Explanation

(b) For maximum power $R_{ext} = R_{int}$
When batteries are connected in parallel.
$R_{int} = \frac{1}{2} = 0.5\Omega$
Current $I = \frac{E}{R_{ext} + R_{int}} = \frac{2}{0.5 + 0.5} = 2\Omega$
Maximum power is given by
$\ P_{\max} = i^{2}R_{ext}$
${\ \Rightarrow P}_{\max} = (2)^{2} \times 0.5 =$2 W
If we assume batteries to be connected in series, then
${\ R}_{int} = 1 + 1 = 2\Omega$
Current $\ I = \frac{2E}{R_{ext} + R_{int}} = \frac{2 \times 2}{2 + 2} = 1$ A
So, maximum power is now given by
$\ P_{\max} = i^{2}R_{ext} = 1 \times 2 =$ 2 W
In either case $P_{\max} =$ 2 W