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(D) Let the $EMF$ of each battery be $E$ and internal resistance be $r = 1 \Omega$.For series connection,the total $EMF$ is $2E$ and total internal re

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$4$

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(D) Let the $EMF$ of each battery be $E$ and internal resistance be $r = 1 \Omega$.For series connection,the total $EMF$ is $2E$ and total internal resistance is $2r = 2 \Omega$. The power dissipated in $R$ is $J_1 = (\frac{2E}{R+2})^2 R$.For parallel connection,the total $EMF$ is $E$ and total internal resistance is $r/2 = 0.5 \Omega$. The power dissipated in $R$ is $J_2 = (\frac{E}{R+0.5})^2 R$.Given $J_1 = 2.25 J_2$,we have $(\frac{2E}{R+2})^2 R = 2.25 (\frac{E}{R+0.5})^2 R$.Dividing both sides by $E^2 R$,we get $\frac{4}{(R+2)^2} = 2.25 \frac{1}{(R+0.5)^2}$.Taking the square root of both sides,$\frac{2}{R+2} = \frac{1.5}{R+0.5}$.Cross-multiplying gives $2(R+0.5) = 1.5(R+2)$,which simplifies to $2R + 1 = 1.5R + 3$.Solving for $R$,we get $0.5R = 2$,so $R = 4 \Omega$.

When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$,the rate of heat produced in $R$ is $J_1$. When the same batteries are connected in parallel across $R$,the rate is $J_2$. If $J_1 = 2.25 J_2$,then the value of $R$ in $\Omega$ is

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