text{eq-shunt}}}\right)V_s$

Substitute $240\angle 45°\text{V}$ for $V_s$ and $\left(0.8372-j4.3116\right)\Omega$ for $Z_{\text{eq-shunt}}$ to obtain the value of $V_{18\Omega -\text{series}}$.

$\begin{array}{c}{V}_{18\Omega -\text{series}}=\left[\frac{18\Omega }{18\Omega +\left(0.8372-j4.3116\right)\Omega }\right]\left(240\angle 45°\text{V}\right)\\ =\left[\frac{18\Omega }{19.3243\angle -12.89°\Omega }\right]\left(240\angle 45°\text{V}\right)\\ =223.5527\angle 57.89°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power absorbed by the $18\Omega$ series resistor as follows:

$S_{18\Omega -\text{series}}=V_{18\Omega -\text{series}}{\left(I_s\right)}^{*}$

Substitute $240\angle 45°\text{V}$ for $V_s$ and $12.4195\angle 57.89°\text{A}$ for $I_s$ to obtain the complex power absorbed by the $18\Omega$ series resistor.

$\begin{array}{c}{S}_{18\Omega -\text{series}}=\left(223.5527\angle 57.89°\text{V}\right){\left(12.4195\angle 57.89°\text{A}\right)}^{*}\\ =\left(223.5527\angle 57.89°\text{V}\right)\left(12.4195\angle -57.89°\text{A}\right)\\ =2776.4127\angle 0°\text{VA}\\ \cong 2.78\angle 0°\text{kVA}\end{array}$

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

$V_1=\left(\frac{Z_{\text{eq-shunt}}}{18\Omega +Z_{\text{eq-shunt}}}\right)V_s$

Substitute $240\angle 45°\text{V}$ for $V_s$ and $\left(0.8372-j4.3116\right)\Omega$ for $Z_{\text{eq-shunt}}$ to obtain the value of $V_1$.

$\begin{array}{c}{V}_1=\left[\frac{\left(0.8372-j4.3116\right)\Omega }{18\Omega +\left(0.8372-j4.3116\right)\Omega }\right]\left(240\angle 45°\text{V}\right)\\ =\left[\frac{4.4\angle -79°\Omega }{19.3243\angle -12.89°\Omega }\right]\left(240\angle 45°\text{V}\right)\\ =54.6462\angle -21.11°\text{V}\end{array}$

Use voltage division rule and find the voltage across $18\Omega$ shunt resistor as follows:

$V_{18\Omega -\text{shunt}}=\left(\frac{18\Omega }{18\Omega +j10\Omega }\right)V_1$

Substitute $54.6462\angle -21.11°\text{V}$ for $V_1$ to obtain the voltage across $18\Omega$ shunt resistor.

$\begin{array}{c}{V}_{18\Omega -\text{shunt}}=\left(\frac{18\Omega }{18\Omega +j10\Omega }\right)\left(54.6462\angle -21.11°\text{V}\right)\\ =\left(\frac{18\Omega }{20.5912\angle 29.05°\Omega }\right)\left(54.6462\angle -21.11°\text{V}\right)\\ =47.7695\angle -50.16°\text{V}\end{array}$

Use current division rule and find the current through first shunt branch (through $18\Omega$ shunt resistor) as follows:

$I_{18\Omega -\text{shunt}}=\left[\frac{Z_{\text{eq-shunt}}}{\left(18+j10\right)\Omega }\right]I_s$

Substitute $12.4195\angle 57.89°\text{A}$ for $I_s$ and $\left(0.8372-j4.3116\right)\Omega$ for $Z_{\text{eq-shunt}}$ to obtain the value of current through first shunt branch.

$\begin{array}{c}{I}_{18\Omega -\text{shunt}}=\left[\frac{\left(0.8372-j4.3116\right)\Omega }{\left(18+j10\right)\Omega }\right]\left(12.4195\angle 57.89°\text{A}\right)\\ =\left(\frac{4.4\angle -79°\Omega }{20.5912\angle 29.05°\Omega }\right)\left(12.4195\angle 57.89°\text{A}\right)\\ =2.6538\angle -50.16°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power absorbed by the $18\Omega$ shunt resistor as follows:

$S_{18\Omega -\text{shunt}}=V_{18\Omega -\text{shunt}}{\left(I_{18\Omega -\text{shunt}}\right)}^{*}$

Substitute $47.7695\angle -50.16°\text{V}$ for $V_{18\Omega -\text{shunt}}$ and $2.6538\angle -50.16°\text{A}$ for $I_{18\Omega -\text{shunt}}$ to obtain the complex power absorbed by the $18\Omega$ shunt resistor.

$\begin{array}{c}{S}_{18\Omega -\text{shunt}}=\left(47.7695\angle -50.16°\text{V}\right){\left(2.6538\angle -50.16°\text{A}\right)}^{*}\\ =\left(47.7695\angle -50.16°\text{V}\right)\left(2.6538\angle 50.16°\text{A}\right)\\ =126.77\angle 0°\text{VA}\end{array}$

Use voltage division rule and find the voltage across $j10\Omega$ inductive reactance as follows:

$V_{j10\Omega }=\left(\frac{j10\Omega }{18\Omega +j10\Omega }\right)V_1$

Substitute $54.6462\angle -21.11°\text{V}$ for $V_1$ to obtain the voltage across $j10\Omega$ inductive reactance.

