Chapter 11, Problem 16P

For the circuit in Fig. 11.47, find the value of Z_L that will receive the maximum power from the circuit. Then calculate the power delivered to the load Z_L.

To determine

Find the value of the load impedance $Z_L$ and the power delivered to the load $Z_L$ in the given Figure 11.47.

Answer to Problem 16P

The value of load impedance $Z_L$ is $4.0125\angle -65.4°\Omega$ and the maximum average power delivered to the load $Z_L$ is $264.28\text{W}$.

Explanation of Solution

Given data:

Refer to Figure 11.47 in the textbook.

The inductance $L$ is $1\text{H}$.

The capacitance C is $0.05\text{F}$

The source voltage is

$v(t)=50\sin 4t\text{V}$ (1)

Formula used:

Write the general expression for the instantaneous voltage.

$v(t)=V_m\sin \left(\omega t\right)$ (2)

Write the expression to find the maximum average power.

$P_{\max }=\frac{1}{2}{\left(\frac{V_{\text{Th}}}{Z_{\text{Th}}+Z_L}\right)}^2{R}_L$ (3)

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 expression for $Z_{\text{Th}}$,

$Z_L=R_L+j{X}_L$ (4)

Calculation:

On comparing equation (1) and (2), the angular frequency is,

$\omega =4$

Write the expression for the reactance of the inductance.

$X_L=j\omega L$

Substitute $1\text{H}$ for L and $4$ for $\omega$ to find the reactance of the inductance in ohms.

$\begin{array}{c}{X}_L=j\times 4\times 1\\ =j4\Omega \end{array}$

Write the expression for the reactance of the capacitance.

$X_C=\frac{1}{j\omega C}$

Substitute $0.05\text{F}$ for C and $4$ for $\omega$ to find the reactance of the capacitance in ohms.

$\begin{array}{c}{X}_C=\frac{1}{j\times 4\times 0.05}\\ =-j5\Omega \end{array}$

Refer to Figure 11.47 in the textbook.

To find the Thevenin equivalent the given Figure is modified as shown in Figure 1.

In Figure 1, apply Kirchhoff’s currrent law at node voltage $v_1$.

$\begin{array}{c}\left[\frac{50-v_1}{2}\right]=\left[\frac{v_1}{-j5}\right]+\left[0.5v_1\right]+\left[\frac{v_1-v_2}{4}\right]\\ 25-0.5v_1=j0.2v_1+0.5v_1+0.25v_1-0.25v_2\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}25=j0.2v_1+0.5v_1+0.25v_1-0.25v_2+0.5v_1\\ 25=1.25v_1+j0.2v_1-0.25v_2\end{array}$

$25=v_1\left(1.25+j0.2\right)-0.25v_2$ (5)

In Figure 1, apply Kirchhoff’s currrent law at node voltage $v_2$.

$\begin{array}{l}\left[\frac{v_1-v_2}{4}\right]+0.5v_1=\frac{v_2}{j4}\\ 0.25v_1-0.25v_2+0.5v_1=-j0.25v_2\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}0.25v_1-0.25v_2+0.5v_1+j0.25v_2=0\\ 0.75v_1+v_2\left(-0.25+j0.25\right)=0\\ v_1=\frac{v_2\left(0.25-j0.25\right)}{0.75}\end{array}$

$v_1=\left(0.33333-j0.33333\right)v_2$ (6)

Substitute equation (6) in equation (5).

$\begin{array}{l}25=\left(1.25+j0.2\right)\left(0.33333-j0.33333\right)v_2-0.25v_2\\ 25=\left(0.4833-j0.3499\right)v_2-0.25v_2\\ 25=0.4833v_2-j0.3499v_2-0.25v_2\\ 25=\left(0.2333-j0.3499\right)v_2\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}{v}_2=\frac{25}{\left(0.2333-j0.3499\right)}\\ v_2=\left(32.965+j49.454\right)\text{V}\end{array}$

The voltage voltage $v_2$ is the voltage across the load terminal. Therefore,

$v_2=V_{\text{Th}}=\left(32.965+j49.454\right)\text{V}$

Convert the equation from rectangular to polar form.

