and we still have the other resistor, 4 amps.

So if we find the current, then we see that that is a current division

problem in this case because the two resistors are equal.

The current divides the 3 amps, divides equally across the two branches and

we have that i (0 minus).

Which is just before the switch changes, is half of 3 amps or

3 half amps and because current in inductors is continuous in time.

We also have that this is the same current through the inductor,

just after the switch changes.

After the switch changes, you have a new circuit.

This time you have two ohms there.

You have the inductor and you have four ohms over here.

And what we know,

in the lesson on first order RL students you would learn how to solve this.

But you would need to know that i at 0,

with 0+, is 3/2 amps for this second circuit.

Okay, to sum up.

Good practice is to redraw the circuit twice.

Once before the switch changes and once after the switch changes.

The inductor current or

the capacitor voltage that you calculate before the switch change.

Will become the initial condition for the circuit after the switch change.

You will use those initial conditions when we do first order circuits.

And you need the initial condition to get the solution to those circuits.

Thank you.

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