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Power in the components of an RC circuit as a Function of time

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    A simple RC circuit with an EMF source ε, Resistor R and capacitor C are in series. When the switch is closed find the rate of energy stored in the capacitor, the power dissipated in the resistor, and the power supplied by the emf as functions of time.

    2. Relevant equations
    For the resistor

    P=i2R

    i=(ε/R)e-t/RC


    For the battery

    P=iε

    i=(ε/R)e-t/RC

    For the capacitor

    q=CE(1-e-t/RC)

    U=q2/(2C)




    3. The attempt at a solution

    I feel decent about the expressions I got for the resistor and the battery.

    Pres=[(ε/R)e-t/RC]2R

    Pbat=[(ε/R)e-t/RC


    The problem I'm having is getting the energy rate in the capacitor. I tried to just use the energy equation from above and divide by time but I don't think it's right, It's undefined at t=0, and I need to evaluate these equations at that point in later steps.

    I think I may need to plug it into the energy and then take the time derivative. Its pretty messy though, is there a better way?

    [CE(1-e-t/RC)]2/(2Ct)

    Thanks for any input
     

    Attached Files:

    Last edited: May 12, 2013
  2. jcsd
  3. May 12, 2013 #2

    SammyS

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    There are several ways to find the rate of energy stored in the capacitor.

    You can use conservation of energy. You know the power supplied by the battery (rate of work done by the battery). You know the power dissipated in the resistor. From those it's fairly direct to account for the rest of the energy.

    Alternatively, you can find the voltage drop across the resistor as a function of time. From that you can find the voltage drop across the capacitor as a function of time. Use that with U = (1/2)V2C.

    You can integrate the current to find the charge on the capacitor.

    ...
     
  4. May 12, 2013 #3
    I see, because what's not dissipated by the resistor is stored in the capacitor. I think I will take that route.

    So its just

    Pemf-Presistor=Pcapacitor

    My other equations look alright I'm assuming?


    Thanks a bunch.
     
  5. May 12, 2013 #4

    SammyS

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    Yes.

    They look right.
     
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