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

AI Thread Summary
The discussion revolves around calculating the estimated capacitance of a capacitor used in a photographic flash unit, which is discharged at an average power of 1000 W over 0.040 seconds. Two methods were attempted to find the capacitance, yielding different results: 80 x 10-6 F and 40 x 10-6 F. The correct approach involves calculating the total energy output from the flash and using the formula for stored energy in a capacitor, which reveals that the first answer (80 x 10-6 F) is accurate. A misunderstanding arose from assuming the voltage remained constant during discharge, which is incorrect as it decreases with charge. The discussion concludes with a clarification on the correct method for determining capacitance.
songoku
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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
 
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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
 
Hi AM

Thanks a lot ! :smile:
 
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