An electronic flash unit for a camera contains a capacitor with a capacitance of
ID: 1459648 • Letter: A
Question
An electronic flash unit for a camera contains a capacitor with a capacitance of 850 µF. When the unit is fully charged and ready for operation, the potential difference between the capacitor plates is 330 V.
a) What is the magnitude of the charge on each fully charged capacitor?
b) Find the energy stored in the charged up flash unit
c) the flash unit itself has a resistance of .50 ohms, what is the time constant for this circuit?
d) How long does it take for voltage across the capacitor to drop to 1/5 of its original voltage?
Please show work, thank you.
Explanation / Answer
part A : Full charge Q = CV
Q = 850 e -6 * 330 = 0.280 C
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ENErgy U = 0.5CV^2
U = 0.5 * 850 e -6 * 330 * 330
U = 46.28 J
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time constant T = RC
T = 0.5 * 850 e -6
T = 4.25 e -4 secs
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use the eqn V = Vo (1- e^-t/RC)
1-e^-t/RC = 1/5 = 0.2
e^-t/RC = 1-0.2 = 0.8
-t/RC = ln (0.8)
t = 0.223 * 4.25 e -4
t = 94.77 us
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