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Two cells of the same e.m.f. 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 found to be zero. The external resistance R [Fig.] is

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a

r_{1 /}r₂

b

r_{1 +} r₂

c

r₁- r₂

d

r₁r₂

answer is A.

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Detailed Solution

Current i in the circuit
$= [2 e / (r_{1} + r_{2} + R)]$
Voltage drop across Ist cell $= V_{1} = e - {ir}_{1}$
$\begin{array}{l} ∴ V_{1} = e - \frac{2 {er}_{1}}{R + r_{1} + r_{2}} = 0 \\ ∴ e = \frac{2 {er}_{1}}{R + r_{1} + r_{2}} or R + r_{1} + r_{2} = 2 r_{1} \end{array}$
$R = r_{1} - r_{2}$

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Two cells of the same e.m.f. 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 found to be zero. The external resistance R [Fig.] is