- #1
SimonZ
- 25
- 0
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?