John-

I was afraid that things were getting long and confusing. The "Part 1" and
"Part 2" comments in my reply refer to the posting immediately below mine.

But to answer your fundamental objection. It is indeed true that if we
place a resistive and capacitive load, measure voltage amplitude AND phase
of the resulting voltage relative to the source (absolute phase), we can
then calculate the source resistance and (inductive) reactance of the
voltage source.

However, if we cannot measure absolute phase of the resulting (loaded)
voltage (phase relative to the source voltage), then we are stuck with
magnitude-only calculations. Obtaining absolute phase needs some sort of
power system clock or stable time reference. Measuring the phase between
the current and voltage at the load tells us only about the phase
characteristics of the load, and nothing about the voltage source.

With the magnitude-only calculations, the resulting voltage equation for a
resistive load contains terms in Rs²,  Xs², and Rs (Rs = source resistance,
Xs = source reactance). For pure reactive loads, the resulting voltage
equations contain terms in Rs², Xs², and Xs, with the coefficients for Rs²
and Xs² in each equation equal. This last property allows one of the two
voltage magnitude equations to be scaled such that when the two equations
are subtracted from the other, the Rs² and Xs² terms cancel, leaving an
equation with Xs as the only unknown. From there, Rs many be found.

On the other hand, the voltage magnitude equation for the resistive load
does NOT have equal coefficients for the Rs² and Xs² terms. Even if the
coefficients were equal, the equation has the Rs term. Remember that the
voltage magnitude equiation for the capacitive load contains an Xs term.
Scaling and subtrating these two equations would yield an equation with Rs
and Xs terms (two unknowns and one equation).

Don Borowski
Schweitzer Engineering Labs
Pullman, WA





John Woodgate <j...@jmwa.demon.co.uk>@majordomo.ieee.org on 11/05/2002
01:23:33 PM

Please respond to John Woodgate <j...@jmwa.demon.co.uk>

Sent by:    owner-emc-p...@majordomo.ieee.org


To:    emc-p...@majordomo.ieee.org
cc:
Subject:    Re: Measuring AC Line Impedance



I read in !emc-pstc that don_borow...@selinc.com wrote (in
<of32a3a512.366695dc-on88256c67.00802...@selinc.com>) about 'Measuring
AC Line Impedance' on Tue, 5 Nov 2002:
>Part 1. This measurement will NOT give just the resistive part of the
>source impedance.

Your post is so long that it is not clear what you are trying to
correct. Also, by top-posting you make it more difficult to understand
how your comments relate to previous texts.

If we measure the voltage developed across a resistive load, we have one
equation with three unknowns, e.m.f., source inductance and source
resistance. If we measure the voltage developed across a pure capacitive
load (or one with known losses), we also have one equation with three
unknowns. If we measure the *open-circuit* voltage, we immediately have
the value of one unknown, the e.m.f. or Thevenin equivalent source
voltage. That leaves us with two equations and two unknowns, which are
soluble.
--
Regards, John Woodgate, OOO - Own Opinions Only.
http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

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