Electronic flash units for cameras contain a capacitor for storing the energy us

Question

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for 1/650 of a second with an average light power output of 2.80 x 105 W.

A.If the conversion of electrical energy to light is 88% efficient, how much energy must be stored in the capacitor for one flash?
(include units with answer)
Hint #1:Find the input power first, then energy.
B.The capacitor has a potential difference between its plates of 135 V when the stored energy equals the value calculated in part A. What is the capacitance?
(include units with answer)

Explanation / Answer

a) Total light energy Ulight = Plight * delta T

electrical energy to light is 88% efficient, 0.88 Uc = Ulight

Uc = Ulight / 0.88 = (Plight * delta T) / 0.88

= (2.80 * 105 * (1/650) ) / 0.88

energy stored in the capacitor for one flash = 489.5 J

b) Uc = 1/2 C V2

C = 2 * Uc / V2 = (2 * 489.5) / (135)2

capacitance = 0.0537 F = 53.7 mF

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