Explanation

(c) We can simplify the network as shown
So, net resistance,
R=2.4+1.6=4.0$\Omega$
Therefore, current from the battery.
$i = \frac{V}{R} = \frac{4}{4} = 1A$
Now, from the circuit (b),
4I’ =6I
$\Longrightarrow \ \ I^{'} = \frac{3}{2}I$
But $i$=I+I’
$\ \ \ \ \ \ \ \ \ \ \ \ = I + \frac{3}{2}I = \frac{5}{2}I$
$\therefore\ \ \ \ 1 = \frac{5}{2}I$
$\Longrightarrow \ \ I = \frac{2}{5} = 0.4A$