Chapter 4, Problem 74P

For the bridge circuit shown in Fig. 4.140, find the load R_L for maximum power transfer and the maximum power absorbed by the load.

Figure 4.140

To determine

Find the maximum power transferred and maximum power absorbed by the load resistor $R_L$.

Answer to Problem 74P

The maximum power transferred and maximum power absorbed by the load resistor $R_L$ is $\frac{v_s^2{\left[\left(R_2{R}_3\right)-\left(R_1{R}_4\right)\right]}^2}{4\left(R_1+R_2\right)\left(R_3+R_4\right)\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}$.

Explanation of Solution

Given data:

Refer to Figure 4.140 in the textbook.

The voltage source is $v_s$.

Formula used:

Write the expression to find the power delivered to the resistor.

$p_{\max }=\frac{V_{\text{Th}}^2}{4R_{\text{Th}}}$        (1)

Here,

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

$R_{\text{Th}}$ is the Thevenin resistance.

Calculation:

In the given circuit, find the Thevenin resistance by turning off the voltage source $v_s$ (replacing with a short circuit) and remove the load resistor $R_L$.

The modified circuit is shown in Figure 1.

In Figure 1, the Thevenin resistance is,

$\begin{array}{c}{R}_{\text{Th}}=\left(R_1\left|\right|R_2\right)+\left(R_3\left|\right|R_4\right)\\ =\frac{\left(R_1{R}_2\right)}{\left(R_1+R_2\right)}+\frac{\left(R_3{R}_4\right)}{\left(R_3+R_4\right)}\\ =\frac{\left(R_1{R}_2\right)\left(R_3+R_4\right)+\left(R_3{R}_4\right)\left(R_1+R_2\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\\ =\frac{\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\end{array}$

Refer to Figure 4.139 in the textbook.

The given circuit is modified as shown in Figure 2 to find the Thevenin voltage.

In Figure 2, the voltage $v_a$ is calculated by using voltage division principle as follows.

$v_a=\left(v_s\right)\left(\frac{R_2}{R_1+R_2}\right)$

In Figure 2, the voltage $v_b$ is calculated by using voltage division principle as follows.

$v_b=\left(v_s\right)\left(\frac{R_4}{R_3+R_4}\right)$

In Figure 2, apply Kirchhoff’s voltage law for the right bottom loop as follows.

$\begin{array}{l}-v_a+V_{\text{Th}}+v_b=0\\ V_{\text{Th}}=v_a-v_b\end{array}$

Substitute $\left(v_s\right)\left(\frac{R_2}{R_1+R_2}\right)$ for $v_a$ and $\left(v_s\right)\left(\frac{R_4}{R_3+R_4}\right)$ for $v_b$ to find the Thevenin voltage in volts.

$\begin{array}{c}{V}_{\text{Th}}=\left(v_s\right)\left[\left(\frac{R_2}{R_1+R_2}\right)-\left(\frac{R_4}{R_3+R_4}\right)\right]\\ =\left(v_s\right)\left[\frac{R_2\left(R_3+R_4\right)-R_4\left(R_1+R_2\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\right]\\ =\left(v_s\right)\left[\frac{\left(R_2{R}_3\right)+\left(R_2{R}_4\right)-\left(R_1{R}_4\right)-\left(R_2{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\right]\\ =\left(v_s\right)\left[\frac{\left(R_2{R}_3\right)-\left(R_1{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\right]\end{array}$

Substitute $\left(v_s\right)\left[\frac{\left(R_2{R}_3\right)-\left(R_1{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\right]$ for $V_{\text{Th}}$ and $\frac{\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}$ for $R_{\text{Th}}$ in equation (1) to find the power delivered to the variable resistor in watts.

$\begin{array}{c}{p}_{\max }=\frac{{\left(\left(v_s\right)\left[\frac{\left(R_2{R}_3\right)-\left(R_1{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\right]\right)}^2}{4\left(\frac{\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}{\left(R_1+R_2\right)\left(R_3+R_4\right)}\right)}\\ =\left[\frac{v_s^2{\left[\left(R_2{R}_3\right)-\left(R_1{R}_4\right)\right]}^2}{{\left[\left(R_1+R_2\right)\left(R_3+R_4\right)\right]}^2}\right]\left[\frac{\left(R_1+R_2\right)\left(R_3+R_4\right)}{4\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}\right]\\ =\frac{v_s^2{\left[\left(R_2{R}_3\right)-\left(R_1{R}_4\right)\right]}^2}{4\left(R_1+R_2\right)\left(R_3+R_4\right)\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}\end{array}$

Conclusion:

Thus, the maximum power transferred and maximum power absorbed by the load resistor $R_L$ is $\frac{v_s^2{\left[\left(R_2{R}_3\right)-\left(R_1{R}_4\right)\right]}^2}{4\left(R_1+R_2\right)\left(R_3+R_4\right)\left(R_1{R}_2{R}_3+R_1{R}_2{R}_4+R_1{R}_3{R}_4+R_2{R}_3{R}_4\right)}$.