Electronic flash units for cameras contain a capacitor for storing the en- ergy

Question

Electronic flash units for cameras contain a capacitor for storing the en-
ergy used to produce the flash. In one such unit, the flash lasts for 1/
675 s
with an average light power output of 2.7×105W.
(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 125 V when the stored energy equals the value calculated in
part (a). What is the capacitance?

Explanation / Answer

A) The power output is 2.7*10^5W and power is work (or energy dissipated) over time: P = E/t 2.7*10^5W = E/(1/675 s) E = 400J That is the light output of the flash which is 95% of the energy stored in the capacitor. The total energy must then be: Ec = 400J/95% = 421J Energy stored in a capacitor equals Ec = (0.5)CV^2 So, 421J = (0.5)C(125V)^2 C = 0.05389F

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