In the circuit shown, the switch $S$ is shifted to position $2$ from position $1$ at $t = 0$, having been in position for a long time. Find the current in the circuit as a function of time. 

Let $I_0$ be the initial current in the steady state condition before shifting the switch.

$\therefore I_0=\frac{E}{R}$

Let $I_0$ be the current in the circuit at the time $t=0$ when the switch is shifted to position $2$.

Since the flux associated with inductors will be same just before and just after the shifting of switch otherwise the induced emf would be infinite.

$\varphi =I_{0}L=I{\text{'}}_0(L+2L)$

Or, $I_0=\frac{I_0}{3}=\frac{E}{3R}$

If $I$ is the instantaneous current in the circuit at time $t$, then

$E-3L\frac{dI}{dt}-IR=0$

Or, $3L\frac{dI}{dt}=E-IR$

Or, ${\int }_{I{\text{'}}_0}^I\frac{dI}{E-IR}=\frac{2}{3L}{\int }_0^{t}dt$

Or, $\ln \left(\frac{E-IR}{E-I{\text{'}}_{0}R}\right)=-\frac{2Rt}{3L}$

Or, $\frac{E-IR}{E-\frac{E}{3R}R}=e^{-2Rt/3L}$

Or $E-IR=\frac{2E}{3}{e}^{-2Rt/3L}$

Or, $I=\frac{E}{R}\left(1-\frac{2}{3}{e}^{-2Rt/3L}\right)$