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Energy dissipated by a resistor in an RC circuit

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a series circuit with a capacitor of capacitance C, a resistor of resistance R, a DC voltage source with a voltage V, and an open switch (by open I mean not connected so no current flows through). At t=0 the switch is closed. How much energy is dissapated by the resistor at t= infinity


    2. Relevant equations
    C=Q/V
    PR=iV
    V=iR
    i=dq/dt


    3. The attempt at a solution
    First off my conceptual understanding of the problem is such: Initially, the voltage source provides a voltage that charges up the capacitor. Eventually, the capacitor charges all of the way up and the current goes to 0. So what I need to do is integrate the power that the resistor dissipated from t=0 to t=infinity
    Using PR=iV and subbing in i=dq/dt.
    The voltage can factor our of the integral since it is constant, and the dt's cancel
    I end up with V times the integral of dq which equals VQ
    Since Q=CV the answer I have is CV2

    Now I have no idea if this is correct since it is a test review question so I was wondering if this looks correct to all of you. If not, as always, any input would be greatly appreciated.
    Thanks!
     
    Last edited: Apr 23, 2009
  2. jcsd
  3. Apr 23, 2009 #2

    berkeman

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    Staff: Mentor

    You could do it a second way to check your answer. Solve for v(t) of the capacitor, and then write the expression for the voltage drop across the resistor as a function of time. Then integrate that voltage difference across the resistor from 0 to infinity. Do you get the same answer?
     
  4. Apr 23, 2009 #3

    Cyosis

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    Homework Helper

    The voltage across the resistor is not constant! V(resistor)=iR. This gives P=i^2 R.
     
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