Two identical cells each of emf $E$ and internal resistance $r$ are connected in parallel with an external resistance $R$ . To get maximum power developed across $R$ , the value of $R$ is :

R = \frac{r}{2}

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R = r

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R = \frac{r}{3}

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R = 2 r

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Solution

The correct option is A $R = \frac{r}{2}$
Equivalent resistance $R_{e q} = \frac{r}{2} + R = \frac{r + 2 R}{2}$
$∴ I = \frac{2 E}{r + 2 R}$
For maximum power consumption $I$ should be maximum so denominator is minimum. For this
$r + 2 R = {(\sqrt{r} - \sqrt{2 R})}^{2} + 2 \sqrt{r} \sqrt{2 R}$
$\Rightarrow \sqrt{r} - \sqrt{2 R} = 0$
$\Rightarrow R = \frac{r}{2}$

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