> It's got something to do with P really not being U * I but there are
> integrals involved and capacitance and inductance turn the phase
> differences between U and I around, which makes for differences under
> idle vs. load conditions.

Right.

Power actually consumed is the RMS value of E(t)I(t).  (RMS = "root
mean square"; power consumed is sqrt(mean((E(t)I(t))^2)).)

But if you measure voltage by itself, your voltmeter will (or should)
display RMS V(t), and an ammeter will display RMS I(t).  If V(t) and
I(t) are precisely in phase, the product of RMS volts and RMS amps will
equal the RMS power.  This is what you get for a perfectly resistive
load (a baseboard heater or an incandescent light bulb are examples of
pure resistive loads - or at least close enough for 50-60Hz work).  A
more complex load may appear purely resistive if the capacitive
impedance is exactly matched by the inductive impedance.

But for loads not purely resistive, RMS volts times RMS amps will
exceed the RMS value over time of the (time-varying) instantaneous
product, because of the phase shift between current and voltage.  If
that phase shift reaches 90° in either direction, the power consumed
will drop to zero (because the integral of sin(t)cos(t) is zero in the
long run).  This would be the case for a purely capacitive or purely
inductive load (which, like absolute zero, are idealizations that can
be approached but never actually achieved).  This is why a motor with
light or no load can draw significant current but not actually consume
correspondingly large amounts of power.

If the voltage or current is not a pure sine wave (fluorescent lights,
motors with triacs for speed control, that sort of thing), things get a
whole lot more interesting, and it's no longer as simple as just a
phase shift.

In electrical engineering, this is all done with complex numbers,
related to the exp(i x) = cos(x) + i sin(x) identity.  I never actually
studied it and never really got a handle on that part, though....

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