d
Growth in current in \(L R_{2}\) branch when switch is closed is given by 

\(i=\frac{E}{R_{2}}\left[1-e^{-R_{2} t / L}\right] \Rightarrow \frac{d i}{d t}=\frac{E}{R_{2}} \cdot \frac{R_{2}}{L} e^{-R_{2} t / L}=\frac{E}{L} e^{-\frac{R_{2} t}{L}}\)

Hence, potential drop across

\(\mathrm{L}=\left(\frac{E}{L} e^{-R_{2} t / L}\right) L=E e^{-R_{2} t / L}=12 e^{-\frac{2 t}{400 \times 10^{-3}}}\)

\(=12 \mathrm{e}^{-5 \mathrm{t}}\, \mathrm{V}\)