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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
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