Explanation

(b) In the given circuit, resistors 4R and 2R are connected in parallel while resistance R is connected in series to it.
Hence, equivalent resistance is
$\frac{1}{R^{'}} = \frac{1}{4R} + \frac{1}{2R}$
$\ \ \ \ \ \ = \frac{6R}{8R^{2}}$
$R^{'} = \frac{8}{6}R$
$\ \ \ \ \ \ = \frac{4}{3}R\ $
$\Longrightarrow R^{''} = R + \frac{4}{3}R = \frac{7R}{3}$
Given, emf is E volts, therefore
$i = \frac{E}{R} = \frac{3E}{7R}$
Potential difference across R is
$V = ir = \frac{3R}{7R} \times R = \frac{3E}{7}$
Potential difference across 2R is
$V^{'} = E - \frac{3E}{7} = \frac{4E}{7}$