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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.

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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.

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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.

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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,

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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.