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 :
Correct Option: , 4
$P_{R}=0.5 W$
$\Rightarrow \mathrm{i}^{2} \mathrm{R}=0.5 \mathrm{~W}$
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.
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