Two sources of equal emf E are connected in s | vedclass

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(C) Let the emf of each source be $E$. The total emf of the circuit is $E + E = 2E$.The total resistance of the circuit is $R_{total} = R + R_1 + R_2$

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$R = R_2 - R_1$

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(C) Let the emf of each source be $E$. The total emf of the circuit is $E + E = 2E$.The total resistance of the circuit is $R_{total} = R + R_1 + R_2$.The current $I$ flowing through the circuit is $I = \frac{2E}{R + R_1 + R_2}$.The potential difference $V_2$ across the source with internal resistance $R_2$ is given by $V_2 = E - I R_2$.Given that $V_2 = 0$,we have $E = I R_2$.Substituting the value of $I$,we get $E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$.Dividing both sides by $E$,we get $1 = \frac{2 R_2}{R + R_1 + R_2}$.$R + R_1 + R_2 = 2 R_2$.$R = 2 R_2 - R_1 - R_2 = R_2 - R_1$.

Two sources of equal emf $E$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then the value of $R$ is:

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