# Find the capacitance

1. Jun 16, 2009

### SimonZ

1. The problem statement, all variables and given/known data
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?

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

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 16, 2009

### turin

Your basic reasoning appears to be sound.

3. Jun 17, 2009

### SimonZ

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

4. Jun 17, 2009

### 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.

5. Jun 17, 2009

### SimonZ

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).