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

Answer:

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

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time $t = \frac{1}{675}$

Average power $P = 2.7 \times 10^{5}$ W

From power equation,

   $P= \frac{E}{t}$

So energy in one light is given by,

   $E = Pt$

  $E = 2.7 \times 10^{5} \times \frac{1}{675} = 400$ J

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

$E_{tot} = \frac{400}{0.95} = 421$ J

(B)

From the formula of energy stored in capacitor,

 $E = \frac{1}{2}C V^{2}$

Where $E = E_{tot}$ and $V = 125$ V

 $C = \frac{2E}{V^{2} }$

 $C = \frac{2 \times 421}{15625}$

 $C = 0.0537 F$