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Potentiometer

Two cells of ...

Question

Two cells of same emf E but internal resistance $r_{1}$ and $r_{2}$ are connected in series are connected in series with 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

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

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