Engineering Solve RL Circuit Problem: Homework Statement & Equations

[Duderonimous]

Homework Statement

The attachment shows the circuit in question. Pardon the crude drawing. The top circuit is the circuit drawn in the text. The middle circuit is how the circuit is supposed to behave for t < 0 (switched closed). The bottom circuit is how the circuit is supposed to behave for t > 0 (switch open). The current source in the top circuit is constant.The switch is closed for a very long time and then at t = 0 it opens.
Derive an expression for i(t) for t > 0.

Homework Equations

V=iR
τ=L/R
i(t)=i(0)e^-(t/τ)

The Attempt at a Solution

Looking at the top circuit:
I handled the case for t<0 to find the Initial current through the inductor (is the initial current in the equation the initial current through the inductor or is it the current flowing through the resistor?) by closing the switch for a longtime and then did a source transformation transforming the current source with a resistor in parallel to a voltage source with a resistor in series.
v_sc=(15A)(24Ω)=360VNow I'm at the middle circuit:
I understand that direct current shorts an inductor.
This in turn shorts out the 5ohm resistor ( Why? I know that current takes the path of least resistance but...)

So I can combine the 12Ω and 8Ω resistor and add it to the 24Ω resistor.
R_eq=(12||8)+24=36.5Ω

Therefore I_eq=360/36.5=9.86A

Then I used current division across the 12Ω and 8Ω resistors in parallel.
I_ο=((8)/(12+8))(9.86A)=3.94A

After finding the initial current I open the switch and thus arrive at the bottom circuit:
From here I find R_L and then find τ which should be my final answer.

The answer is 5e^(-2t) A. Could someone please help? Thank you.

 

[axmls]

The issue is with how you're drawing the middle circuit.

The 5 Ohm resistor is still there. The thing is, there's no voltage difference across it, because the end of the other two resistors are both connected to ground (so the voltage on the left is 0 and the voltage on the right is 0). So you can conclude that no current passes through the 5 Ohm resistor. So now, you know you only have to find the current through the 12 Ohm resistor to find the current through the inductor (which is short circuited since the switch has been closed for a long time).

I found it easier to leave the current source in and use current division twice, but to each his own.

 

[Duderonimous]

Thank you axmls.