Average power when capacitor discharges

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SUMMARY

The discussion centers on calculating the average power output of a flashbulb in a circuit consisting of a 9 V battery, a 50 kΩ resistor, and a 140 μF capacitor. The capacitor discharges over a time span of 3.3 μs. The average power is determined using the formula U = (1/2)CV² to find energy and P = V/T for power, resulting in an average power output of 1718.1818 W during the discharge period.

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Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1202/summer/homework/Ch-21-DC-Circuits/flashbulb_circuit/light-bulb.jpg

The above shows a simplified circuit for a photographer's flash unit. The circuit consists of a V = 9 V battery, connected to a R = 50 kΩ resistor, a C = 140 μF capacitor and two switches. Initially, the capacitor is uncharged and the two switches are open. To charge the unit, switch S1 is closed. To fire the flash, S2 (which is connected to the camera's shutter) is also closed.

Assume that capacitor discharges itself in 3.3 μsec. What average power put out by the flashbulb during this time? 2. The attempt at a solution
I have tried different power equation but i need to include the time so I am doing it completely wrong
 
Last edited by a moderator:
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nvm i figured it out
U=(1/2)CV^2
P=V/T
answer was 1718.1818
 

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