Aug8-11, 01:23 PM
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
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
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.
Aug8-11, 04:42 PM
the apparent power is VrmsIrms … it's voltage times current
(but you need to know them because they come up in exam questions! )
from the pf library on impedance …
|Register to reply|
|Electric Power measurement Current Transformers and reactive power||Electrical Engineering||8|
|reactive power||Electrical Engineering||1|
|reactive power injection||Electrical Engineering||0|
|Reactive power||Engineering, Comp Sci, & Technology Homework||5|
|Reactive Power||Electrical Engineering||6|