$V_{\text{Th}}=59.433\angle 56.31°\text{V}$

The Thevenin voltage is,

$V_{\text{Th}}=\text{|}{V}_{\text{Th}}\text{|}=59.433\text{V}$

In Figure 1, to calculate the Thevenin impedance $Z_{\text{Th}}$, connect a $1\text{A}$ current source between the load terminals and replacing the $50\text{V}$ voltage source with a short circuit. The modified circuit is shown in Figure 2.

In Figure 2, apply Kirchhoff’current law at node voltage $v_1$ as follows.

$\begin{array}{l}\left[\frac{v_1}{2}\right]+\left[\frac{v_1}{-j5}\right]+\left[0.5v_1\right]+\left[\frac{v_1-v_2}{4}\right]=0\\ 0.5v_1+j0.2v_1+0.5v_1+0.25v_1-0.25v_2=0\\ 1.25v_1+j0.2v_1-0.25v_2=0\\ \left(1.25+j0.2\right)v_1-0.25v_2=0\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}{v}_1=\frac{0.25v_2}{\left(1.25+j0.2\right)}\\ v_1=\frac{0.25v_2}{1.2659\angle 9.09°}\end{array}$

$v_1=\left(0.197488\angle -9.09°\right)v_2$ (7)

In Figure 2, apply Kirchhoff’current law at node voltage $v_2$ as follows.

$\begin{array}{l}1+\left[\frac{v_1-v_2}{4}\right]+\left[0.5v_1\right]=\left[\frac{v_2}{j4}\right]\\ 1+0.25v_1-0.25v_2+0.5v_1=-j0.25v_2\end{array}$

Rearrange the equation as follows,

$0.75v_1+\left(-0.25+j0.25\right)v_2=-1$ (8)

Substitute equation (7) in equation (8).

$\begin{array}{l}0.75\left(0.197488\angle -9.09°\right)v_2+\left(-0.25+j0.25\right)v_2=-1\\ 0.146256v_2-j0.0234v_2-0.25v_2+j0.25v_2=-1\\ \left(-0.103744+j0.2266\right)v_2=-1\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}{v}_2=\frac{-1}{\left(-0.103744+j0.2266\right)}\\ v_2=\left(1.67033+j3.6483\right)\text{V}\end{array}$

Convert the equation from rectangular to polar form.

$v_2=4.0125\angle 65.4°\text{V}$

The Thevenin impedance $Z_{\text{Th}}$ is,

$Z_{\text{Th}}=\frac{v_2}{1}$

Substitute $4.0125\angle 65.4°\text{V}$ for $v_2$ in the equation to find the Thevenin impedance $Z_{\text{Th}}$ in ohms.

$\begin{array}{c}{Z}_{\text{Th}}=\frac{4.0125\angle 65.4°}{1}\\ =4.0125\angle 65.4°\Omega \end{array}$

Convert the equation from polar to rectangular form.

$Z_{\text{Th}}=\left(1.67033+j3.6483\right)\Omega$

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,

$\begin{array}{c}{Z}_L=Z_{\text{Th}}^{\text{*}}\\ =\left(1.67033+j3.6483\right){\Omega }^{*}\end{array}$

$Z_L=\left(1.67033-j3.6483\right)\Omega$ (9)

Convert the equation from rectangular to polar form.

$Z_L=4.0125\angle -65.4°\Omega$

On comparing the equation (9) with equation (4).

$R_L=1.67033\Omega$

Substitute $59.433\text{V}$ for $V_{\text{Th}}$, $1.67033\Omega$ for $R_L$, $\left(1.67033-j3.648\right)\Omega$ for $Z_L$, and $\left(1.67033+j3.648\right)\Omega$ for $Z_{\text{Th}}$ in equation (3) to find the maximum average power absorbed by load $Z_L$ in watts.

$\begin{array}{c}{P}_{\max }=\frac{1}{2}{\left(\frac{59.433\text{V}}{\left(1.67033-j3.648\right)\Omega +\left(1.67033+j3.648\right)\Omega }\right)}^{2}1.67\Omega \\ =\frac{1}{2}\left(\frac{3532.281{\text{V}}^2}{11.16\Omega }\right)1.67\Omega \\ =264.28\frac{{\text{V}}^2}{\Omega }\\ =264.28\text{W}\left\{\because 1\text{W}=\frac{{\text{V}}^2}{\Omega }\right\}\end{array}$

Conclusion:

Thus, the value of load impedance $Z_L$ is $4.0125\angle -65.4°\Omega$ and the maximum average power absorbed by $Z_L$ is $264.28\text{W}$.