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Homework Help: Voltage through a steady state capacitor.

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    According to my notes when a circuit is in DC steady state inductors are short circuited and capacitors are open circuited. This leads me to believe that the voltage across the capacitor is basically the voltage looking in through the terminal of the open circuit:

    -----------R------o +

    ------------------o _

    if it's an open circuit then no current flows through R.

    (7sin12t - Vc)/R = 0

    Vc = 7sin12t

    This doesn't seem right to me, can anyone tell me where i'm going wrong?

    thanks in advance.
    Last edited: Nov 19, 2011
  2. jcsd
  3. Nov 19, 2011 #2


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    Staff: Mentor

    Your circuit isn't in DC steady state. It's being driven by a sinusoidal source, so it's in AC steady state. The inductor and capacitor will have reactances (impedances) at the given driving frequency. Find expressions for them and then do the voltage divider thing (or find the current first, then the voltage across the cap.).
  4. Nov 19, 2011 #3
    This is an AC circuit with a supply voltage given by V = 7Sin12t.
    This means that ω = 12.
    I don't know what 'steady state' means in this context and would assume it means find the voltage (AC) across the capacitor.
    I attempted to find the reactance of L and C and got the following:
    Xl = ωL = 12 x (2 +β) = (24 + 12β)Ω
    Xc =1/ωC = 1/(12 x 3x 10^-3) = 28Ω
    The resistance = (12-ε)Ω
    The next logical thing is to get an expression for the impedance, Z,
    using Z^2 = [(Xl-Xc)]^2 + R^2
    Then use this to get an expression for the current and therefore an expression for Vc
    Sorry !!!!! Hope you get somewhere with it
  5. Nov 19, 2011 #4

    The phasor voltage of 7sin12t is 7

    The impedance of the resistor is the resistance

    the impedance of the cap is 1/j(12)(3mF)

    impedance of the inductor is j(12)(L)
  6. Nov 19, 2011 #5


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    Staff: Mentor

    What happened to the ε and β?
  7. Nov 19, 2011 #6
    i'm assuming they're 0 for now, they are just variables with given values, sorry i didn't clarify that.
  8. Nov 19, 2011 #7

    thanks, sorry i didnt mention that epsilon and beta are variables, my bad.

    i have basically the same work as you, i used voltage division to get the voltage through the capacitor and then reverted back from phasor to time domain form.

    thanks for all the help
  9. Nov 19, 2011 #8
    It is much easier now.
    I got Vc =15.5V, Vl =13.2V and Vr = 6.6V
  10. Nov 19, 2011 #9
    should the answer not be in the form xsin(wt + y)?

    for the voltage across the capacitor i got

    vc(t) = 9.2186sin(12t + 0.288)

    does this look correct?

    edit: i used 6 for epsilon and 2 for beta, but the answer should be roughly there or thereabouts i hope.
  11. Nov 19, 2011 #10
    I thought ε and β were being taken to = 0 ?
    You are right, my voltages should be given as
    Vc = 15.5 Sin(12t)... Strictly should be 15.5Sin(12t - ∏/2) because Vc is ∏2 behind I
    Vl = 13.2 Sin(12t).... strictly speaking should be 13.2 Sin(12t + ∏/2) because Vl is ∏/2 ahead of I
    Vr = 6.6 Sin(12t)...this is OK as it is because Vr is inphase with I

    A good check is that Vs^2 = Vr^2 + [Vl - Vc]^2

    ie 7^2 = 6.6^2 + [13.2 - 15.5]^2
    49 = 43.6 + 5.3 = 48.9 (near enough)
  12. Nov 19, 2011 #11

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