Discharging a Capacitor: Calculating Energy Dissipation

[apamirRogue]

Homework Statement

A 0.25μF capacitor is charged to 50V. It is then connected in series with a 25Ω resistor and a 100Ω resistor and allowed to discharge completely. What is the energy dissipated by the 25Ω resistor in Joules?

Homework Equations

I'm assuming I need:
P=V^2/R
I(t)=I₀*e^-(t/τ)
τ=RC

The Attempt at a Solution

So I used P=V^2/R to get that the power output of the 25Ω resistor is 100W. Then all I need is the time to find out how much energy was released. I know I cannot solve I(t) for when it is 0 without getting ∞. So I calculated the time it would take for the capacitor to be at an I=.01(I₀), got a t=1.43*10^-4 secs. Then multiplying by the power, I got an energy release of .0144 J which is not correct. I'm not sure how else I would do this.

 

[gneill]

Homework Statement

A 0.25μF capacitor is charged to 50V. It is then connected in series with a 25Ω resistor and a 100Ω resistor and allowed to discharge completely. What is the energy dissipated by the 25Ω resistor in Joules?

Homework Equations

I'm assuming I need:
P=V^2/R
I(t)=I₀*e^-(t/τ)
τ=RC

The Attempt at a Solution

So I used P=V^2/R to get that the power output of the 25Ω resistor is 100W. Then all I need is the time to find out how much energy was released. I know I cannot solve I(t) for when it is 0 without getting ∞. So I calculated the time it would take for the capacitor to be at an I=.01(I₀), got a t=1.43*10^-4 secs. Then multiplying by the power, I got an energy release of .0144 J which is not correct. I'm not sure how else I would do this.

Surely the power dissipated by the resistors changes over time, since the current and hence the voltage drops change over time. So "100W" for the 25Ω resistor doesn't seem to be meaningful.

You might also want to take note of the fact that while P = V²/R, it is also P = I²R.

e⁰ = 1, so there's no problem solving for I when t = 0.

You should think about integrating the instantaneous power over time to find the total energy.