A photographic flash unit consists of a capacitor previously charged by a 100V source. The average power delivered to the flash tube is 2000W and the flash lasts 0.040s. The capacitance of the capacitor is:
A 40 x 10^-6 F
B 80 x 10^-6 F
C 160 x 10^-6 F
D 80 x 10^-3 F
E 160 x 10^-3 F

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A photographic flash unit consists of a capacitor previously discharged by a 100V source. The average power delivered to the flash tube is 2000W and the flash lasts 0.040s. The capacitance of the capacitor is?

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Originally Posted by anishsaro

A photographic flash unit consists of a capacitor previously discharged by a 100V source. The average power delivered to the flash tube is 2000W and the flash lasts 0.040s. The capacitance of the capacitor is?

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So...what's your answer?

I'm sure your text has an equation for energy stored in a capacitor as a function of its capacitance C and voltage V. Hint: Remember that power times time is equal to energy.

NOTE: it is 1000V.. not 100V

Actually, I found the way out! I only didn't think of the equation for energy... Energy = 0.5*C*(V)^2

i.e; using W = Pt to find work done in joules
then equating it to Energy = 0.5*C*(V)^2 as work done = energy
therefore, C = Pt/(0.5*(V)^2)

C = 160 x 10^-6 F

Quote:

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Originally Posted by anishsaro

NOTE: it is 1000V.. not 100V

...

C = 160 x 10^-6 F

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Good! I was a bit worried because none of the answers worked for the data you originally provided.