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Homework Help: Archived Maximum Charge on a Capacitor, Help

  1. Apr 20, 2008 #1
    Hi, thanks in advance to all who help

    a) Determine the time constant for charging the capacitor in the circuit given.
    b) what is the maximum charge on the capacitor?
    | | |
    | < |
    E R2 > C
    | | |

    E = emf and C is the capacitor, sorry for the poor illustration.

    Mentor's edit: Here's a better illustration:

    For a, I've determined the equivalent resistance of parallel resistors to be (1/R1 + 1/R2)^-1 = R1R2/(R1+R2), thus the time constant would be tau = CReq = CR1R2/(R1+R2)

    For b) I have a feeling Qmax for the capacitor will be Qmax = CV where V is the voltage across R2 since R2 and C are parallel, but I can't figure out V across R2 in terms of E, R1, R2, C. Can anyone help? Thanks again.
    Last edited by a moderator: Feb 8, 2016
  2. jcsd
  3. Feb 10, 2016 #2


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    I got the same result after solving with differential equations.
    Let the time constant be T.
    Current supplied by the source is,
    From this equation, it can be verified that at t=0, I=E/R1(capacitor acts as a short at t=0) and at t=∞, I=E/(R1+R2)..(capacitor acts as open circuit at t=∞).
    In steady state, the capacitor will act as an open circuit. Hence, current will flow though the series combination of R1 and R2.
    So, Voltage across C=Voltage across R2 in steady state= E*R2/(R1+R2)
    Final charge on the capacitor=Q=CV= ECR2/(R1+R2).
  4. Feb 11, 2016 #3


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    The above approach involves some mathematical calculations and a generalized expression for current I(t). Instead, I believe the best and simplest approach to solve these two questions a and b is Thevenin's theorem.
  5. Feb 13, 2016 #4
  6. Feb 13, 2016 #5
    Why do you regard R1 and R2 as being in parallel?
    When C is fully charged then the current through the capacitor is ?

    Which makes the fial voltage across C ? ?

    Hence the max charge will be ? ? ?
  7. Feb 14, 2016 #6


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    Are those your own queries or are you trying to give hints to the OP?
  8. Feb 15, 2016 #7
    Vc=VR2 where R1 & R2 are voltage divider.
    I = E/(R1+R2) and VR2=IR2, so Vc =E[R2/(R1+R2)]
  9. Feb 16, 2016 #8
    Since C=Q/Vc, Q =C Vc, so the charge Q depends on the value of capacitance you put on your circuit and magnitude of Vc
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