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## Homework Statement

## Homework Equations

## The Attempt at a Solution

When the t < 0, the switch is open, the inductor acts like a short therefore the circuit looks like

The current is then 20V / 4 ohms = 5 A

Since the inductor has been in series with the battery for a long time, it is safe to assume it is fully charged? So the current across the inductor is 5 A?

Assuming this is correct, i

_{o}= 5 A

If we close the switch, R2 becomes shorted, then the current in the circuit would then be 20V / 1 ohm = 20 A

So the change is in current is 20A -5A = 15A

I don't know if we need this value, but inductors resist change in current so it probably is important?

____________________________

when t = 0, switch is closed and battery becomes shorted and R2 becomes shorted.

Current in circuit is now 20A.

To find voltage change in inductor, we will probably need the change in current and the time for current change.

I'm guessing to find the time, we will need I(t) = Io e^-t/(L/R))

time constant when switch is closed is L/R = 0.08H / 1 ohm = 0.08 sec

20 A = 5A (e^-t/(0.08))

4 = e

^{-12.5t}= e

^{1/12.5t}

ln 4 = 1/12.5t

1.386* 12.5= 17.32 = 1/t

t= 0.0577 sec

Am I on the right path so far?