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

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(B) The total $e.m.f.$ of the two cells connected in series is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{total} = R + r_1 + r_2

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$r_1 - r_2$

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(B) The total $e.m.f.$ of the two cells connected in series is $E_{eq} = E + E = 2E$.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 across the first cell 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 the equation: $E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$.Dividing both sides by $E$: $1 = \frac{2r_1}{R + r_1 + r_2}$.Rearranging the terms: $R + r_1 + r_2 = 2r_1$.Solving for $R$: $R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

Two cells of equal $e.m.f.$ $E$ and of internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ are connected in series. Now they are connected 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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