Q78 / 84 · JEE Advanced 2010 (Paper 1) · Physics
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
Correct answer: 4
Solution
In series, $i=\frac{2 E}{2+R}$
$$
\therefore J_1=i^2 R=\left(\frac{2 E}{2+R}\right)^2 \cdot R
$$
In parallel, $i=\frac{E}{0.5+R}$
$$
\begin{aligned}
\therefore \quad J_2 & =i^2 R=\left(\frac{E}{0.5+R}\right)^2 \cdot R \\
\frac{J_1}{J_2} & =2.25=\frac{4(0.5+R)^2}{(2+R)^2}
\end{aligned}
$$
or $1.5=\frac{2(0.5+R)}{(2+R)}$
Solving we get, $R=4 \Omega$
$\therefore$ The answer is 4 .
Hence, the correct answer is 4.