Below is part of what I was wondering about: Doesn't that mean,
in reality, that, unless the PF is corrected, that the customer
is the one with the advantage? As in, "free" power? IE of
shifted waveforms is going to be less than in-phase IE over time.
Therefore, the correction equipment is to "correct" the numbers
so the power company isn't delivering power it isn't charging
for? If so, why would anyone voluntarily install a capacitor
system?

Pop


wrote in message
oups.com...
: "The physics of reactive power requires calculus to understand.
"
:
: It's really pretty simple to understand the basics.
Instantaneous
: power is always voltage times current. With an AC circuit and
a purely
: resistive load, the voltage and current are always in phase
with each
: other. Place a graph of voltage over a graph of current and
they line
: up perfectly. So simply multiplying RMS Voltage times RMS
Current
: gives power.
:
: With a load that has capacitance or inductance in addition to
: resistance (eg a motor), the voltage and current are no longer
in
: phase. Place a graph of one over the other and they appear
shifted.
: So when voltage is at it's peak, current is not, hence the
power
: consumed will be less. How much less depends on how far out of
phase
: voltage and current are. With a purely capacitive load or a
purely
: inductive load, the power will be zero.
: