maths question

**kmatthews** · 16-06-2010

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.

**topquark** · 16-06-2010

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.

**kmatthews** · 16-06-2010

looks good to me. cheers

can you take a look at the other question i posted

thanks

**pushrod** · 16-06-2010

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!

**lee sandland** · 16-06-2010

Originally Posted by pushrod

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!

Ohm's law lol V=IXR
I=V/R
R=V/I
BUT DID HE THANK YOU