When two cells of emfs E_1 and E_2 and differ | vedclass

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(C) Let the total internal resistance be $r = r_1 + r_2$. When the cells are in series with the same polarity,the total emf is $E_1 + E_2$. The curren

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$\frac{7}{3}$

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(C) Let the total internal resistance be $r = r_1 + r_2$. When the cells are in series with the same polarity,the total emf is $E_1 + E_2$. The current $I_1$ is given by $I_1 = \frac{E_1 + E_2}{R + r} = 5 \ A$. So,$E_1 + E_2 = 5(R + r) \quad (1)$. When the polarity of $E_2$ is reversed,the total emf becomes $E_1 - E_2$ (assuming $E_1 > E_2$). The current $I_2$ is given by $I_2 = \frac{E_1 - E_2}{R + r} = 2 \ A$. So,$E_1 - E_2 = 2(R + r) \quad (2)$. Dividing equation $(1)$ by equation $(2)$,we get: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{5}{2}$. Applying componendo and dividendo: $\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{5 + 2}{5 - 2}$. $\frac{2E_1}{2E_2} = \frac{7}{3}$. Therefore,$\frac{E_1}{E_2} = \frac{7}{3}$.

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