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Energy dissipated by a resistor in an RC circuit

  • Thread starter ehilge
  • Start date
  • #1
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Homework Statement


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


Homework Equations


C=Q/V
PR=iV
V=iR
i=dq/dt


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:

Answers and Replies

  • #2
berkeman
Mentor
56,448
6,366
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?
 
  • #3
Cyosis
Homework Helper
1,495
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The voltage across the resistor is not constant! V(resistor)=iR. This gives P=i^2 R.
 

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