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?

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Solution

Steady state current through inductor
$i_{0} = \frac{E}{R_{2}} = \frac{12}{2} = 6 A$
(From top to bottom through inductor)
Now this current decrease exponentially to zero through $R_{1}$ and $R_{2}$
$∴$ $i = i_{0} e^{- t / τ_{L}}$
where $τ_{L} = \frac{L}{R_{1} + R_{2}} = \frac{0.4}{4} = 0.1 s$
$∴$ $i = 6 e^{- t / 0.1} = 6 e^{- 10 t}$ (clockwise)

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