YOKOGAWA WT500 User Manual page 336

Appendix 3 Power Basics (Power, harmonics, and AC RLC circuits)
App-16
AC Power
AC power cannot be determined as easily as DC power, because of the phase difference
between the voltage and current caused by load.
If the instantaneous voltage u = U
the instantaneous AC power p is as follows:
p = u × i = U sin t × I
ω
m
U and I represent the rms voltage and rms current, respectively.
p is the sum of the time-independent term UIcos
voltage or current at twice the frequency –UIcos(2
AC power refers to the average power over 1 period. When the average over 1 period is
taken, AC power P is as follows:
P = UIcos [W]
f
Even if the voltage and current are the same, the power varies depending on the phase
difference . The section above the horizontal axis in the figure below represents positive
f
power (power supplied to the load), and the section below the axis represents negative
power (power fed back from the load). The difference between the positive and negative
powers is the power consumed by the load. As the phase difference between the voltage
and current increases, the negative power increases. At
powers are equal, and the load consumes no power.
When the phase difference between voltage and current is 0
Positive
power
u
0
i
When the phase difference between voltage and current is
u
Positive power
p
i
0
Negative
power
When phase difference between voltage and current is
u
p
i
2
0
The positive and negative powers are the same
sin t and the instantaneous current i = I
ω
m
sin( t – ) = UIcos – UIcos(2
ω f f
m
f
p
Average power
P=UI
2 t
Average power
P=UIcos
2
t
Average power
P=UIcos
2
t
2
t – )
ω f
and the AC component term of the
t – )."
ω f
= /2, the positive and negative
f π
2
= 0
sin( t – ),
ω f
m
IM 760201-01E