Find Capacitance for Camera Flash Unit

[SimonZ]

Homework Statement

The flash unit in a camera uses a special circuit to “step up” the V = 3.1 V from the batteries to V' = 300 V, which charges a capacitor The capacitor is then discharged through a flashlamp. The discharge takes t = 12 μs, and the average power dissipated in the flashlamp is P = 12 W. What is the capacitance of the capacitor?

Homework Equations

Energy = power * time
energy stored in a capacitor = CV²/2

The Attempt at a Solution

Energy dissipated by the flashlamp = Pt
This energy comes from the decreasing energy stored in the capacitor = CV’²/2 – CV²/2
so Pt = CV’²/2 – CV²/2
then
C = 2Pt/(V’² – V²) = 3.2*10^-9 F

Anything wrong?

 

[turin]

Your basic reasoning appears to be sound.

 

[SimonZ]

but the website says ir is incorrect!
I don't know why.

 

[ideasrule]

What's "ir"? I thought you were finding the capacitance.

Anyways, if the capacitor completely discharges (which it does), then in your equation:

CV’²/2 – CV²/2

V would be zero, not the battery's voltage.

 

[SimonZ]

What's "ir"? I thought you were finding the capacitance.

Anyways, if the capacitor completely discharges (which it does), then in your equation:

CV’²/2 – CV²/2

V would be zero, not the battery's voltage.

The question is on www.masteringphysics.com
You're right, V = 0, but it won't affect the numerical result (only 2 significant figures needed).