Transient Circuit Analysis: Finding v0(t) for t>0 - Simple Inductance Problem

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Homework Statement


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

Homework Equations


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)

thanks for your time.
 
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Discussion

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

thanks for your time
 
qpham26 said:
For the polarities, when the switch is open, the inductor become a current source?
Yup.
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)?
Yup.
thanks for your time
No problem. Glad to help.