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Reactive Power Apparent Power Issue 
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#1
Aug811, 01:23 PM

P: 152

This is not a HW question.
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*tphi) 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 the Average value of real power and Peak value of Reactice power and call it the apparent power. Am I misunderstanding this? Shouldn't we be adding the PEAK real power and PEAK reactive 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 


#2
Aug811, 04:42 PM

Sci Advisor
HW Helper
Thanks
P: 26,157

hi azaharak!
the apparent power is V_{rms}I_{rms} … it's voltage times current (but you need to know them because they come up in exam questions! ) from the pf library on impedance … Power: 


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