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Power Delivered and Heat Dissipated in a Circuit

A cell of con...

Question

A cell of constant emf first connected to a resistance $R_{1}$ and then connected to a resistance $R_{2}$ . If power delivered in both cases is same, then the internal resistance of the cell is

\sqrt{R_{1} R_{2}}

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\sqrt{\frac{R_{1}}{R_{2}}}

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\frac{R_{1} - R_{2}}{2}

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\frac{R_{1} + R_{2}}{2}

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Solution

The correct option is A $\sqrt{R_{1} R_{2}}$

Given, same power is delivered for both load resistances $R_{1}$ and $R_{2}$

$∴ i_{1}^{2} R_{1} = i_{2}^{2} R_{2}$

$\Rightarrow {(\frac{E}{R_{1} + r})}^{2} R_{1} = {(\frac{E}{R_{2} + r})}^{2} R_{2}$

$R_{1} {(R_{2} + r)}^{2} = R_{2} {(R_{1} + r)}^{2}$

$R_{1} (R_{2}^{2} + r^{2} + 2 R_{2} r) = R_{2} (R_{1}^{2} + r^{2} + 2 R_{1} r)$

$R_{1} {R_{2}}^{2} + R_{1} r^{2} + 2 R_{1} R_{2} r = R_{2} {R_{1}}^{2} + R_{2} r^{2} + 2 R_{1} R_{2} r$

$R_{1} R_{2}^{2} - R_{2} R_{1}^{2} = r^{2} (R_{2} - R_{1})$

$R_{1} R_{2} (R_{2} - R_{1}) = r^{2} (R_{2} - R_{1})$

$\Rightarrow r^{2} = R_{1} R_{2}$

$∴ r = \sqrt{R_{1} R_{2}}$

Hence, option $(a)$ is correct.

Key concept
power = ${(current)}^{2} \times Resistance$

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