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

[qpham26]

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
i_L(0⁺) = 3 X 4/8 = 1.5A (current divider rule)

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

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

thanks for your time.

 

[gneill]

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

 

[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)?

thanks for your time

 

[gneill]

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.