Chapter 11, Problem 13E

(a)

To determine

Calculate the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $9\Omega$.

Answer to Problem 13E

The average power delivered by the current source of $4-j2\text{A}$ to the impedance of $9\Omega$ is $\underset{\_}{90\text{W}}$.

Explanation of Solution

Given data:

The value of current source $\left(I\right)$ is $4-j2\text{A}$.

The value of impedance $\left(Z\right)$ is $9\Omega$.

Formula used:

Write the expression for average power delivered to the resistor.

$P_R=\frac{1}{2}{I}_m^{2}R$        (1)

Here,

$I_m$ is the maximum value of the current, and

$R$ is the resistance of the resistor.

Calculation:

Convert the rectangular form of the current value into polar form as follows:

$\begin{array}{c}I=4-j2\text{A}\\ \text{=}\sqrt{20}\angle -26.6°\text{A}\end{array}$

Therefore, the maximum value of the current $\left(I_m\right)$ is $\sqrt{20}\text{A}$ and phase angle of current $\left(\varphi \right)$ is $-26.6°$.

The given value of impedance has purely resistive. Therefore, the value of R is equal to Z.

$R=9\Omega$

Modify equation (1) for impedance of $9\Omega$ as follows:

$P_{9\Omega }=\frac{1}{2}{I}_m^{2}R$

Substitute $\sqrt{20}\text{A}$ for $I_m$ and $9\Omega$ for $R$ to find $P_R$.

$\begin{array}{c}{P}_{9\Omega }=\frac{1}{2}{\left(\sqrt{20}\text{A}\right)}^2\left(9\Omega \right)\\ =90\text{W}\end{array}$

Conclusion:

Thus, the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $9\Omega$ is $\underset{\_}{90\text{W}}$.

(b)

To determine

Calculate the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $-j1000\Omega$.

Answer to Problem 13E

The average power delivered by the current source of $4-j2\text{A}$ to the impedance of $-j1000\Omega$ is $\underset{\_}{0\text{W}}$.

Explanation of Solution

Given data:

The value of current source $\left(I\right)$ is $4-j2\text{A}$.

The value of impedance $\left(Z\right)$ is $-j1000\Omega$.

Formula used:

Write the expression for average power delivered.

$P=\frac{1}{2}{V}_m{I}_m\cos \left(\theta -\varphi \right)$        (2)

Here,

$V_m$ is the maximum value of the current,

$I_m$ is the maximum value of the current,

$\theta$ is the phase angle of voltage,

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

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

Write the expression for voltage.

$V=ZI$        (3)

Here,

$Z$ is impedance of the circuit, and

$I$ is applied current.

Calculation:

Convert the value of $-j1000\Omega$ in to polar form.

$\begin{array}{c}Z=-j1000\Omega \\ =1000\angle 270°\Omega \end{array}$

Modify equation (3) for impedance of $-j1000\Omega$ as follows:

$V_{-j1000\Omega }=ZI$

Substitute $1000\angle 270°\Omega$ for $Z$ and $\sqrt{20}\angle -26.6°\text{A}$ for $I$ to find the value of $V_{-j1000\Omega }$.

$\begin{array}{c}{V}_{-j1000\Omega }=\left(1000\angle 270°\Omega \right)\left(\sqrt{20}\angle -26.6°\text{A}\right)\\ =1000\sqrt{20}\angle \left(270°-26.6°\right)\text{V}\\ \text{=}1000\sqrt{20}\angle 243.4°\text{V}\end{array}$

Therefore, the maximum value of the voltage $\left(V_{m\left(-j1000\right)}\right)$ is $1000\sqrt{20}\text{V}$ and phase angle of voltage $\left({\theta }_{-j1000}\right)$ is $243.4°$.

Modify equation (2) for impedance of $-j1000\Omega$ as follows:

$P_{-j1000}=\frac{1}{2}{V}_{m\left(-j1000\right)}{I}_m\cos \left({\theta }_{-j1000}-\varphi \right)$

Substitute $1000\sqrt{20}\text{V}$ for $V_{m\left(-j1000\right)}$, $\sqrt{20}\text{A}$ for $I_m$, $243.4°$ for ${\theta }_{-j1000}$ and $-26.6°$ for $\varphi$ to find $P_{-j1000}$.

