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?

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?

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

NOTE: it is 1000V.. not 100V

...

C = 160 x 10^-6 F

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