Chapter 11, Problem 15P

In the circuit of Fig. 11.46, find the value of Z_L that will absorb the maximum power and the value of the maximum power.

To determine

Find the value of the load impedance $Z_L$ and the maximum average power absorbed by $Z_L$.

Answer to Problem 15P

The value of load impedance $Z_L$ is $\left(0.5-j0.5\right)\Omega$ and the maximum average power absorbed by $Z_L$ is $90\text{W}$.

Explanation of Solution

Given data:

Refer to Figure 11.46 in the textbook.

The inductance $L$ is $j\Omega$.

The capacitance C is $-j\Omega$

The source voltage is $12\text{V}$.

Formula used:

Write the expression to find the maximum average power.

$P_{\max }=\frac{1}{2}{\left(\frac{V_{\text{Th}}}{Z_{\text{Th}}+Z_L}\right)}^2{R}_L$ (1)

Here,

$R_L$ is the load resistor,

$Z_L$ is the load impedance,

$V_{\text{Th}}$ is the Thevenin voltage, and

$Z_{\text{Th}}$ is the Thevenin impedance.

Write the general expression for load impedance $Z_L$.

$Z_L=R_L+j{X}_L$ (2)

Calculation:

Refer to Figure 11.46 in the textbook.

In the circuit, to calculate the Thevenin impedance $Z_{\text{Th}}$, connect a $1\text{A}$ current source between the load terminals and replacing the $12\text{V}$ voltage source with a short circuit. The modified circuit is shown in Figure 1.

In Figure 1, apply Kirchhoff’s currrent law at node voltage $V_o$.

$\begin{array}{l}\left[\frac{V_o}{1}\right]+\left[\frac{V_o}{j}\right]=\left[\frac{V_2-V_o}{-j}\right]\\ \left[\frac{j{V}_o+V_o}{j}\right]=\left[\frac{V_2-V_o}{-j}\right]\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}-j\left(j{V}_o+V_o\right)=j\left(V_2-V_o\right)\\ V_o-j{V}_o=j{V}_2-j{V}_o\end{array}$

$V_o=j{V}_2$ (3)

In Figure 1, apply Kirchhoff’s currrent law at node voltage $V_2$.

$\begin{array}{l}1+2V_o=\left[\frac{V_2-V_o}{-j}\right]\\ 1+2V_o=j\left(V_2-V_o\right)\\ 1+2V_o=j{V}_2-j{V}_o\end{array}$

Rearrange the equation as follows,

$1=j{V}_2-j{V}_o-2V_o$

$1=j{V}_2-\left(2+j\right)V_o$ (4)

Substitute equation (3) in equation (4) to find $V_2$.

$\begin{array}{l}1=j{V}_2-\left(2+j\right)j{V}_2\\ 1=j{V}_2-\left(2+j\right)j{V}_2\\ 1=j{V}_2-2j{V}_2+V_2\\ 1=\left(1-j\right)V_2\end{array}$

Rearrange the equation as follows,

$V_2=\frac{1}{\left(1-j\right)}$

The Thevenin impedance is,

$Z_{\text{Th}}=\frac{V_2}{1}$

Substitute $\frac{1}{\left(1-j\right)}$ for $V_2$ in the equation to find the Thevenin impedance $Z_{\text{Th}}$ in ohms.

$\begin{array}{c}{Z}_{\text{Th}}=\frac{1}{1-j}\\ =\left(\frac{1}{1-j}\right)\left(\frac{1+j}{1+j}\right)\\ =\frac{1+j}{1+1}\left\{\because j^2=-1\right\}\\ =\frac{1+j}{2}\Omega \end{array}$

Simplify the equation as follows,

$\begin{array}{c}{Z}_{\text{Th}}=\left(\frac{1}{2}+\frac{j}{2}\right)\Omega \\ =\left(0.5+j0.5\right)\Omega \end{array}$

For maximum average power transfer, the load impedance $Z_L$ must be equal to the complex conjugate of the Thevenin impedance $Z_{\text{Th}}$. Therfore,

$\begin{array}{c}{Z}_L=Z_{\text{Th}}^{\text{*}}\\ ={\left(0.5+j0.5\right)}^{*}\Omega \\ =\left(0.5-j0.5\right)\Omega \end{array}$

On comparing the equation with equation (2),

$R_L=0.5\Omega$

The given circuit of Figure 11.46 is modified as shown in Figure 2.

In Figure 2, apply Kirchhoff’s current law at node voltage $V_o$ as follows.

$\begin{array}{l}-2V_o+\frac{V_o-12}{1}+\frac{V_o}{j}=0\\ -2V_o+V_o-12-j{V}_o=0\\ -V_o-12-j{V}_o=0\\ -V_o\left(1+j\right)-12=0\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}-V_o\left(1+j\right)=12\\ V_o=\frac{-12}{1+j}\text{V}\end{array}$

In Figure 2, apply Kirchhoff’current law at node voltage $V_2$ as follows.

$-V_o-\left(-j\times 2V_o\right)+V_2=0$

Rearrange the equation as follows,

$\begin{array}{l}{V}_2=V_o+\left(-j\times 2V_o\right)\\ V_2=V_o-j2V_o\\ V_2=\left(1-j2\right)V_o\end{array}$

Substitute $\frac{-12}{1+j}\text{V}$ for $V_o$ to find the node voltage $V_2$.

$\begin{array}{l}{V}_2=\left(1-j2\right)\left(\frac{-12}{1+j}\right)\text{V}\\ V_2=\left(6+j18\right)\text{V}\end{array}$

The voltage $V_2$ is the Thevenin voltage across the load. Therefore,

$V_{\text{Th}}=V_2=\left(6+j18\right)\text{V}$

Convert the equation from rectangular to polar form.

$V_{\text{Th}}=18.97\angle 71.57°\text{V}$

The Thevenin voltage,

$V_{\text{Th}}=\text{|}{V}_{\text{Th}}\text{|}=18.97\text{V}$

Substitute $\left(6+j18\right)\text{V}$ for $V_{\text{Th}}$, $0.5\Omega$ for $R_L$, $\left(0.5-j0.5\right)\Omega$ for $Z_L$, and $\left(0.5+j0.5\right)\Omega$ for $Z_{\text{Th}}$ in equation (1) to find the maximum average power absorbed by load $Z_L$ in watts.

$\begin{array}{c}{P}_{\max }=\frac{1}{2}{\left(\frac{18.97\text{V}}{\left(0.5+j0.5+0.5-j0.5\right)\Omega }\right)}^{2}0.5\Omega \\ =\frac{1}{2}{\left(\frac{18.97\text{V}}{1\Omega }\right)}^{2}0.5\Omega \\ =90\frac{{\text{V}}^2}{\Omega }\\ =90\text{W}\left\{\because 1\text{W}=\frac{{\text{V}}^2}{\Omega }\right\}\\ \end{array}$

Conclusion:

Thus, The value of load impedance $Z_L$ is $\left(0.5-j0.5\right)\Omega$ and the maximum average power absorbed by $Z_L$ is $90\text{W}$.