The topic of this problem is Energy Storage Elements and we're going to work with Circuits with Switches. The problem is to find the current i sub L immediately before the switch is thrown and at time t = infinity a long time after the switch has been thrown. We know that when we're working with inductors, that the current through an inductor cannot change instantaneously. So when we throw the switch at t = 0, the current through the inductor cannot change instantaneously. It will go from one steady state value which is the value before the switch is thrown, to another steady state value a long time after the switch is thrown. And it's going to have some transient period in which the current changes from that first steady state condition before the switch is thrown to the second steady state condition after the switch has been thrown. So in order to solve this circuit, we have to use our properties of the inductor that we know. That is that in a steady state condition, the inductor looks like a short circuit. So we draw our circuit, first of all, before the switch is thrown, so this is at t = 0- before the switch is thrown. And we're going to use this to find our current i sub L before the switch is thrown. So we have our 2 amp source which is in our circuit. We have an inductor which acts as a short circuit. We have a resistor, R2. We have another resistor, R1. We have i sub L At t = 0-. And that's what our circuit looks like before the switch is thrown, before this circuit is complete on the right-hand side of our 2 amp source. So to find i sub L at t = 0-, we simply use current division between R1 and R2. We know that the 2 amp source is going to be split between R1 and R2 depending on the values of R1 and R2. So i sub L, at (t = 0-) = 2 amps, and again it's divided between R1 and R2. So it's going to be 2 amps (R1/R1 + R2). That's our current i sub L at t = 0-. So the second part of this question is to find i sub L at t = infinity, in other words, a long time after the switch has been thrown. So in order to find that, we have to first of all, we draw our circuit for that condition. So we have our 2 amp source which hasn't changed, we have our switch which is thrown in this case, so it's closed, we have a closed switch. We still have our resistor R1, our inductor acts as a short circuit in a steady state, and we have a resistor R sub 2. And so we have a circuit, That has a short, In parallel with a resistor R1 and another resistor R2. So if we're looking for i sub L at t = infinity, and we know that all this current is going to flow through this short circuit, obviously not a good circuit to work with. If we in practice had this situation, the 2 amp source will eventually burn up due to the short which have been in placed across it. But in this condition, we know that the current i sub L at t = infinity, is going to be equal to 0 because no current is going to flow through R2. All the 2 amps is going to flow through the short circuit.