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(B) Let the $EMF$ of each battery be $E$ and internal resistance be $r = 1 \Omega$.In series connection,total $EMF$ $= 2E$ and total internal resistan

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$4$

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(B) Let the $EMF$ of each battery be $E$ and internal resistance be $r = 1 \Omega$.In series connection,total $EMF$ $= 2E$ and total internal resistance $= 2r = 2 \Omega$.The current $I_s = \frac{2E}{R + 2}$.The power $P_1 = I_s^2 R = \left(\frac{2E}{R + 2}\right)^2 R$.In parallel connection,total $EMF$ $= E$ and total internal resistance $= \frac{r}{2} = 0.5 \Omega$.The current $I_p = \frac{E}{R + 0.5}$.The power $P_2 = I_p^2 R = \left(\frac{E}{R + 0.5}\right)^2 R$.Given $P_1 = 2.25 P_2$,so $\left(\frac{2E}{R + 2}\right)^2 R = 2.25 \left(\frac{E}{R + 0.5}\right)^2 R$.Dividing both sides by $E^2 R$,we get $\frac{4}{(R + 2)^2} = 2.25 \frac{1}{(R + 0.5)^2}$.Taking square root on both sides: $\frac{2}{R + 2} = \frac{1.5}{R + 0.5}$.$2(R + 0.5) = 1.5(R + 2) \implies 2R + 1 = 1.5R + 3$.$0.5R = 2 \implies R = 4 \Omega$.

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