Explanation

(b) Switch $S_{2}$ is open so capacitor is not in circuit
Current through $3\Omega$ resistor $= \frac{24}{3 + 3} = 4A$
Let potential of point $'O'$ shown in fig. is $V_{O}$
Then using ohm’s law
$V_{O} - V_{a} = 3 \times 4 = 12V$ …(i)
Now current through $5\Omega$ resistor $\frac{24}{5 + 1} = 4A$
So $V_{0} - V_{b} = 4 \times 1 = 4V$ …(ii)
From equation (i) and (ii) $V_{b} - V_{a} = 12 - 4 = 8V$