$\begin{array}{c}{P}_{-j1000}=\frac{1}{2}\left(1000\sqrt{20}\text{V}\right)\left(\sqrt{20}\text{A}\right)\cos \left(243.4°-\left(-26.6°\right)\right)\\ =10000\cos \left(270°\right)\text{W}\\ =10000\left(0\right)\text{W}\\ =0\text{W}\end{array}$

Conclusion:

Thus, the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $-j1000\Omega$ is $\underset{\_}{0\text{W}}$.

(c)

To determine

Calculate the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $1-j2+j3\Omega$.

Answer to Problem 13E

The average power delivered by the current source of $4-j2\text{A}$ to the impedance of $1-j2+j3\Omega$ is $\underset{\_}{10\text{W}}$.

Explanation of Solution

Given data:

The value of current source $\left(I\right)$ is $4-j2\text{A}$.

The value of impedance $\left(Z\right)$ is $1-j2+j3\Omega$.

Calculation:

Convert the value of $1-j2+j3\Omega$ in to polar form.

$\begin{array}{c}Z=1-j2+j3\Omega \\ =1+j1\Omega \\ =\sqrt{2}\angle 45°\Omega \end{array}$

Modify equation (3) for impedance of $1-j2+j3\Omega$ as follows:

$V_{1-j2+j3\Omega }=ZI$

Substitute $\sqrt{2}\angle 45°\Omega$ for $Z$ and $\sqrt{20}\angle -26.6°\text{A}$ for $I$ to find the value of $V_{1-j2+j3\Omega }$.

$\begin{array}{c}{V}_{1-j2+j3\Omega }=\left(\sqrt{2}\angle 45°\Omega \right)\left(\sqrt{20}\angle -26.6°\text{A}\right)\\ =\sqrt{2}\sqrt{20}\angle \left(45°-26.6°\right)\text{V}\\ \text{=}\sqrt{2}\sqrt{20}\angle 18.4°\text{V}\end{array}$

Therefore, the maximum value of the voltage $\left(V_{m\left(1-j2+j3\Omega \right)}\right)$ is $\sqrt{2}\sqrt{20}\text{V}$ and phase angle of voltage $\left({\theta }_{1-j2+j3\Omega }\right)$ is $18.4°$.

Modify equation (2) for impedance of $1-j2+j3\Omega$ as follows:

$P_{1-j2+j3\Omega }=\frac{1}{2}{V}_{m\left(1-j2+j3\Omega \right)}{I}_m\cos \left({\theta }_{1-j2+j3\Omega }-\varphi \right)$

Substitute $\sqrt{2}\sqrt{20}\text{V}$ for $V_{m\left(1-j2+j3\Omega \right)}$, $\sqrt{20}\text{A}$ for $I_m$, $18.4°$ for ${\theta }_{1-j2+j3\Omega }$ and $-26.6°$ for $\varphi$ to find $P_{1-j2+j3\Omega }$.

$\begin{array}{c}{P}_{1-j2+j3\Omega }=\frac{1}{2}\left(\sqrt{2}\sqrt{20}\text{V}\right)\left(\sqrt{20}\text{A}\right)\cos \left(18.4°-\left(-26.6°\right)\right)\\ =14.142\cos \left(45°\right)\text{W}\\ =9.99\text{W}\\ \cong 10\text{W}\end{array}$

Conclusion:

Thus, the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $1-j2+j3\Omega$ is $\underset{\_}{10\text{W}}$.

(d)

To determine

Calculate the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $6\angle 32°\Omega$.

Answer to Problem 13E

The average power delivered by the current source of $4-j2\text{A}$ to the impedance of $6\angle 32°\Omega$ is $\underset{\_}{50.88\text{W}}$.

Explanation of Solution

Given data:

The value of current source $\left(I\right)$ is $4-j2\text{A}$.

The value of impedance $\left(Z\right)$ is $6\angle 32°\Omega$.

Calculation:

Modify equation (3) for impedance of $6\angle 32°\Omega$ as follows:

$V_{6\angle 32°\Omega }=ZI$

Substitute $6\angle 32°\Omega$ for $Z$ and $\sqrt{20}\angle -26.6°\text{A}$ for $I$ to find the value of $V_{6\angle 32°\Omega }$.

$\begin{array}{c}{V}_{6\angle 32°\Omega }=\left(6\angle 32°\Omega \right)\left(\sqrt{20}\angle -26.6°\text{A}\right)\\ =6\sqrt{20}\angle \left(32°-26.6°\right)\text{V}\\ \text{=6}\sqrt{20}\angle 5.4°\text{V}\end{array}$

Therefore, the maximum value of the voltage $\left(V_{m\left(6\angle 32°\Omega \right)}\right)$ is $6\sqrt{20}\text{V}$ and phase angle of voltage $\left({\theta }_{6\angle 32°\Omega }\right)$ is $5.4°$.

