Hi all, functioning on the last trouble in a homework collection and it's obtained me a bit stumped. Not certain if there's a theoretical problem or if I'm simply integrating incorrectly.

You are watching: How much energy is dissipated by the 25ω resistor?


Here's the problem: A 0.25μF capacitor is fee to 60 V . The is then associated in series with a 45Ω resistor and also a 140 Ω resistor and allowed to discharge completely. Just how much power is dissipated by the 45Ω resistor?


Here's my effort at a solution:

P = I2R

dU/dt = I2R

dU = I2R dt

ΔU = ∫I2R dt

ΔU = ∫(Ie-t/RC)2R dt

ΔU = I2R∫e-2t/RC dt

ΔU = I2R<-RC/2*e-2t/RC> from t = 0 to infinity

ΔU = I2R2C/2


From below I simply plugged in the values and also got 26.6 μJ, however it's no correct. Any type of help?

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level 1
· 5y
I don't check out where you obtained an ns from, to plug in.

Power = VI, right? So energy is the time integral of that: E = QV (charge times volts).

That is, if you have a capacitor already charged come V volts, and you shot to roll one more bit dQ of charge up the potential slope, to include to the charge on the capacitor, you need V dQ joules.

This is really like the amount of potential energy stored in a spring: the pressure required because that a little bit more stretch is proportional to the stretch currently there. So the formula because that the power stored in the capacitor is proportional come the capacitance and also the square the the voltage across it.

Once you recognize the full energy in the capacitor, you understand that 45/(45+140) of that is dissipated in the 45 ohm resistor, for this reason you deserve to do this there is no actually functioning out the existing at each time.

level 2
Op · 5y

I think I watch what my problem was. In my final answer, I2R2 is the same as V2, so all I finished up doing to be calculating the potential energy across the capacitor 1/2CV2. When I multiply this through 45/(45+140) I gained the potential power dissipated through the 45 ohm resistor specifically and also got the correct answer. Thanks for that, for some factor I was absent the totality proportional component of the trouble lol.

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