In Fig. 27-50, two batteries with an emf$\mathit{\epsilon }\mathbf{=}\mathbf{12}\mathbf{.0}\mathbf{\text{\hspace{0.17em}V}}$and an internal resistance $\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.200}\mathbf{\text{\hspace{0.17em}Ω}}$are connected in parallel across a resistance R. (a) For what value of Ris the dissipation rate in the resistor a maximum? (b) What is that maximum?

The EMF is,$\epsilon =12.0\text{\hspace{0.17em}V}$

From the given information, it is clear that the batteries are parallel, and hence the potential different across them is the same. The current between them is also the same. So we can find the power dissipated. To find the maximum power, we can find its derivative and equate it to zero.

Formula:

Current in block,

$\mathit{i}\mathbf{=}\frac{\mathbf{V}}{\mathbf{R}}$

Power dissipated,$\mathit{P}\mathbf{=}{\mathit{i}}^{\mathbf{2}}\mathit{R}$

Let $i$be the current in both the batteries; then by junction rule, the current in Ris .$2i$

If we apply the loop rule, we can write

$\begin{array}{c}\epsilon -ir-2iR=0\\ i=\frac{\epsilon }{r+2R}\end{array}$

The formula for power dissipation is

$P=i^{2}R$

So, power dissipated in R is

$P={\left(2i\right)}^{2}R$

Substitute the value of$i$in the above equation.

$P=\frac{4{\epsilon }^{2}R}{{(r+2R)}^2}$

Now, for calculating the maximum power dissipated across R, we need to find the derivate P with respect to R.

$\begin{array}{c}\frac{dP}{dR}=\frac{4{\epsilon }^2}{{(r+2R)}^3}-\frac{16{\epsilon }^{2}R}{{(r+2R)}^3}\\ \frac{dP}{dR}=\frac{4{\epsilon }^2(r-2R)}{{(r+2R)}^3}\end{array}$

For maximum power, the derivative vanishes.

$\begin{array}{c}\frac{dP}{dR}=0\\ \frac{4{\epsilon }^2(r-2R)}{{(r+2R)}^3}=0\\ 4{\epsilon }^2(r-2R)=0\\ (r-2R)=0\\ r=2R\\ R=\frac{r}{2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}R=\frac{0.300\text{\hspace{0.17em}Ω}}{2}\\ =0.150\text{\hspace{0.17em}Ω}\end{array}$

Hence the value of R for which the dissipation rate in the resistor is maximum is, $R=0.150\text{\hspace{0.17em}Ω}$

Maximum power dissipated is

$P_{\max }=\frac{4{\epsilon }^{2}R}{{(r+2R)}^2}$

If we put,$R=\frac{r}{2}$, then the above equation becomes

$P_{\max }=\frac{4{\epsilon }^2\left(\frac{r}{2}\right)}{{(r+2\left(\frac{r}{2}\right))}^2}$

After solving this, we get

$P_{max}=\frac{{\epsilon }^2}{2r}$

$\begin{array}{c}{P}_{max}=\frac{{\left(12.0\text{\hspace{0.17em}Ω}\right)}^2}{2\times 0.300\text{\hspace{0.17em}Ω}}\\ P_{max}=240\text{\hspace{0.17em}W}\end{array}$

Hence, the maximum value of power dissipated is,$240\text{\hspace{0.17em}W}$ .