Two cells of equal e.m.f. E and internal resi | vedclass

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(B) The total $e.m.f.$ of the series combination is $E_{eq} = E + E = 2E$.The total internal resistance is $r_{eq} = r_1 + r_2$.The total resistance o

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$r_1 - r_2$

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(B) The total $e.m.f.$ of the series combination is $E_{eq} = E + E = 2E$.The total internal resistance is $r_{eq} = r_1 + r_2$.The total resistance of the circuit is $R_{total} = R + r_1 + r_2$.The current $i$ flowing through the circuit is given by $i = \frac{2E}{R + r_1 + r_2}$.The potential difference $V_1$ across the first cell (with internal resistance $r_1$) is given by $V_1 = E - i r_1$.Given that $V_1 = 0$,we have $E - i r_1 = 0$,which implies $E = i r_1$.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 of equal $e.m.f.$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ 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

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