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RC circuit with complex exponential

  1. Apr 2, 2005 #1
    I am stuck on this...
    Given a circuit: current source (Is(t)), R , C - all parallel; Is(t) = e^jt, Vs(t) = 223.6e^j(t - 63.43), Vs(t) is voltage across the current source, which I assume to be the same across R and C since they are ||.
    Find R and C. (ans: 500 Ohm, 4mF)

    My attemp was to use source transformation and get R from it, but it did not match the answer.
    I.e. I tried to make a circuit all series by source transformation: V = IR, so given Vs(t) and Is(t) , R = 223.6e^(-j63.43) and that does not turn out to be 500 Ohm :grumpy:

    Another way I am thinking is that Vs(t) is the response on the C, so I will have to work backwards from the equations for phase and Vm (response magnitude).

    Any help on this is GREATLY appreciated...this is as clear as mudd... :cry:
    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 3, 2005 #2


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    The impedance of the parallel RC circuit is Z = Vs(t)/Is(t).
    But Z = R + jC. The real part of Z is R and the imaginary part is C.
  4. Apr 3, 2005 #3
    Thanks for helping, SGT, although we have not talked about impedance.
    So, I got Z = 223.6e^-j63.43, and corresponding representation as 223.6cos(-63.43) + j 223.6 sin(-63.43) and I still do not get 500 for the real part :frown:
    Could you explain in little more detail, I do not think I am quite getting it.
  5. Apr 3, 2005 #4


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    My mistake.
    The admittance of the parallel RC circuit is Y = 1/R+jωC (it is the complex equivalent of conductance).
    Where ω is the frequency of the voltage/current. In your example ω=1.
    But Y = Is(t)/Vs(t) = 1/223.6et+63.43
    So, 1/R is the real part and C the imaginary part of Y.
  6. Apr 3, 2005 #5
    Thanks, that worked! I will have to look a little ahead to understand it better.

    Umm, I have a related question though, different problem, didn't want to start another thread:
    if I have input excitation Vs(t) = 100sin(10t) and I see all examples in the book with cosines, so converting it I get 100cos(10t - 90). Then I want to be sure if this is correct to assume:
    1. Vs(t) = 100sin(10t) = 100cos(10t - 90) = Re[Vs e^j(10t - 90)]
    2. Vc(t) = Vs cos(10t + phita) or is it (10t + phita - 90)?

    And let's say I need IL(t) and not Vc(t) in diff equation for the circuit, is the response form different for current? is Vs just a const that marks amplitude/magnitude?
    Appreciate your help very much.
    Last edited: Apr 3, 2005
  7. Apr 3, 2005 #6


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    Vs is the voltage of your source, VC the voltage across the capacitor and VL the voltage across the inductor.
    In a series circuit all elements have the same current and different voltages. In a parallel circuit all elements have the same voltage and different currents.
    If you have an RC series circuit, the impedance Z = R - jXc, where Xc = 1/(ωC).
    The current across the elements is I = V/Z and the voltage across the capacitor is Vc = IXc.
  8. Apr 3, 2005 #7
    Well, the circuit is RLC series, so I have 2-order diff equation and we have to use IL as a variable in the equation.
    The input is Vs(t) = 100 sin(10t) with LRC given, we need to find IL(t) and then Vc(t), so in this case if IL is the variable, does my diff equation end like this:

    ... + (R/L)deriv[IL(t)] + IL/(LC) = deriv(Vs)/L?

    but then it is in sinus form, so I converted it in cos(10t - 90) and then I do not know how response looks like.

    Thanks again.
    P.S. we did not start phasors yet.

    <edit> I think I got it </edit>
    Last edited: Apr 4, 2005
  9. Apr 5, 2005 #8


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    You have a 2nd order non-homogeneous differential equation. It's solution is the sum of the solution to the homogeneous equation and a particular solution.
    You should not derive the voltage of the source. If you have a series circuit you should use voltages:
    vL + vR + vC = Vs
    But vR = R.i and vL = Ldi/dt and i = CdvC/dt.
    If you replace the values in your equation you will have a 2nd order non-homogeneous differential equation in vC.
    The solution of the homogeneous equation depends on the relative values of R, L and C.
    The particular solution is of the form Acos(10t+φ)
    If you replace this solution and it's two first derivatives in your equation you can solve for A and φ.
  10. Apr 5, 2005 #9
    I think you made a typo that can be misleading...should be Z = R + jX, where X = -1/(w*C) or w*L (w is omega = 2*pi*f).
    Last edited: Apr 5, 2005
  11. Apr 5, 2005 #10
    I should have read all the posts...you caught that one.
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