0:03

The topic of this problem is Energy Storage Elements and

Â we're going to work with Circuits with Switches.

Â The problem is to find the current i sub L immediately before the switch

Â is thrown and at time t = infinity a long time after the switch has been thrown.

Â 0:28

So when we throw the switch at t = 0,

Â the current through the inductor cannot change instantaneously.

Â It will go from one steady state value which is the value before the switch

Â is thrown, to another steady state value a long time after the switch is thrown.

Â 1:02

So in order to solve this circuit,

Â we have to use our properties of the inductor that we know.

Â That is that in a steady state condition, the inductor looks like a short circuit.

Â So we draw our circuit, first of all, before the switch is thrown, so

Â this is at t = 0- before the switch is thrown.

Â 1:52

We have i sub L At t = 0-.

Â And that's what our circuit looks like before

Â the switch is thrown, before this circuit is complete

Â on the right-hand side of our 2 amp source.

Â So to find i sub L at t = 0-, we simply use current division between R1 and R2.

Â We know that the 2 amp source is going to be split between R1 and

Â R2 depending on the values of R1 and R2.

Â So i sub L, at (t = 0-) = 2 amps,

Â and again it's divided between R1 and R2.

Â So it's going to be 2 amps (R1/R1 + R2).

Â That's our current i sub L at t = 0-.

Â So the second part of this question is to find i sub L at t = infinity,

Â in other words, a long time after the switch has been thrown.

Â So in order to find that,

Â we have to first of all, we draw our circuit for that condition.

Â So we have our 2 amp source which hasn't changed,

Â we have our switch which is thrown in this case,

Â so it's closed, we have a closed switch.

Â We still have our resistor R1, our inductor acts as a short

Â circuit in a steady state, and we have a resistor R sub 2.

Â And so we have a circuit,

Â 4:28

But in this condition, we know that the current i sub L at t = infinity,

Â is going to be equal to 0 because no current is going to flow through R2.

Â All the 2 amps is going to flow through the short circuit.

Â