# Solve this following

Question:

A battery of $3.0 \mathrm{~V}$ is connected to a resistor dissipating $0.5 \mathrm{~W}$ of power. If the terminal voltage of the battery is $2.5 \mathrm{~V}$, the power dissipated within the internal resistance is :

1. $0.50 \mathrm{~W}$

2. $0.125 \mathrm{~W}$

3. $0.072 \mathrm{~W}$

4. $0.10 \mathrm{~W}$

Correct Option: , 4

Solution:

$P_{R}=0.5 W$

$\Rightarrow \mathrm{i}^{2} \mathrm{R}=0.5 \mathrm{~W}$

Also, $\mathrm{V}=\mathrm{E}-\mathrm{ir}$

$2.5=3-\mathrm{ir}$

$\Rightarrow$ ir $=0.5$

Power dissipated across $^{\prime} r^{\prime}: P_{r}=i^{2} r$

Now $\mathrm{iR}=2.5$

ir $=0.5$

On dividing $: \frac{\mathrm{R}}{\mathrm{r}}=5$

Now $\frac{\mathrm{P}_{\mathrm{R}}}{\mathrm{P}_{\mathrm{r}}}=\frac{\mathrm{i}^{2} \mathrm{R}}{\mathrm{i}^{2} \mathrm{r}} \Rightarrow \frac{\mathrm{P}_{\mathrm{R}}}{\mathrm{P}_{\mathrm{r}}}=\frac{\mathrm{R}}{\mathrm{r}} \Rightarrow \frac{\mathrm{P}_{\mathrm{R}}}{\mathrm{P}_{\mathrm{r}}}=5$

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

$\Rightarrow \mathrm{P}_{\mathrm{r}}=\frac{0.50}{5} \Rightarrow \mathrm{P}_{\mathrm{r}}=0.10 \mathrm{~W}$

option (4) is correct.