A battery of 3.0  V is connected to a resistor dissipating 0.5  W of power. If the terminal…

Given,

Emf of the battery, $\mathrm{E}=3 \mathrm{V}$

Potential difference across resister $R$ is ${\mathrm{V}}_{\mathrm{R}}=2.5 \mathrm{V}$

Power dissipation in the resister is $P=0.5 \mathrm{W}$

By using Kirchhoff's voltage law is given by,

$V_{\mathit{r}}+V_{\mathit{R}}=E ...\left(1\right)$

$V_r$ is voltage across internal resister.

Using equation (1) and substitute the given values,

$\Rightarrow V_{\mathit{r}}+2.5=3$

$\Rightarrow V_{\mathit{r}}=0.5$

Now the ratio of potential across $R$ and $r$ is given by,

$\frac{{\mathit{V}}_{\mathit{R}}}{{\mathit{V}}_{\mathit{r}}}=\frac{\mathit{I}\mathit{R}}{\mathit{I}\mathit{r}}=\frac{2.5}{0.5}=5$

$\Rightarrow \frac{\mathit{R}}{\mathit{r}}=5$

Now the power dissipation across the internal resistance is given by

$\frac{{\mathit{P}}_{\mathit{R}}}{{\mathit{P}}_{\mathit{r}}}=\frac{{\mathit{I}}^{\mathit{2}}\mathit{R}}{{\mathit{I}}^{\mathit{2}}\mathit{r}}=\frac{\mathit{R}}{\mathit{r}}$

$\Rightarrow \frac{{\mathit{P}}_{\mathit{R}}}{{\mathit{P}}_{\mathit{r}}}=5$

$\Rightarrow P_{\mathit{r}}=\frac{P_{\mathit{R}}}{5}$

$\Rightarrow P_r=\frac{0.5}{5}=0.1\mathrm{W}$