An inductor of inductance $L = 400 mH$ and resistors of resistances $R_1=2 \Omega$ and $R_2=2 \Omega$ are connected to a battery of emf $\xi =12 V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$ .

What is the potential drop across $L$ as a function of time ? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through $R_1$ as a function of time?

By loop rule, we have

                         $\xi -i_1{R}_1=0$    …(i)

and  $\xi --L\frac{d{i}_2}{dt}-i_2{R}_2=0$     …(ii)

After simplifying equations (i) and (ii), we get

The potential difference across inductor

                                 $$\begin{gathered} V_L=-L\frac{d{i}_2}{dt} \\ =-L\frac{d}{dt}\left[\left(\frac{\xi }{R}\left(1-e^{\frac{-R_{2}t}{L}}\right)\right)\right] \\ =\xi e^{-\frac{R_{2}t}{L}}=12e^{-5t} \end{gathered}$$

At steady state,         $i_1=\frac{\xi }{R_1}=\frac{12}{2}=6 A$

and                            $i_2=\frac{\xi }{R_2}=\frac{12}{2}=6 A$

 When switch $S$ is opened, the inductor opposes the decay of current, so direction of current in $R_2$ remains same while the direction of current in $R_1$ is reversed. If $i_3$ be the current then 

     $L\frac{d{i}_3}{dt}+i_3{R}_1+i_3{R}_2=0$

After simplifying, we get

                                $$\begin{gathered} i_3=i_3{e}^{-t\frac{\left(R_1+R_2\right)}{L}} \\ =6e^{-t\frac{\left(R_1+R_2\right)}{L}} \\ =6e^{-t\frac{\left(2+2\right)}{0.4}} \\ =6e^{-10t} \end{gathered}$$