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Potentiometer

Two sources o...

Question

Two sources of equal emf are connected to a resistance R. The internal resistance of these sources are $r_{1}$ and $r_{2} (r_{1} > r_{2})$ . If the potential difference across the source having internal resistance $r_{2}$ is zero, then

R = \frac{r_{1} r_{2}}{r_{2} - r_{1}}

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R = r_{2} (\frac{r_{1} + r_{2}}{r_{2} - r_{1}})

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R = (\frac{r_{1} r_{2}}{r_{2} + r_{1}})

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R = r_{2} - r_{1}

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Solution

The correct option is D $R = r_{2} - r_{1}$
Let the emf of each source be E. When they are connected in series, the current in the circuit
$I = \frac{E_{t o t}}{R_{t o t}} = \frac{E + E}{r_{1} + r_{2} + R} = \frac{2 E}{r_{1} + r_{2} + R}$
$∴$ potential drop across the cell of internal resistance
$r_{2}, (\frac{2 E}{r_{1} + r_{2} + R}) r_{2}$
Hence, $E - \frac{2 E}{(r_{1} + r_{2} + R)} r_{2} = 0$
$r_{1} + r_{2} + R = 2 r_{2}$
$\Rightarrow R = r_{2} - r_{1}$

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