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Two cells of ...

Question

Two cells of same emf $E$ but internal resistance $r_{1}$ and $r_{2}$ are connected in series to an external resistor $R$ (figure). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero.

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Solution

Current flowing through the circuit,

$I = \frac{E_{e q}}{R + r_{1} + r_{1}}$

$I = \frac{E + E}{R + r_{1} + r_{1}}$

Potential difference across the first cell,

$V_{1} = E - I r_{1} = E - \frac{2 E}{r_{1} + r_{2} + R} r_{1} = 0$

$E = \frac{2 E r_{1}}{r_{1} + r_{2} + R}$

$1 = \frac{2 r_{1}}{r_{1} + r_{2} + R}$

$r_{1} + r_{2} + R = 2 r_{1}$

$R = r_{1} - r_{2}$

$Final answer : r_{1} - r_{2}$

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