Answer
a.) 1 Amp
b.) 0 Amps
c.) 3 Amps
Work Step by Step
a.) Immediately when the switch is closed we can say that the inductor acts like a piece of open-wire that doesn't complete the circuit. This is because at $t=0$ the rate of change of current is still zero so $\Delta V_L = -L(0) = 0$ Thus, we can completely ignore the inductor when using Kirchoff's loop law:
$\displaystyle \varepsilon - IR_1-IR_2 = 0$
$I = \displaystyle \frac{\varepsilon}{R_1+R_2} = \frac{30}{10+20} = 1$ A
The current in the circuit, which passes through the $20$ $\Omega$ resistor, is 1 Amp.
b.) When the switch has been open for a long time, we can say that there is a steady current in the circuit such that $\displaystyle \frac{dI}{dt} = 0$. Thus, the inductor just acts like a piece of wire and the circuit becomes "shorted" as the current will completely bypass the $20$ $\Omega$ resistor since it's in parallel with the inductor. Therefore, no current passes through the $20$ $\Omega$ resistor (0 Amp)
c.) You might now think that because the switch is now opened, that this is the same question as part a.); that is not true. The difference between part a.) and part c.) is that in part a.) it states that the switch has been open for a long time meaning that no current has run through the circuit. In part c.), the circuit was closed and then opened, meaning that there was a current running through the system.
In part a.) the inductor has no stored energy.
In part c.) the inductor HAS stored energy in its magnetic field.
So how do we find the current through the $20$ $\Omega$ resistor?
When the switch was closed in part b.), even though there wasn't a current through the $20$ $\Omega$ resistor, there was still current through the system that was $\varepsilon - IR_1 = 0\qquad\rightarrow \qquad\displaystyle \frac{\varepsilon}{R_1} = \frac{30}{10} = 3$ A $= I$
Immediately after the switch is opened, this "initial" current that ran through the circuit when the inductor acted as a piece of shorted wire now travels through the $20$ $\Omega$ resistor. Thus, there is a 3 A current running through the $20$ $\Omega$ resistor immediately after the switch is reopened.
