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Byju's Answer

Standard VIII

Physics

Heating Effect of Electric Current

A battery is ...

Question

A battery is delivering same power to resistance $R_{1}$ and $R_{2}$ . Then find the internal resistance of battery :

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

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R_{1} + R_{2}

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

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\sqrt{R_{1} + R_{2} / 2}

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Solution

The correct option is A $\frac{R_{1} - R_{2}}{2}$
$\begin{matrix} P = \frac{V^{2}}{R^{2} + r} - - - - - - - (1) P = \frac{V^{2}}{R^{2} - r} - - - - - - - (2) s o l v e e q a t i o n (1) a n d (2) = \frac{R_{1} - R_{2}}{2} \end{matrix}$
Hence,
option $(A)$ is correct answer.

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A battery is delivering same power to resistance R1 and R2 . Then find the internal resistance of battery :

The correct option is A R1−R22P=V2R2+r−−−−−−−(1)P=V2R2−r−−−−−−−(2)solveeqation(1)and(2)=R1−R22Hence,option (A) is correct answer.

A battery is delivering same power to resistance $R_{1}$ and $R_{2}$ . Then find the internal resistance of battery :

The correct option is A $\frac{R_{1} - R_{2}}{2}$
$\begin{matrix} P = \frac{V^{2}}{R^{2} + r} - - - - - - - (1) P = \frac{V^{2}}{R^{2} - r} - - - - - - - (2) s o l v e e q a t i o n (1) a n d (2) = \frac{R_{1} - R_{2}}{2} \end{matrix}$
Hence,
option $(A)$ is correct answer.