Explanation

(a) In the circuit shown total external resistance $R = 2\Omega +$parallel combination of two 4$\ \Omega$ resistors + parallel combination of three 15$\ \Omega$ resistors
$= 2 + \frac{4}{2} + \frac{15}{3} = 2 + 2 + 5 = 9\Omega$
As $E = 10V\ $and $i = 1A$, hence internal resistance $r$ of the cell should have a value given by
$E = i(R + r)\ or\ r = \frac{E}{i} - R = \frac{10}{1} - 9 = 1\Omega$
If $4\Omega$ resistors are replaced by $2\Omega$ resistors , then as before
$R^{'} = 2 + \frac{2}{2} + \frac{15}{3} = 2 + 1 + 5 = 8\Omega$
$\therefore$ New circuit current $i^{'} = \frac{E}{R^{'} + r} = \frac{10}{8 + 1} = 1.11A$