In the circuit shown in figure, switch $S_1$  is initially closed and $S_{2 }$  is open. Find  $V_b-V_a$

Switch $S_2$  is open so capacitor is not in circuit  

Current through  $3\mathrm{\Omega }$ resistor  $=\frac{24}{3+3}=4\mathrm{A}$

Let potential of point ‘O’ shown in fig. is  ${\mathrm{V}}_0$ then using ohm’s law

${\mathrm{V}}_0-{\mathrm{V}}_{\mathrm{a}}=3\times 4=12\mathrm{V}$.....(i)

Now current through   $5\mathrm{\Omega }$   resistor $=\frac{24}{5+1}=4\mathrm{A}$

So,  ${\mathrm{V}}_0-{\mathrm{V}}_{\mathrm{b}}=4\times 1=4\mathrm{V}$ ....(ii)

From equation (i) and (ii)  ${\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}}=12-4=8\mathrm{V}$