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(B) The two cells are connected in series. The total $emf$ of the circuit is $E + E = 2E$.The total resistance of the circuit is $R + R_1 + R_2$.The c

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$R = R_2 - R_1$

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(B) The two cells are connected in series. The total $emf$ of the circuit is $E + E = 2E$.The total resistance of the circuit is $R + R_1 + R_2$.The current $i$ flowing through the circuit is given by $i = \frac{2E}{R + R_1 + R_2}$.The potential difference $V$ across the cell with internal resistance $R_2$ is given by $V = E - i R_2$.Given that $V = 0$,we have $0 = E - i R_2$,which implies $E = i R_2$.Substituting the value of $i$ into this equation:$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 = R_2 - R_1$.

Two sources of equal $emf$ are connected 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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