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RC Circuit, Charging, Question

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the circuit below, where C1=80mF, C2=30mF, R=50ohms and E=5V.
    At t=0, both capacitors are uncharged - so it's charging.

    The circuit is in series, with E ---> C1 ---> C2 ---> R ---> (back to) E

    a) Find the time constant.
    b) The initial potential difference across R.
    c) The current at t=3s.
    d) The potential difference across C2, at t=3s.

    2. Relevant equations

    C=Q/V
    V=IR
    X=Xmax(e^(-t/RC)) (where X is I, V or Q and is decreasing with time)
    X=Xmax(1 - e(-t/RC)) (where X is I, V or Q and is increasing with time)

    3. The attempt at a solution

    Here are my answers, can someone check if I'm right?
    a)
    time constant = RC = (50 ohms)(21.8 mF) = 1.09 s
    b)
    V=5.0 ohms (same as E)
    c)
    V=RI -> 5V=(50 ohms)I -> I=0.1 A

    I=Imax(e-t/RC)
    I=(0.1 A)[/SUB](e(-3s/1.09s))
    I=6.37x10-3 A

    d)
    VCeq= Vmax(1-e-t/RC)
    VCeq= (5 V)(1-e-3s/1.09s)
    VCeq= 4.68 V

    V=C/Q -> 4.68V=30mF/Q -> Q=6.4x10-3 C
     
  2. jcsd
  3. Oct 30, 2012 #2

    lewando

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    τ(seconds) = resistance(ohms) * capacitance(Farads). Your result is a bit too big.
     
  4. Oct 30, 2012 #3

    lewando

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    Nevermind,τ looks good.
     
  5. Oct 30, 2012 #4
    So are the other answers good to?
     
  6. Oct 30, 2012 #5

    lewando

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    b) I know you meant V not ohms
    c) good
    d) looking for a voltage across C2, VCeq is right though.
     
  7. Oct 30, 2012 #6
    Thanks. And OK, you're right, I fixed it.

    V=Q/Ceq
    4.68=Q/21.8mF
    Q=1.02 C, which is the same across each capacitor since they're in series.

    And can you please, since you seem to know what you're talking about when it comes to RC circuits, answer this question of mine as well?

    https://www.physicsforums.com/showthread.php?t=648367
     
  8. Oct 30, 2012 #7

    lewando

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    for d) you need to determine the voltage across C2. It is done the same way as if you were going a resistor divider.

    On your other question, gneill is on it (and will advise you well).
     
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