Two cells,having the same e.m.f. E,are connec | vedclass

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(C) The total $e.m.f.$ of the series combination is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{eq} = R + r_1 + r_2$.The current

Vedclass · Accepted Answer

$r_1 - r_2$

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(C) The total $e.m.f.$ of the series combination is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{eq} = R + r_1 + r_2$.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 expression for $I$:$E = \left( \frac{2E}{R + r_1 + r_2} ight) r_1$.Dividing both sides by $E$ (assuming $E 
eq 0$):$1 = \frac{2r_1}{R + r_1 + r_2}$.$R + r_1 + r_2 = 2r_1$.$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.Thus,the correct option is $C$.

Two cells,having the same $e.m.f.$ $E$,are connected in series through an external resistance $R$. The cells have internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively. When the circuit is closed,the potential difference across the first cell is zero. The value of $R$ is

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