A capacitor of 2 F (practically not possible to have a capacity of 2 F) is charged by a battery of 6 V. The battery is removed and circuit is made as shown. Switch is closed at time t = 0. Choose the correct options. 

(a) At t = 0, emf of the circuit= PD across the capacitor = 6V

$\therefore$ $\mathrm{i}=\frac{6}{1+2}=2\mathrm{A}$

Half-life of the circuit 

= (In 2) ${\mathrm{\tau }}_{\mathrm{c}}$ = (In 2) CR= (6 ln 2) s

In half-life time all values get halved. For example, 

$$\begin{gathered} {\mathrm{V}}_{\mathrm{c}}=\frac{6}{2}=3\mathrm{V} \\ \mathrm{i}=\frac{2}{1}=1\mathrm{A} \\ \therefore {\mathrm{V}}_{1\mathrm{\Omega }}=\mathrm{iR}=1\mathrm{V} \\ {\mathrm{V}}_{2\mathrm{\Omega }}=\mathrm{iR}=2\mathrm{V} \\ \mathrm{Power} \mathrm{generated} \mathrm{in} 1\mathrm{\Omega } =1^2\times 1\mathrm{W} = 1 \mathrm{W} \\ \mathrm{Power} \mathrm{generated} \mathrm{in} 2 \mathrm{\Omega } = 1^2\times 2 \mathrm{W} =2 \mathrm{W} \end{gathered}$$