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.

 

[rezeew]

I would like to point out that your solution is correct. However, I would also like to provide a more detailed explanation for the steps you took to arrive at your solution.

Firstly, it is important to note that the circuit is at steady state at t<0, which means that all the capacitors in the circuit are fully charged and behave like open circuits. This means that the 2Ω resistor can be ignored and the 6Ω and 12Ω resistors can be replaced by a single equivalent resistance of 4Ω using the parallel resistance formula.

Next, the 12V voltage source can be replaced by a current source of 12/4 = 3A, as you correctly stated. This current will flow through the 4Ω resistor, splitting into two branches of 1.5A each due to the current divider rule.

At t=0+, the switch is opened and the inductor is disconnected from the circuit. This means that the current through the inductor will remain constant at 1.5A. However, since the inductor is disconnected, the current will flow through the 4Ω resistor and the 2Ω resistor in series. This results in a voltage drop of 3V across the 2Ω resistor, which will be the initial voltage across the inductor.

Now, for t>0, the 2Ω resistor is removed from the circuit, leaving only the 6Ω and 12Ω resistors in series. This means that the inductor will discharge through the 6Ω and 12Ω resistors. The equivalent resistance seen by the inductor will be 6Ω+12Ω = 18Ω. Using the time constant formula, we can calculate the time constant as T = L/R = 2H/18Ω = 1/9 seconds.

Finally, we can use the formula for the voltage across an inductor in a discharging circuit, which is Vo(t) = Vo(0)e^(-t/T), where Vo(0) is the initial voltage across the inductor at t=0+. Substituting the values we calculated, we get Vo(t) = 3e^(-9t). Multiplying this by the factor of 4, we get the final solution of Vo(t) = 12e^(-9t).

In conclusion,