Two cells with the same e.m.f. E and differen | vedclass

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(C) The total e.m.f. of the two cells connected in series is $E + E = 2E$.The total resistance of the circuit is $R + r_1 + r_2$.The current $I$ flowi

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$r_1-r_2$

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(C) The total e.m.f. of the two cells connected in series is $E + E = 2E$.The total resistance of the circuit is $R + r_1 + r_2$.The current $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_1 + r_2}$.The potential difference $V$ across the first cell (with internal resistance $r_1$) is given by $V = E - Ir_1$.We are given that the potential difference across the first cell is zero, so $E - Ir_1 = 0$, which implies $E = Ir_1$.Substituting the value of $I$ into this equation:$E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$$1 = \frac{2r_1}{R + r_1 + r_2}$$R + r_1 + r_2 = 2r_1$$R = 2r_1 - r_1 - r_2$$R = r_1 - r_2$

Two cells with the same e.m.f. $E$ and different internal resistances $r_1$ and $r_2$ are connected in series to an external resistance $R$. The value of $R$ so that the potential difference across the first cell is zero is

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