Posted by Mike on .
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.
0.000775 Farads= Coulumbs / 330V
Now here I'm not sure what to do next.
But I push on:
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)