Average power when capacitor discharges

AI Thread Summary
The discussion focuses on calculating the average power output of a flashbulb in a circuit involving a 9V battery, a 50 kΩ resistor, and a 140 μF capacitor. The capacitor discharges over a period of 3.3 μsec, and the user initially struggled with the power equations but eventually determined the correct approach. Using the formula for energy stored in a capacitor, U=(1/2)CV^2, and the power formula P=V/T, the calculated average power output was found to be approximately 1718.18 watts. This calculation highlights the importance of understanding capacitor discharge in electrical circuits. The discussion concludes with a successful resolution of the problem.
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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
 
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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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