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(A) The total $emf$ of the two sources connected in series is $\varepsilon_{eq} = \varepsilon + \varepsilon = 2\varepsilon$.The total resistance of th

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$R = R_2 - R_1$

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(A) The total $emf$ of the two sources connected in series is $\varepsilon_{eq} = \varepsilon + \varepsilon = 2\varepsilon$.The total resistance of the circuit is $R_{total} = R + R_1 + R_2$.The current $I$ flowing through the circuit is given by $I = \frac{2\varepsilon}{R + R_1 + R_2}$.The potential difference $V$ across the source with internal resistance $R_2$ is given by $V = \varepsilon - I R_2$.Given that $V = 0$, we have $\varepsilon - I R_2 = 0$, which implies $\varepsilon = I R_2$.Substituting the value of $I$:$\varepsilon = \left( \frac{2\varepsilon}{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 $emf$ $(\varepsilon)$ 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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