Electronic flash unit cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for 1/675 s with an average light power output of 2.7❝105 W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy) how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125V when the stored energy equals the value calculated in part (a). What is the capacitance?

(A) 421 J energy stored in the capacitor for one flash.

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time

Average power W

From power equation,

So energy in one light is given by,

  J

Since efficiency is 95 % so we can write, energy stored in one flash,

J

(B)

From the formula of energy stored in capacitor,

Where and V

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west