When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a…
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
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