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Homework Help: RC circuit with AC - finding current

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I am given an RC circuit with an alternating current. The circuit contains a capacitor and a resistor in parallel. Part (a) says "Use KCL to find a differential equation for I in terms of V." Part (b) says "For an applied voltage V = Voexp(iwt), find the current I."



    2. Relevant equations

    ITotal = I1 + I2
    Z = impedance
    ZR = R
    ZC = (1/wiC)
    V = IZ


    3. The attempt at a solution

    I understand how KCL works, but I'm not sure how to get it in a differential equation. Ignoring this and using KCL anyway, I get:

    V - I1ZR = V - I1R = 0
    V - I2ZC = V - I2(1/iwC) = 0
    I1 = ViwC
    I2 = (V/R)
    ITotal = ViwC + (V/R)

    Am I on the right track, or does the fact that I need a differential equation mean I need to do something completely different?
     
    Last edited: Sep 22, 2010
  2. jcsd
  3. Sep 22, 2010 #2

    Delphi51

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    Looks like you are supposed to use the differential equation method (which is more general than the impedance method you started).
    The current through a capacitor is related to the voltage across it by i=C*dV/dt. Apply KCL to the node where the current splits, part through R and part through C. You should have only the combined i, the common V and of course R and C in your equation. And a derivative.
     
  4. Sep 22, 2010 #3
    I've gotten as far as:

    i = C*dV/dt + V/R

    Is it mathematically correct to say that V is common to both terms on the right, and therefore:

    i = V(C*d/dt + 1/R) ?

    If so, I would plug in the V value given in step (b):

    i = Voeiwt(C*d/dt + 1/R)

    How would I go about integrating this (or should I have integrated before plugging in V)?
     
  5. Sep 22, 2010 #4

    Delphi51

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    The first formula looks great, but you can't factor the V out of the derivative. For the (b) part you'll have to find the derivative of the V function. After that, likely you'll be able to take out a common factor because the derivative of an exponential function is the same function multiplied by some constants.
     
  6. Sep 23, 2010 #5
    Is the i in the voltage function V = Voeiwt the current I want? If so, I've been spending MUCH longer on this than I needed to... haha

    In that case, it'd just be:

    V = Voeiwt

    dV/di = Vo(iwt)eiwt
     
  7. Sep 23, 2010 #6

    Delphi51

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    dV/dt = iwVo*e^(iwt)
     
  8. Sep 24, 2010 #7
    RIGHT that's what I meant.

    Is that my answer (solving for i) or do I set that equal to I/C and solve for I?

    I guess I'm getting hung up on the fact that there's an instantaneous current (i) and an average/overall current (I), and I don't know which to solve for.
     
  9. Sep 24, 2010 #8

    Delphi51

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    You can't solve it for i.
    The answer is your i = C*dV/dt + V/R with that expression for dV/dt subbed in. Oh, and replace V with its given value. Maybe you can simplify it. Factoring out the exponential would be nice.

    The average current will always be zero with AC.
     
  10. Sep 24, 2010 #9
    Ah, that makes sense! Thank you!

    Just to double check, I got for an answer:

    i = Voeiwt (iwC + 1/R)

    Does it matter that there's also an i in the exponent?
     
    Last edited: Sep 24, 2010
  11. Sep 24, 2010 #10

    Delphi51

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    Looks good!
     
  12. Sep 25, 2010 #11
    Excellent! Thanks for your help.
     
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