Modify equation (2) for impedance of $6\angle 32°\Omega$ as follows:

$P_{6\angle 32°\Omega }=\frac{1}{2}{V}_{m\left(6\angle 32°\Omega \right)}{I}_m\cos \left({\theta }_{6\angle 32°\Omega }-\varphi \right)$

Substitute $6\sqrt{20}\text{V}$ for $V_{m\left(6\angle 32°\Omega \right)}$, $\sqrt{20}\text{A}$ for $I_m$, $5.4°$ for ${\theta }_{6\angle 32°\Omega }$ and $-26.6°$ for $\varphi$ to find $P_{6\angle 32°\Omega }$.

$\begin{array}{c}{P}_{6\angle 32°\Omega }=\frac{1}{2}\left(6\sqrt{20}\text{V}\right)\left(\sqrt{20}\text{A}\right)\cos \left(5.4°-\left(-26.6°\right)\right)\\ =60\cos \left(32°\right)\text{W}\\ =50.88\text{W}\end{array}$

Conclusion:

Thus, the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $6\angle 32°\Omega$ is $\underset{\_}{50.88\text{W}}$.

(e)

To determine

Calculate the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$.

Answer to Problem 13E

The average power delivered by the current source of $4-j2\text{A}$ to the impedance of $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$ is $\underset{\_}{4.7\text{kW}}$.

Explanation of Solution

Given data:

The value of current source $\left(I\right)$ is $4-j2\text{A}$.

The value of impedance $\left(Z\right)$ is $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$.

Calculation:

Convert the value of $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$ in to polar form.

$\begin{array}{c}Z=\frac{1.5\angle -19°}{2+j}\text{k}\Omega \\ =\frac{1.5\angle -19°}{2.24\angle 26.57°}\text{k}\Omega \\ =0.6696\angle 45.57°\text{k}\Omega \end{array}$

Modify equation (3) for impedance of $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$ as follows:

$V_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }=ZI$

Substitute $0.6696\angle 45.57°\text{k}\Omega$ for $Z$ and $\sqrt{20}\angle -26.6°\text{A}$ for $I$ to find the value of $V_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }$.

$\begin{array}{c}{V}_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }=\left(0.6696\angle 45.57°\text{k}\Omega \right)\left(\sqrt{20}\angle -26.6°\text{A}\right)\\ =0.6696\sqrt{20}\angle \left(45.57°-26.6°\right)\text{kV}\\ \text{=}0.6696\sqrt{20}\angle 18.97°\text{kV}\end{array}$

Therefore, the maximum value of the voltage $\left(V_{m\left(\frac{1.5\angle -19°}{2+j}\text{k}\Omega \right)}\right)$ is $0.6696\sqrt{20}\text{kV}$ and phase angle of voltage $\left({\theta }_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }\right)$ is $18.97°$.

Modify equation (2) for impedance of $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$ as follows:

$P_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }=\frac{1}{2}{V}_{m\left(\frac{1.5\angle -19°}{2+j}\text{k}\Omega \right)}{I}_m\cos \left({\theta }_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }-\varphi \right)$

Substitute $0.6696\sqrt{20}\text{kV}$ for $V_{m\left(\frac{1.5\angle -19°}{2+j}\text{k}\Omega \right)}$, $\sqrt{20}\text{A}$ for $I_m$, $18.97°$ for ${\theta }_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }$ and $-26.6°$ for $\varphi$ to find $P_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }$.

$\begin{array}{c}{P}_{\frac{1.5\angle -19°}{2+j}\text{k}\Omega }=\frac{1}{2}\left(0.6696\sqrt{20}\text{kV}\right)\left(\sqrt{20}\text{A}\right)\cos \left(18.97°-\left(-26.6°\right)\right)\\ =6.696\cos \left(45.57°\right)\text{kW}\\ =4.68\text{kW}\\ \cong 4.7\text{kW}\end{array}$

Conclusion:

Thus, the average power delivered by the current source of $4-j2\text{A}$ to the impedance of $\frac{1.5\angle -19°}{2+j}\text{k}\Omega$ is $\underset{\_}{4.7\text{kW}}$.