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(A) The total emf of the circuit is $2E$ and the total resistance is $R + r_{1} + r_{2}$.The current $I$ in the circuit is given by:$I = \frac{2E}{R +

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$r_{2} - r_{1}$

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(A) The total emf of the circuit is $2E$ and the total resistance is $R + r_{1} + r_{2}$.The current $I$ in the circuit is given by:$I = \frac{2E}{R + r_{1} + r_{2}} \quad \dots (i)$The potential difference across the second cell (with emf $E$ and internal resistance $r_{2}$) is given by $V = E - Ir_{2}$.Given that the potential difference across the second cell is zero:$E - Ir_{2} = 0 \Rightarrow I = \frac{E}{r_{2}} \quad \dots (ii)$Comparing the values of $I$ from equations $(i)$ and $(ii)$:$\frac{E}{r_{2}} = \frac{2E}{R + r_{1} + r_{2}}$$R + r_{1} + r_{2} = 2r_{2}$$R = 2r_{2} - r_{2} - r_{1}$$R = r_{2} - r_{1}$

Two cells of same emf $E$ but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with a resistance $R$. The value of resistance $R$,for which the potential difference across the second cell is zero,is

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