# In Diar Need of HELP! PLEASE!!!!!(Current phase)

Discussion in 'Homework Help' started by lsecrease, Sep 4, 2008.

1. ### lsecrease Thread Starter New Member

Sep 4, 2008
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0
Ok, I realize this to be true but how do you prove that a circuit whose current is 90 degrees out of phase with the driving voltage consumes no power, averaged over an entire cycle???

Thanks a MILLION!!!

2. ### Ratch New Member

Mar 20, 2007
1,068
3
lsecrease,

Why, it is childishly simple. Assuming sinusoidal waveforms, we get the average power by taking the average of the instantaneous power over a complete period. The instantaneous power of a "90 degrees out of phase" voltage*current is proportional to sin(x)*cos(x). The integral of sin(x)*cos(x) is (sin^2(x))/2. Pluging in the limits of the period 0 to 2*pi gives 0. Dividing by the period to get the average is still 0. Doing the same with and in-phase voltage*current means using either sin^2(x) or cos^2(x), which gives an answer of 1/2.

Ratch

Last edited: Sep 4, 2008
3. ### Papabravo Expert

Feb 24, 2006
10,179
1,800
Can you do it without using calculus? I know it, but maybe the OP doesn't.

4. ### Ratch New Member

Mar 20, 2007
1,068
3
Papabravo,

Until the OP says otherwise, I will assume he does.

Ratch