1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Am I doing this correctly? (s domain analysis)

  1. Feb 27, 2006 #1
    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.

    Attached Files:

  2. jcsd
  3. Feb 27, 2006 #2
    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

    [tex]I_0 = \frac {V_a}{R}[/tex]
  4. Feb 27, 2006 #3
    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

    [tex]I_0(s) = \frac {V_a}{sR}[/tex]

    Last edited: Feb 27, 2006
  5. Feb 27, 2006 #4
    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

    [tex]LI_0 = L \frac {Va}{R}[/tex]

    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

    [tex]\frac {Va}{s} + 2R I(s) + sL I(s) + L \frac {Va}{R} = 0[/tex]
  6. Feb 27, 2006 #5
    So I factor, move the I(s) term over, then divide, is this correct?

    [tex]\frac{VaR}{sR} + \frac{SLVa}{sR} = -I(s)(2R + sL)

    I(s) = \frac{-Va}{R} \ast \frac{(R + sL)}{s(2R + sL)} [/tex]
    Last edited: Feb 27, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Am I doing this correctly? (s domain analysis)