Am I doing this correctly? (s domain analysis)

[seang]

The switch in the following circuit has been close for a long time and is opened at t = 0. Transform the circuit into the s doman and solve for Isubl(s) and Isubl(t) in symbolic form.

I've only found Isubl(s) so far, and I want to see how I'm doing before I convert it back to the time domain. I hope my work is clear; I used the t < 0 circuit to find out the inductors initial current, then used superposition to find the final current.

How am I doing? Thanks alot.

 

[Corneo]

I think you found the intial current incorrectly. Since the switched has been closed for a long time, then we can safely assume that the circuit is in steady state. This means that the inductor is acts like a perfect conductor or simply a wire. Then the intial current is actually

I_0 = \frac {V_a}{R}

 

[seang]

Ok, so is it customary to find hte IC before converting the circuit to the s domain? Because I intuitively knew that an inductor was a short circuit in a DC setting, but regardless, its impedance in the s domain is Ls, right? How does this work?

EDIT: I guess it shouldn't matter, right?

So the formula you presented is in the time domain, right?
its s domain equivalent would be

I_0(s) = \frac {V_a}{sR}

yay?

 

[Corneo]

Yes you need to find the intial conditions before transforming the circuit into the s domain. The equation I wrote is in s domain already. No need to modify it. The battery that is introduced in the s domain should be

LI_0 = L \frac {Va}{R}

The impedance of the inductor in the s domain is indeed sL.

After you transform the circuit into the s domain, write a KVL to obtain

\frac {Va}{s} + 2R I(s) + sL I(s) + L \frac {Va}{R} = 0

 

[seang]

So I factor, move the I(s) term over, then divide, is this correct?\frac{VaR}{sR} + \frac{SLVa}{sR} = -I(s)(2R + sL)I(s) = \frac{-Va}{R} \ast \frac{(R + sL)}{s(2R + sL)}