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Energy dissipated in a resistor during a time interval?

  1. Nov 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG

    2. Relevant equations
    P = I*V energy dissipated = wr = ∫ ( P *dt) .... t is from 0 to .4 seconds vs = 400t^2 = 400 * (.4^2) = 64 V

    3. The attempt at a solution
    Using KVL I said Vs ( 64 V) = 100 * i .... I found that I was 64/100 = .64 amps. I then said power is .64 amps ^ 2 * 100 ohms = 40.96 Watts. I then integrated 40.96 Watts with respects to time ... Energy dissipated = wr = ∫ ( 40.96 Watts *dt ) from 0 to .4 seconds = 40.96 * .4 = 16.384 J.


    Where did I go wrong?
     
    Last edited by a moderator: Nov 9, 2017
  2. jcsd
  3. Nov 9, 2017 #2

    BvU

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    The 40.96 is not the power at ##t<0.4## s. You calculate as if the voltage was a constant 64 V from ## t= 0 ## to ##t=0.4 ## s.
     
  4. Nov 9, 2017 #3
    So then if the voltage is changing I could integrate with respect to time: Vs = ∫ 400t^2 dt from 0 to .4 s which is equal to 8.533 V then divide by 100 ohms to find the current .08533 amps. Then integrate power with respect to time: Wr= ∫ ( .08533^2 * 100) dt from 0 to .4 sec ?
     
  5. Nov 9, 2017 #4

    BvU

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    Aren't you doing something similar again ? Why integrate twice ? Use your own relevant equation ##\displaystyle \int_0^{0.4 {\rm s}}\; P(t) \;dt ## where you substitute an appropriate expression for ##P(t)## :cool:
     
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