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(D) The total $emf$ 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

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$R = R_2 - R_1$

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(D) The total $emf$ 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 in the circuit is given by $i = \frac{2E}{R + R_1 + R_2}$.The potential difference $V$ across a source with $emf$ $E$ and internal resistance $r$ is given by $V = E - ir$. For the source with internal resistance $R_2$,the potential difference is zero:$0 = E - i R_2$$E = i R_2$Substituting the value of $i$:$E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$$1 = \frac{2 R_2}{R + R_1 + R_2}$$R + R_1 + R_2 = 2 R_2$$R = 2 R_2 - R_2 - R_1$$R = 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:

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