In Fig. 27-48,${\mathit{R}}_{\mathbf{1}}\mathbf{=}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathit{\Omega }$, and the ideal battery has emf.$\mathit{\epsilon }\mathbf{=}\mathbf{12}\mathbf{}\mathbf{\text{V}}$

(a) What value of${\mathbf{\text{R}}}_{\mathbf{3}}$maximizes the rate at which the battery supplies energy and (b) what is that maximum rate?

Consider the voltage .$V=12V$

Consider the resistances .$R_1=R_2=10Ohm$

Use theconcept of maximum rate in which.$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{R}}\mathbf{=}\mathbf{0}$ For this, find the unknown resistance. Then use the formula for maximum rate in terms of voltage and resistance.

Consider the formula for the equivalent resistance:

$$\begin{gathered} R_{eq}=R_A+R_B \\ \frac{1}{R_{eq}}=\frac{1}{R_A}+\frac{1}{R_B} \\ V=IR \end{gathered}$$

To find equivalent resistance$\left(R_{eq}\right):$

We have, parallel combination of$R_2\&R_3$is in series with$R_1$

Therefore, equivalent resistance of circuit is given by

$$\begin{gathered} R_{eq}=\frac{R_2{R}_3}{R_2+R_3}+R_1 \\ {R}_{eq}=\frac{R_1{R}_2+R_1{R}_3+R_2{R}_3}{R_2+R_3} \end{gathered}$$

By ohm’s law,

Current $\left(i\right)$

$\begin{array}{c}\left(i\right)=\frac{V}{R_{eq}}\\ =\frac{V(R_2+R_3)}{R_1{R}_2+R_1{R}_3+R_2{R}_3}\end{array}$

The rate at which battery supplies energy is as follows:

$\begin{array}{c}P=Vi\\ =\frac{V^2(R_3+R_2)}{R_1{R}_2+R_1{R}_3+R_2{R}_3}\end{array}$

To find the value of$R_3$which gives maximum rate of energy:

Differentiate P with respect to$R_3$

$\frac{dP}{d{R}_3}=-\frac{R_2{V}^2}{(R_1{R}_2+R_1{R}_3+R_2{R}_3)2}$

For maximum rate$\frac{dP}{d{R}_3}=0,$

So

$0=\frac{R_2^2{V}^2}{{(R_1{R}_2+R_1{R}_3+R_2{R}_3)}^2}$

So

$$\begin{gathered} R_3=0ohm \\ {\text{R}}_3=0\left(\text{atmaximumrateofenergy}\right) \end{gathered}$$

Step 3:(b) Determine the maximum rate

Now maximum rate is as follows:

As $R_3=0ohm$so equivalent resistance is as follows:

$R_{eq}=\frac{R_1{R}_2+R_2{R}_3+R_1{R}_3}{R_3+R_2}$

So

$\begin{array}{c}{R}_{eq}=\frac{10\times 10}{10}\\ =10ohm\end{array}$

So maximum rate is obtained as:

$\begin{array}{c}{P}_{\max }=\frac{V^2}{R}\\ =\frac{{12}^2}{10}\\ =14.4W\end{array}$

$\text{Maximumrateofenergy}\left(\text{P}\right)=14.4\text{W}$