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(C) The total $emf$ of the circuit is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{eq} = R + r_1 + r_2$.The current in the circuit

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$R = r_2 - r_1$

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(C) The total $emf$ of the circuit is $E_{eq} = E + E = 2E$.The total resistance of the circuit is $R_{eq} = R + r_1 + r_2$.The current in the circuit is $I = \frac{2E}{R + r_1 + r_2}$.The potential difference across the internal resistance $r_1$ is given by $V_{r1} = I r_1$.According to the problem,$V_{r1} = E$,so $I r_1 = E$.Substituting the value of $I$,we get $\left( \frac{2E}{R + r_1 + r_2} \right) r_1 = E$.Dividing both sides by $E$,we get $\frac{2r_1}{R + r_1 + r_2} = 1$.$2r_1 = R + r_1 + r_2$.$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

If the potential difference across the internal resistance $r_1$ is equal to the $emf$ $E$ of the battery,then

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