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(B) The current through the circuit before the battery of emf $E_2$ is short-circuited is given by:$I_1 = \frac{E_1 + E_2}{R + r_1 + r_2}$After short-

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$E_1r_2 > E_2(R + r_1)$

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(B) The current through the circuit before the battery of emf $E_2$ is short-circuited is given by:$I_1 = \frac{E_1 + E_2}{R + r_1 + r_2}$After short-circuiting the battery of emf $E_2$,the current through the resistance $R$ becomes:$I_2 = \frac{E_1}{R + r_1}$For the current to increase,we must have $I_2 > I_1$:$\frac{E_1}{R + r_1} > \frac{E_1 + E_2}{R + r_1 + r_2}$Cross-multiplying gives:$E_1(R + r_1 + r_2) > (E_1 + E_2)(R + r_1)$$E_1R + E_1r_1 + E_1r_2 > E_1R + E_1r_1 + E_2R + E_2r_1$Subtracting $E_1R + E_1r_1$ from both sides:$E_1r_2 > E_2(R + r_1)$

Under what conditions can the current passing through a resistance $R$ be increased by short-circuiting the battery of emf $E_2$? The internal resistances of the two batteries are $r_1$ and $r_2$ respectively.

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