Electronic Flash Unit
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Energy Stored in a Capacitor?
Electronic flash attachments for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for 1/100 s with an average light power output of 600 W.
a) If the conversion of electrical energy to light energy is 95.0% efficient
(the rest goes to thermal energy), how much energy must be stored in the capacitor for one flash?
b) If the capacitance of the capacitor in the unit is 0.832 mF, what is the potential
difference between its plates when the capacitor has stored in it the amount of energy calculated in part (a)?
Power = energy/time
P = E/t
E = Pt = 600W)(0.01seconds) = 6 Joules
Since you lose 5% of the total energy and it takes 6 J to do the flash, then the total energy would be:
6J/0.95 = 6.32J
So the first answer is 6.32J
Now for the 2nd part:
E capacitor = ½CV² Where C = capacitance in Farads and V = voltage in volts
we know E and C so we need to solve for V
2E = CV²
V² = 2E/C
V = √2E/C
V = √ 2(6.32)/(0.832E-6)
V = 3897.73 volts
a very good example of why you don't touch capacitors. You don't know they may be charged as such!
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