# SImple inductance problem

1. Nov 3, 2012

### qpham26

1. The problem statement, all variables and given/known data
In the following transient circuit, assume at t<0, the circuit is at steady state.
Find v0(t) for t>0

http://sphotos-b.xx.fbcdn.net/hphotos-prn1/c0.0.277.277/p403x403/550205_509330739086446_705354858_n.jpg [Broken]

2. Relevant equations

3. The attempt at a solution
short out the inductor, the total resistance will be : [(6||12)||4] + 2 = 4Ω?

if the above is correct then the 12V will deliver a current source of: 12/4 = 3A
current of the inductor at t= 0+ will be
iL(0+) = 3 X 4/8 = 1.5A (current divider rule)

iL(∞) = 0 (because the circuit is opened?)

t>0, the 2Ω is out Rth as seen from the inductor will be (6||12) + 4 = 8Ω ?
time constant T = 2H/ 8Ω
iL(t) = 1.5e(-4t) => Vo(t) = 4 x iL(t) = 1.5e^(-4t) = 6e(-4t)

Last edited by a moderator: May 6, 2017
2. Nov 3, 2012

### Staff: Mentor

Your approach looks good. However, pay attention to the current direction and the indicated polarity of measurement for Vo.

3. Nov 3, 2012

### qpham26

For the polarities, when the switch is open, the inductor become a current source?
and the current originally was going down, so the bottom of the 4Ω will be the (+) for t>0
so when KVL is applied Vo(t) will be -V of 4Ω? which should be -6e^(-4t)?