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Non-Ideal Battery

Two cells of ...

Question

Two cells of same emf $E$ but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is zero. Then $R$ is

r_{1} + r_{2}

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r_{1} - r_{2}

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r_{2} - r_{1}

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r_{1} r_{2}

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Solution

The correct option is B $r_{1} - r_{2}$
Equivalent emf of the circuit $E_{e q} = E + E = 2 E$
Equivalent resistance of the circuit $R_{e q} = r_{1} + r_{2} + R$ (series connection)
Thus current flowing through the circuit $i = \frac{2 E}{r_{1} + r_{2} + R}$
Potential drop across the first cell $V_{1} = E - i r_{1}$
But $V_{1} = 0$
$∴$ $E - \frac{2 E r_{1}}{r_{1} + r_{2} + R} = 0$
Or $1 - \frac{2 r_{1}}{r_{1} + r_{2} + R} = 0$
Or $R + r_{2} - r_{1} = 0$
$⟹$ $R = r_{1} - r_{2}$

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Two cells of same emf E but different internal resistances r1 and r2 are connected in series with an external resistance R. The potential drop across the first cell is zero. Then R is

Two cells of same emf $E$ but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is zero. Then $R$ is