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Two cells, having the same e.m.f., are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2$ ($r_1 \gt r_2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is :-
A
$r_1 - r_2$
B
$\dfrac{r_1 + r_2}{2}$
C
$\dfrac{r_1 - r_2}{2}$
D
$r_1 + r_2$
Detailed Solution
The two cells of emf $E$ each are in series, so the total emf is $2E$ and the total resistance is $r_1 + r_2 + R$.
Current in the circuit: $I = \dfrac{E + E}{r_1 + r_2 + R} = \dfrac{2E}{r_1 + r_2 + R}$
Terminal potential difference across the first cell: $V_1 = E - Ir_1$
According to the question, $E - Ir_1 = 0 \Rightarrow I = \dfrac{E}{r_1}$
$\therefore \dfrac{E}{r_1} = \dfrac{2E}{r_1 + r_2 + R}$
$\Rightarrow r_1 + r_2 + R = 2r_1$
$\Rightarrow R = r_1 - r_2$
Current in the circuit: $I = \dfrac{E + E}{r_1 + r_2 + R} = \dfrac{2E}{r_1 + r_2 + R}$
Terminal potential difference across the first cell: $V_1 = E - Ir_1$
According to the question, $E - Ir_1 = 0 \Rightarrow I = \dfrac{E}{r_1}$
$\therefore \dfrac{E}{r_1} = \dfrac{2E}{r_1 + r_2 + R}$
$\Rightarrow r_1 + r_2 + R = 2r_1$
$\Rightarrow R = r_1 - r_2$