What is the estimated capacitance of the capacitor in a photographic flash unit?

[songoku]

Homework Statement

A capacitor, previously charged by a 1000 V source, is discharged through a photographic flash unit. The capacitor delivered an average power of 1000 W and the flash lasts 0.040 s. What is the estimated capacitance of the capacitor

a. 40 x 10^-6 F
b. 20 x 10^-6 F
c. 60 x 10^-6 F
d. 80 x 10^-6 F
e. 80 x 10^-3 F

Homework Equations

Q = CV
W = 1/2 QV
P = VI
Q = I t

The Attempt at a Solution

I got two different answers...

1.
P = W/t = CV² / 2t

C = 2 Pt / V² = 80 x 10^-6 F (d)


2.
P = VI = VQ/t = CV² / t

C = 40 x 10^-6 F (a)

Something's definitely wrong with my concept but I don't know what...

Thanks

 

[Andrew Mason]

First, find the energy stored in the capacitor. (hint: an average power of 1000 J/sec for .04 seconds provides a total energy of ____ J.).

Then use your formula for stored energy in the capacitor (E = CV^2/2) to determine the capacitance.

Your first answer was correct. In your second attempt, you were assuming that the voltage is a constant 1000 V as the capacitor discharges. It decreases as charge decreases.

AM

 

[songoku]

Hi AM

Thanks a lot !