Chapter 11, Problem 55P

Using Fig. 11.74, design a problem to help other students better understand the conservation of AC power.

To determine

Design a problem to make the better understand of conservation of AC power.

Explanation of Solution

Problem design:

In Figure 11.74, consider the value of capacitive reactance $X_C$ is $20\Omega$, the resistance R is $20\Omega$, the inductive reactance $X_L$ is $10\Omega$, the voltage source $V_1$ is $40\angle 0°\text{V}\text{rms}$, and the voltage source $V_{\text{2}}$ is $50\angle 90°\text{V}\text{rms}$. Find the complex power absorbed by each of the five elements shown in the circuit of Figure 11.74.

Formula used:

Write the expression to find the complex power.

$S=V{I}^{*}$ (1)

Here,

$V$ is the voltage, and

$I$ is the current.

Write the expression for the impedance $Z$.

$Z=R+jX$ (2)

Here,

R is the value of resistance, and

X is the value of reactance.

Calculation:

Refer to Figure 11.74 in the textbook.

The given circuit is modified as shown in Figure 1.

In Figure 1, apply Kirchhoff’s voltage law for loop 1 as follows.

$\begin{array}{l}40=\left(20-j20\right)I_1-20I_2\\ 40=20\left[\left(1-j\right)I_1-I_2\right]\end{array}$

$2=\left(1-j\right)I_1-I_2$ (3)

In Figure 1, apply Kirchhoff’s voltage law for loop 2 as follows.

$\begin{array}{l}-j50=\left(20+j10\right)I_2-20I_1\\ -j50=10\left[\left(2+j\right)I_2-2I_1\right]\end{array}$

$-j5=-2I_1+\left(2+j\right)I_2$ (4)

Put equation (3) and (4) in matrix form as follows,

$\left[\begin{array}{l}2\\ -j5\end{array}\right]=\left[\begin{array}{l}1-j-1\\ -22+j\end{array}\right]\left[\begin{array}{l}{I}_1\\ I_2\end{array}\right]$ (5)

Evaluate the determinant $\Delta$ in equation (5) as follows.

$\begin{array}{c}\Delta =\left[\begin{array}{l}1-j-1\\ -22+j\end{array}\right]\\ =\left[\left(1-j\right)\left(2+j\right)-\left(-1\right)\left(-2\right)\right]\\ =\left[2+j-j2+1-2\right]\\ =1-j\end{array}$

Evaluate the determinant ${\Delta }_1$ in equation (5) as follows.

$\begin{array}{c}{\Delta }_1=\left[\begin{array}{l}2-1\\ -j52+j\end{array}\right]\\ =\left[4+j2-j5\right]\\ =4-j3\end{array}$

Evaluate the determinant ${\Delta }_2$ in equation (5) as follows.

$\begin{array}{c}{\Delta }_2=\left[\begin{array}{l}-12\\ 2+j-j5\end{array}\right]\\ =\left[-j5-5+4\right]\\ =-1-j5\end{array}$

In Figure 1, the current $I_1$ is,

$I_1=\frac{{\Delta }_1}{\Delta }$

Substitute $1-j$ for $\Delta$ and $4-j3$ for ${\Delta }_1$ in the equation to find the current $I_1$ in amperes.

$\begin{array}{c}{I}_1=\frac{4-j3}{1-j}\\ =3.5+j0.5\\ =\text{3}\text{.535}\angle \text{8}\text{.13}°\text{A}\end{array}$

The magnitude of current $I_1$ is,

$I_1=\left|I_1\right|=3.535\text{A}$

In Figure 2, the current $I_2$ is,

$I_2=\frac{{\Delta }_2}{\Delta }$

Substitute $1-j$ for $\Delta$ and $-1-j5$ for ${\Delta }_2$ in the equation to find the current $I_2$ in amperes.