$\begin{array}{c}{V}_{j10\Omega }=\left(\frac{j10\Omega }{18\Omega +j10\Omega }\right)\left(54.6462\angle -21.11°\text{V}\right)\\ =\left(\frac{10\angle 90°\Omega }{20.5912\angle 29.05°\Omega }\right)\left(54.6462\angle -21.11°\text{V}\right)\\ =26.5386\angle 39.84°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power absorbed by the $j10\Omega$ inductive reactance as follows:

$S_{j10\Omega }=V_{j10\Omega }{\left(I_{18\Omega -\text{shunt}}\right)}^{*}$

Substitute $26.5386\angle 39.84°\text{V}$ for $V_{j10\Omega }$ and $2.6538\angle -50.16°\text{A}$ for $I_{18\Omega -\text{shunt}}$ to obtain the complex power absorbed by the $j10\Omega$ inductive reactance.

$\begin{array}{c}{S}_{j10\Omega }=\left(26.5386\angle 39.84°\text{V}\right){\left(2.6538\angle -50.16°\text{A}\right)}^{*}\\ =\left(26.5386\angle 39.84°\text{V}\right)\left(2.6538\angle 50.16°\text{A}\right)\\ =70.43\angle 90°\text{VA}\end{array}$

Use current division rule and find the current through second shunt branch (through $-j5\Omega$ capacitive reactance) as follows:

$I_{-j5\Omega }=\left[\frac{Z_{\text{eq-shunt}}}{-j5\Omega }\right]I_s$

Substitute $12.4195\angle 57.89°\text{A}$ for $I_s$ and $\left(0.8372-j4.3116\right)\Omega$ for $Z_{\text{eq-shunt}}$ to obtain the value of current through second shunt branch.

$\begin{array}{c}{I}_{-j5\Omega }=\left[\frac{\left(0.8372-j4.3116\right)\Omega }{-j5\Omega }\right]\left(12.4195\angle 57.89°\text{A}\right)\\ =\left(\frac{4.4\angle -79°\Omega }{5\angle -90°\Omega }\right)\left(12.4195\angle 57.89°\text{A}\right)\\ =10.9291\angle 68.89°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power absorbed by the $-j5\Omega$ capacitive reactance as follows:

$S_{-j5\Omega }=V_1{\left(I_{-j5\Omega }\right)}^{*}$

Substitute $54.6462\angle -21.11°\text{V}$ for $V_1$ and $10.9291\angle 68.89°\text{A}$ for $I_{-j5\Omega }$ to obtain the complex power absorbed by the $-j5\Omega$ capacitive reactance.

$\begin{array}{c}{S}_{-j5\Omega }=\left(54.6462\angle -21.11°\text{V}\right){\left(10.9291\angle 68.89°\text{A}\right)}^{*}\\ =\left(54.6462\angle -21.11°\text{V}\right)\left(10.9291\angle -68.89°\text{A}\right)\\ \cong 597.23\angle -90°\text{VA}\end{array}$

Use current division rule and find the current through third shunt branch (through $1000\Omega$ resistor) as follows:

$I_{1000\Omega }=\left[\frac{Z_{\text{eq-shunt}}}{1000\Omega }\right]I_s$

Substitute $12.4195\angle 57.89°\text{A}$ for $I_s$ and $\left(0.8372-j4.3116\right)\Omega$ for $Z_{\text{eq-shunt}}$ to obtain the value of current through $1000\Omega$ resistor.

$\begin{array}{c}{I}_{1000\Omega }=\left[\frac{\left(0.8372-j4.3116\right)\Omega }{1000\Omega }\right]\left(12.4195\angle 57.89°\text{A}\right)\\ =\left(\frac{4.4\angle -79°\Omega }{1000}\right)\left(12.4195\angle 57.89°\text{A}\right)\\ =0.0546\angle -21.11°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power absorbed by the $1000\Omega$ resistor as follows:

$S_{1000\Omega }=V_1{\left(I_{1000\Omega }\right)}^{*}$

Substitute $54.6462\angle -21.11°\text{V}$ for $V_1$ and $0.0546\angle -21.11°\text{A}$ for $I_{1000\Omega }$ to obtain the complex power absorbed by the $1000\Omega$ resistor.

$\begin{array}{c}{S}_{1000\Omega }=\left(54.6462\angle -21.11°\text{V}\right){\left(0.0546\angle -21.11°\text{A}\right)}^{*}\\ =\left(54.6462\angle -21.11°\text{V}\right)\left(0.0546\angle 21.11°\text{A}\right)\\ \cong 2.98\angle 0°\text{VA}\end{array}$

Conclusion:

Thus, the complex power absorbed by $18\Omega$ series resistor, $18\Omega$ shunt resistor, $j10\Omega$ inductive reactance of the inductor, $-j5\Omega$ capacitive reactance of the capacitor, $1000\Omega$ resistor, and the source power factor are $\underset{\_}{2.78\angle 0°\text{kVA}}$, $\underset{\_}{126.77\angle 0°\text{VA}}$, $\underset{\_}{70.43\angle 90°\text{VA}}$, $\underset{\_}{597.23\angle -90°\text{VA}}$, $\underset{\_}{2.98\angle 0°\text{VA}}$, and 0.975 leading, respectively.