Electronic flash unit cameras contain a capacitor for storing the energy used to

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 2.7105 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?

Explanation / Answer

a) E used = Estored*e

Estored = P T/e = 2.7E5*(1/675)/0.95= 421 J

b) E = 1/2 C V^2

421 = 0.5*C*125^2

C=0.0539 F

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