If a cell of constant E.M.F. produces the same amount of the heat during the same time in two independent resistors $R_{1}$ and $R_{2}$ , when they are separately connected across the terminals of the cell, one after the another, the internal resistance of the cell is $\sqrt{x R_{1} R_{2}}$ . Find $x$ .

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Solution

Here current through the resistance $R_{1}$ is
$I_{1} = \frac{ε}{r + R_{1}}$
and current through the resistance $R_{2}$ is
$I_{2} = \frac{ε}{r + R_{2}}$ .
Also, the heat produced due to passage of current through both the resistances is same.
Hence, $Q = I_{1}^{2} R_{1} = I_{2}^{2} R_{2}$
Using the values of $I_{1}$ and $I_{2}$ in above equation we get ${(\frac{ε}{r + R_{1}})}^{2} R_{1} = {(\frac{ε}{r + R_{2}})}^{2} R_{2}$
where, $ε$ is the emf and $r$ is the internal resistance of battery.
$\frac{ε^{2} R_{1}}{{(r + R_{1})}_{1}^{2}} = \frac{ε^{2} R_{1}}{{(r + R_{1})}_{1}^{2}}$
$r^{2} (R_{1} - R_{2}) = R_{1} R_{2} ((R_{1} - R_{2})$
$r^{2} = R_{1} R_{2}$
$∴ r = \sqrt{R_{1} R_{2}}$

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