A battery of emf $12\text{ V}$ and internal resistance $2\Omega$ is connected to a variable resistor $R$. The power delivered to the resistor is maximum when $R$ is

Solution

1. The power delivered to the load is $$\begin{array}{rl}P& =I^{2}R\\ & =(E/(R+r){)}^{2}R\end{array}$$.
2. To find the maximum power, we set $dP/dR=0$.
3. This condition is satisfied when $R=r$.
4. Given $r=2\Omega$, the power is maximum when $R=2\Omega$.
5. Hence the answer is (B).