Quoting from a previous mail:
> Consider two circuits connected by a pair of wires. Assuming circuits
do not accumulate charge nor radiate, whatever current goes in must
eventually go out, therefore it is sufficient to specify the
instantaneous current I(t) in one wire. If we take one of the wires
as a voltage reference then let U(t) be the instantaneous voltage
difference between the two. The instantaneous power exchanged between
the two circuits circuits is then P(t) = U(t) * I(t). Assuming again
that the system is stationary, each quantity has a DC component and an
AC component: U(t) = U_DC + U_AC(t) and I(t) = I_DC + I_AC(t). It
then follows that
> P(t) = (U_DC + U_AC(t))*(I_DC + I_AC(t))
= U_DC * I_DC + U_DC * I_AC(t) + U_AC(t) * I_DC + U_AC(t) * I_AC(t)
> We have a power meter that measures voltage and current separately to
calculate instantaneous power. If it cannot measure DC currents NOR
voltages, the meter will only be using U_AC(t) and I_AC(t) and the
estimated power will be
> P_est_1(t) = U_AC(t) * I_AC(t)
> and the error will be
> P(t) - P_est_1(t) = U_DC * I_DC + U_DC * I_AC(t) + I_DC * U_AC(t).
But U_DC * I_AC(t) and I_DC * U_AC(t) have no net integrals over time,
so the error is:
> U_DC * I_DC.
In other words DC times AC cross-products do not contribute net power.
--
Berke Durak
