Physics - Current Electricity Question with Solution | TestHub

Given,

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

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

Power dissipation in the resister is $P=0.5\text{W}$By using Kirchhoff's voltage law is given by,

$V_r+V_R=E...(1)$

$V_r$ is voltage across internal resister. Using equation (1) and substitute the given values,

$\Rightarrow V_r+2.5=3$

$\Rightarrow V_r=0.5$

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

$\frac{V_R}{V_r}=\frac{IR}{Ir}=\frac{2.5}{0.5}=5$

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

Now the power dissipation across the internal resistance is given by

$\frac{P_R}{P_r}=\frac{I^{2}R}{I^{2}r}=\frac{R}{r}$

$\Rightarrow \frac{P_R}{P_r}=5$

$\Rightarrow P_r=\frac{P_R}{5}$

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