Electronic – Steady state AC circuit

The current through an inductor is continuous so

$$i_L(0+) = 2A$$

The voltage across a capacitor is continuous so

$$v_C(0+) = 0V$$

Now, the inductor and the resistor are in series so they have identical currents. So, armed with just KVL and Ohm's Law,

$$v_L(0+)=?$$

This is resolved through Ohm's law. If you think of the switch as an infinite resistance, then if any current flows on the wire directly above the switch in the schematic, then an equal amount in the opposite direction must flow in the infinite resistor (switch) because there's no other path for the current to come back from. Any current at all in an infinite resistor implies infinite voltage. You can assume that you never have infinite voltage in any realistic system. Therefore, the current through the infinite resistor must be 0 which implies that the current through the wire above it must also be 0.

Also, a circuit implies closed loop as seen here: "An electrical circuit is a network consisting of a closed loop, giving a return path for the current." http://en.wikipedia.org/wiki/Electrical_circuit

Perhaps I should also mention that Kirkchhoff's Voltage Law would also get interesting if you had current going through the switch. That would imply infinite voltage which would then imply negative infinite voltage across the rest of the circuit going through the top wire due to KVL. Obviously, it's not possible.

The only caveat that I should mention here is something like an antenna where there physically doesn't have to be a closed loop return path. This is resolved through capacitance to ground and the fact that one uses a high frequency wave to send short pulses through the inherent capacitance to "ground". The point I'm making is that through capacitance in a circuit, you can send current through an "open-loop", but that can only be done for very short periods of time. That's a different area of study than what you're doing here. You're doing pulsed-dc analysis whereas AC analysis can get interesting.