$\begin{array}{c}{I}_2=\frac{-1-j5}{1-j}\\ =2-j3\\ =3.605\angle -56.31°\text{A}\end{array}$

The magnitude of current $I_2$ is,

$I_2=\left|I_2\right|=3.605\text{A}$

The current $I_3$ through the 20 ohms resistor is,

$I_3=I_1-I_2$

Substitute $3.5+j0.5$ for $I_1$ and $2-j3$ for $I_2$ in the equation to find the current $I_3$ in amperes.

$\begin{array}{c}{I}_3=\left(3.5+j0.5\right)-\left(2-j3\right)\\ =1.5+j3.5\\ =\text{3}\text{.808}\angle \text{66}\text{.8}°\text{A}\end{array}$

The magnitude of current $I_3$ is,

$I_3=\left|I_3\right|=3.808\text{A}$

The complex power absorbed by the $40\text{V}$ source is,

$S=-V_1{I}_1^{*}$

Substitute $40\text{V}$ for $V_1$ and $\left(3.5+j0.5\right)\text{A}$ for $I_1$ in the equation to find the complex power absorbed by the $40\text{V}$ source in VA.

$\begin{array}{c}S=-\left(40\right){\left(3.5+j0.5\right)}^{*}\text{VA}\\ \text{=}-\left(40\right)\left(3.5-j0.5\right)\text{VA}\\ =\left(-140+j20\right)\text{VA}\end{array}$

From Figure (1), the impedance of capacitance is,

$Z_C=-j20\Omega$

The complex power absorbed by the capacitor is,

$S=|I_1{|}^2{Z}_C$

Substitute $3.535\text{A}$ for $I_1$ and $-j20\Omega$ for $Z_C$ in the equation to find the complex power absorbed by the capacitor in VA.

$\begin{array}{c}S={\left(3.535\text{A}\right)}^2\left(-j20\Omega \right)\\ =\left(12.496\right)\left(-j20\right){\text{A}}^2\cdot \Omega \\ =-j250\text{VA}\left\{\because 1\text{V}=\text{A}\cdot \Omega \right\}\end{array}$

The complex power absorbed by the resistor is,

$S=|I_3{|}^{2}R$

Substitute $3.808\text{A}$ for $I_3$ and $20\Omega$ for $R$ in the equation to find the complex power absorbed by the resistor in VA.

$\begin{array}{c}S={\left(3.808\text{A}\right)}^2\left(20\Omega \right)\\ =\left(14.5\right)\left(20\right){\text{A}}^2\cdot \Omega \\ =290\text{VA}\left\{\because 1\text{V}=\text{A}\cdot \Omega \right\}\end{array}$

From Figure (1), the impedance of inductance is,

$Z_L=j10\Omega$

The complex power absorbed by the inductor is,

$S=|I_2{|}^2{Z}_L$

Substitute $3.605\text{A}$ for $I_2$ and $j10\Omega$ for $Z_L$ in the equation to find the complex power absorbed by the inductor in VA.

$\begin{array}{c}S={\left(3.605\text{A}\right)}^2\left(j10\Omega \right)\\ =\left(13\right)\left(j10\right){\text{A}}^2\cdot \Omega \\ =j130\text{VA}\left\{\because 1\text{V}=\text{A}\cdot \Omega \right\}\end{array}$

The complex power absorbed by the $j50\text{V}$ source is,

$S=V_2{I}_2^{*}$

Substitute $j50\text{V}$ for $V_2$ and $\left(2-j3\right)\text{A}$ for $I_2$ in the equation to find the complex power absorbed by the $j50\text{V}$ source in VA.

$\begin{array}{c}S=\left(j50\right){\left(2-j3\right)}^{*}\text{VA}\\ =\left(j50\right)\left(2+j3\right)\text{VA}\\ =\left(-150+j100\right)\text{VA}\end{array}$

Therefore, the complex power absorbed by the $40\text{V}$ source is $\left(-140+j20\right)\text{VA}$, the capacitor is $-j250\text{VA}$, the resistor is $290\text{VA}$, the inductor is $j130\text{VA}$, and the $j50\text{V}$ source is $\left(-150+j100\right)\text{VA}$.

Conclusion:

Thus, a problem is designed and solved for the better understand of conservation of AC power.