Answer to Question #132394 in Mechanics | Relativity for Jessica

Question

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 an 710-μF capacitor is 370 V. (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for 6.6 ms, find the effective power or "wattage" of the flash.

Answer 1

solution

given data

potential difference(V)=370V

capacitance(C)=710$\mu F$

(a)

energy can be written as

$E=\frac{CV^2}{2}$

by putting the value

$E=\frac{710\times10^{-6}\times(370)^2}{2}=48.6J$

(b)

power is given by

$P=\frac{E}{t}$

by putting the value

$P=\frac{48.6}{6.6\times10^{-3}}=7.36KW$

therefore power is 7.36KW and energy is 48.6J.