Question

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?

Answer 1

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

(B) The value of capacitance is 0.0537 F

Step-by-step explanation:

Given :

(A)

Time
`t = (1)/(675)`

Average power
`P = 2.7 * 10^(5)` W

From power equation,

`P= (E)/(t)`

So energy in one light is given by,

`E = Pt`

`E = 2.7 * 10^(5) * (1)/(675) = 400` J

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

`E_(tot) = (400)/(0.95) = 421` J

(B)

From the formula of energy stored in capacitor,

`E = (1)/(2)C V^(2)`

Where
`E = E_(tot)` and
`V = 125` V

`C = (2E)/(V^(2) )`

`C = (2 * 421)/(15625)`

`C = 0.0537 F`