What is the flash wattage for a camera flash with a 765 µF capacitor at 345 V?

[StudentofPhysics]

The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 765 µF capacitor is 345 V.

(a) Determine the energy that is used to produce the flash in this unit.

Using the formula:
Energy = .5C(V^2) i found the energy to be 45.527 J

(b) Assuming that the flash lasts for 5.0 10-3 s, find the effective power or "wattage" of the flash.


I can not find a formula to convert the given information into wattage however.

 

[fargoth]

i'd calculate the energy difference on the capacitor (before and after the flash) and divide it by the time of the flash...

 

[neutrino]

The defenition of power is the change in energy per unit time. Can you do the problem now?

 

[StudentofPhysics]

how do i know what the energy is before and after the flash? Would it be that energy before then 0 after?

 

[StudentofPhysics]

ok, i undersstand know. I take the energy I found and simply divide it by the time passed. Thank you.