K

kmatthews

a capacitor of 100uf is connected in parallel with a 20ohm pure resistor. the group is connected to a 200v 50hz supply calculate the power dissipated in the combination. the answer is 2000w but i dont know the workings out anyone please?

is it as easy as p=v2/r im sure there is more than that.
 
As they are in parallel you can effectively ignore the capacitor, so you are left with:

V(volts) = I(amps) x R(ohms) therefore I = V / R so I = 200 / 20 = 10amps

and

P(watts) = I(amps) x V(volts) therefore P = 10 * 200 = 2000watts

Hope that helps.
 
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Hey i answered this one late last night with the same answer on your gola question thread when i was the worse for wear from beer and still got no thanks! :rolleyes:
 
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