Energy released, capacitor power output

AI Thread Summary
The discussion centers on calculating energy stored in a capacitor and its power output. For part (a), the energy stored in a 1.25 x 10^-4 F capacitor at 220 V is calculated using the formula PE = 1/2 CV^2, yielding 3.025 J. For part (b), to find the power output when 1.25 J is released in 0.25 seconds, the formula P = E/t is applied, resulting in an average power output. The user expresses uncertainty in setting up part (b) but acknowledges the need to calculate average power over the given time. The conversation emphasizes understanding energy storage and power calculations in electrical devices.
KMc19
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Homework Statement



a) An electrical device stores energy in a 1.25 x 10^-4 F capacitor at 220 V. How much energy can be stored in this device?

b) If 1.25 J were released from capacitor in 0.25 second, what would be its power output?

I believe I solved (a), but I cannot find out how to go about setting up and solving (b). Please help! :)


Homework Equations



PE=1/2 CV^2


The Attempt at a Solution



a) = 1/2( 1.25 x 10^-4)( 220 V) = 3.025 J


b) I'm not sure about howto set this one up, that's where I'm having problems.
 
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Power is the rate at which energy is moved or work done.
P = E/t.
You can only do this as an average over the whole 0.25 s, for which the energy is given as 1.25 J.
 


Thanks Delphi!
 
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