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Homework Help: Transient Analysis of an RLC Circuit

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    For a series circuit with the following components:

    1. Vin(t)
    2. 1K resistor
    3. 100 mH Inductor
    4. 1 uF capacitor

    1. Write down the differential eqn. for Vc in the circuit.
    2. Solve the DE, calculate the natural and forced response with the following initial conditions: Vc(0) = 0 and dVc(t)/dy = 0
    3. Explain the difference between theoretical and experimental results.

    2. Relevant equations

    This was the signal we applied in the lab:

    Vin(t) = (1V)sin(2∏(2 kHz)t)u(t)

    3. The attempt at a solution

    This is what I have for the DE:

    dv(t)/dt + (1/RL)Vl(t) + (C/R)d^2Vc(t)/dt^2 = 0

    This is for a lab where we have to compare theoretical value to measured value, and apparently they are supposed to be different.
    I think it's right... but I'm not positive. I'm a little confused because there are 4 different voltages in the circuit (Vl, Vc, Vr, and Vtot), and I'm getting tripped up as to which one I'm solving for. I know that I should take the DE and set it equal to the initial conditions in order to solve for the forced and natural responses, but I really need a little push here to get started. Any help would be appreciated. Thanks.
  2. jcsd
  3. Dec 9, 2013 #2

    rude man

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    Is Vc the capacitor voltage?
    What is v(t) in your equation? What happened to Vr? Vtot?

    You can'tw write a diff. eq. with just Vc as the dependent variable. Write the d.e. for current i instead.
  4. Dec 9, 2013 #3
    Yeah Vc is the capacitor voltage, Vr is resistor voltage, Vl is inductor voltage, and Vtot is the source voltage. Is finding V(t) the first step of the process? If so, how do I go about that? Thanks again.
  5. Dec 9, 2013 #4

    rude man

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    V(t) = Vtot is the input voltage you were given.
    So your diff. eq should be
    Vtot = Vl + Vc + Vr.
    But you need to substitute the voltage-current relationship for each of Vl, Vc and Vr. Then you wind up with a diff. eq. with current as the dependent variable and time as the independent variable. Then you can get Vc from the i-V relationship of a capacitor.

    Another, simpler approach is by using the complex impedance expressions for V, R, L and C. You set up a voltage divider with R and L in series connecting the input to the output, and C going from output to ground. V(t) is applied from the input to ground. Have you had phasors and complex math?

    But this mthod does not allow you to get the transient response. I suspect that is not relevant for you anyway.

    The transient response depends on at what point in the input voltage cycle the voltage is applied (t=0). You probably have no control over that.

    This last observation is also a hint as to why your results may differ from what your diff. eq. predicts.
  6. Dec 10, 2013 #5
    Hey thanks for the response. I ended up putting everything into the S domain, and doing a voltage divider as you suggested. Once I had this I did an inverse Laplace transformation back into t. This solved my problem, but I still definitely need to work on my 2nd order DEs.
  7. Dec 10, 2013 #6

    rude man

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    That's good. I should have realized you DID want the transient response as well as the steady-state one from the title of your post.

    Did you answer the question as to why the lab results differ from theory? Or, did you run several trials in the lab & got different results?

    Hint: when you applied your step function it was actually not sin(wt) but really sin(wt + phi) where phi is a random variable -180 deg. < phi < +180 deg.

    BTW the best way to solve your linear d.e's with constant coefficients is usually via the Laplace transform. It's a 'one size fits all' mehtod. It automatically includes initial conditions and avoids the silly "guessing" of the classical approach. God Bless Our Laplace Tables! Get a really good one or use Wolfram Alpha.
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