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Heating Effect of Electric Current

A cell of e.m...

Question

A cell of e.m.f. E is connected to a resistance $R_{1}$ for time t and the amount of heat generated in it is H. If the resistance $R_{1}$ is replaced by another resistance $R_{2}$ and is connected to the cell for the same time t, the amount of heat generated in $R_{2}$ is 4H. Then the internal resistance of the cell is

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is D $\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}} - \sqrt{R_{1}}}{\sqrt{R_{2}} - 2 \sqrt{R_{1}}}$
When a resistance $R_{1}$ is connected with the cell of internal resistance $r$ and emf $E$ .
Current through the circuit $i_{1} = \frac{E}{r + R_{1}}$
Heat produced in $R_{1}$ $H_{p} = i_{1}^{2} R_{1} t$
$∴$ $H = \frac{E^{2}}{(r + R_{1})^{2}} R_{1} t$ ........(1)
When a resistance $R_{2}$ is connected with the cell of internal resistance $r$ and emf $E$ .
Current through the circuit $i_{2} = \frac{E}{r + R_{2}}$
Heat produced in $R_{2}$ $H_{p} = i_{2}^{2} R_{2} t$
$∴$ $4 H = \frac{E^{2}}{(r + R_{2})^{2}} R_{2} t$ ........(2)
Dividing $(2)$ by (1) we get $4 = \frac{R_{2} (r + R_{1})^{2}}{R_{1} (r + R_{2})^{2}}$
OR $4 \frac{R_{1}}{R_{2}} = \frac{(r + R_{1})^{2}}{(r + R_{2})^{2}}$
OR $2 \frac{\sqrt{R_{1}}}{\sqrt{R_{2}}} = \frac{(r + R_{1})}{(r + R_{2})}$
OR $2 \sqrt{R_{1}} R_{2} + 2 \sqrt{R_{1}} r = R_{1} \sqrt{R_{2}} + r \sqrt{R_{2}}$
OR $(\sqrt{R_{2}} - 2 \sqrt{R_{1}}) r = 2 \sqrt{R_{1}} R_{2} - R_{1} \sqrt{R_{2}}$
$⟹$ $r = \sqrt{R_{1} R_{2}} (\frac{2 \sqrt{R_{2}} - \sqrt{R_{1}}}{\sqrt{R_{2}} - 2 \sqrt{R_{1}}})$

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