I have a test tomorrow and I am wondering how to do a practice problem I was assigned.
Any help?
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 770 µF capacitor is 325 V.
(a) Determine the energy that is used to produce the flash in this unit(Joule).
(b) Assuming that the flash lasts for 5.0 10-3 s, find the effective power or "wattage" of the flash(Watt).
2007-03-18 02:21:54 · 1 answers · asked by neverconsiderlifeonascale 1 in Science & Mathematics ➔ Physics
Energy = (1/2) Q^2 / C where Q is charge stored
Q = CV
Energy = (1/2) C V^2 = 0.5 * 770 * 10^-6 * (325*325)
= 40.66 joule
b) Power = energy /time = 40.66 /5*10^-3
= 8133.125 watt or joule/sec
2007-03-18 02:45:42 · answer #1 · answered by anil bakshi 7 · 0⤊ 0⤋