This is not a HW question.(adsbygoogle = window.adsbygoogle || []).push({});

Assume you have a sinusoidally varying voltage, such that current lags by some phase angle phi

Then you can write the power as a function of time as

p(t)=i(t)v(t)=2*Vrms*Irms*cos(omega*t)*cos(omega*t-phi)

You can break this up via trig identity into the following

Eq1: p(t)=2*Vrms*Irms*( [1+cos(2*omega*t)]*cos(phi) + sin(2*omega*t)*sin(phi) )

If you average over 1 period, then you obtain that the average REAL power is

P=Vrms*Irms*cos(phi) This is average real power

The peak real power would be twice this value since (take t=0 in Eq1:)

real power is associated with cos(phi) term.

The average reactive power is zero, it bounces back and forth between inductive and capacitive loads (B and E fields).

But the peak reactive power is defined as Q=Irms*Vrm*sin(phi).

HERE IS THE QUESTion.

I see often that the Apparent power S = P + i*Q. Unfortunately it is never throuroughly defined. Why would would be adding theAveragevalue of real power andPeakvalue of Reactice power and call it the apparent power.

Am I misunderstanding this? Shouldn't we be adding thePEAKreal power andPEAKreactive power and call this the apparent power.

Also, other relations are P=(I^2)R Q=(I^2)X S=(I^2)Z

THANK YOU!!

Am i correct in saying that the reason why it is defined this way is so that "you" could say that the phase angle is given by

Phi = arctan(Q/P).

I'm just a little weired out by combining the Average Real power with Peak Reactive power.

Thanks

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# Homework Help: Reactive Power Apparent Power Issue

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