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 a time interval of 1.50×10-3 s with an average light power output of 2.80×105 W
a) If the conversion of electrical energy to light has an efficiency of 90.0 % (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 125 V when the stored energy equals the value calculated in part A. What is the capacitance?

Explanation / Answer

a. .90 x Energy in = Energy out .90(electrical energy) = light energy .90(electrical energy) = Power x time .90(electrical energy) =2.80 x 10^5(1.50 x 10^-3) Electrical energy = 466.67 Joules b.E = .5CV^2 466.67 = .5(125^2)C C= .05973 Farads Hope this helped! Rate lifesaver :)

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