Physics (simple electronics)

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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 775 µF capacitor is 330 V.

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

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

So far:
0.000775 Farads= Coulumbs / 330V
Coulumbs=0.25575C

Now here I'm not sure what to do next.

But I push on:
1A=1C/s

So:
0.25575C/5x10^-3s=51.15A

P=IV

P=51.15A*330V=16879.5W

But, that answer is kicked backed as wrong, and I'm not quite sure how to go about part A.

  • Physics (simple electronics) -

    (a) Energy stored = (1/2)CV^2

    (b) Power = Energy/(Flash duration)

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