Two cells of equal e.m.f. and of internal resistances $\Gamma_1$ and $\Gamma_2(\Gamma_1 > \Gamma_2)$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be
$(a) r_1 + r_2 (b) r_1 - r_2 (c) \frac{r_1 + r_2}{2} (d) \frac{r_1 - r_2}{2}$
Text Solution
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Let the voltage across any one cell is V, then
$V = E - ir = E - r_1 \left(\frac{2E}{r_1 + r_2 + R}\right)$
But V = 0
⇒ ⇒ $E - \frac{2Er_1}{r_1 + r_2 + R} = 0$
⇒ ⇒ $r_{1} + r_{2} + R = 2r_{1}$
⇒ ⇒ $R = r_1 - r_2$
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