Voltage across inductor immediately after the switch is closed (again)

The switch in this circuit has been open for a very long time (such that the circuit is in DC steady state condition).

What is v2?

The correct answer is v2 = 0 V, but how does that make sense? I know that the current through the inductor is 0 A, but why is the voltage 0?

Answer

I understand why you are wondering: the current in an inductor does not change instantly, but the voltage can.

However, there is something else in your schematic that influences the voltage: the capacitor.

After a long time, the capacitor is also discharged and its voltage can not change instantly, but its current can.

So immediately after closing the switch, the voltage over the capacitor can not change and therefore the voltage presented to the further right of your circuit is 0. Before closing the switch that voltage was also 0.

Therefore, there is no reason to have current flow in \$R_3\$. And \$L_1\$ is not seeing any change either. The voltage and the current of \$L_1\$ are 0 just after the switch closing (and that 0V is due to the capacitor)