Explanation

(a) The emf of the circuit is
$E = E_{1} + E_{2}$
$\ \ \ \ \ = 4V + 2V = 6V$
In the given circuit, 3$\Omega$ and 6$\Omega$ are connected in parallel, hence equivalent resistance is
$\frac{1}{R^{'}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$
$\Longrightarrow R^{'} = 2\Omega$
Total resistance of circuit is
$R = 1\Omega + 1\Omega + 2\Omega = 6\Omega$
From Ohm’s law $V = iR$
$\Longrightarrow i = \frac{V}{R} = \frac{6}{6} = 1A$
The 3$\Omega\ and\ 6\Omega$ resistors are in parallel, hence
$\ \ \ \ \ \ \ i_{1}R_{1} = i_{2}R_{2} = V$
$\therefore i_{1} \times 3 = i_{2} \times 6$
$\Longrightarrow i_{1} = 2i_{2}\ \ and\ i_{1} + i_{2} = 1$
$\ \ \ \ \ 2i_{2} + i_{2} = 1$
$3i_{2} = 1 \Longrightarrow i_{2} = \frac{1}{3}A$