On Tue, 09 Dec 2003 09:19:26 -0800, Bill Turner
wrote:

On Tue, 09 Dec 2003 09:59:03 +0200, Paul Keinanen
wrote:

L = Xl / (2 pi f) applies only to _pure_inductive Xl

It does _not_ apply to L = X / (2 pi f) in which X is some combination
of Xl and Xc !

_________________________________________________ ________

And what is the basis for the second statement above? Network analysis
normally combines reactances of opposite signs and treats them as a
single reactance, provided the frequency is constant of course.

Yes, that is true.

For a series resonant circuit X = Xl + Xc and since Xl and Xc have
opposite signs, the total reactance is less than either of them. At
resonance, their magnitude is the same and they cancel each other,
thus, the reactance is zero at resonance.

The situation is a bit more complicated with parallel resonant
circuits (hopefully I got the formula right)

X = 1 / ( 1 / Xl + 1 / Xc)

At resonance 1/Xl and 1/Xc have equal magnitude but opposite sign and
the divisor goes to zero and X to infinity. At resonance X also
changes sign.

Most test equipment simply measure the total reactance
X = 1 / ( 1 / Xl + 1 / Xc) at a specified frequency f.

Using the notations

Xl = 2 pi f L
Xc = -1/(2 pi f C)

Please show _mathematically_ how you from a single measurement of the
total reactance X at frequency f can solve both the inductance L and
capacitance C.

If you can not do it, then the single measurement of the total
reactance X is not sufficient for solving both L and C :-).

You could also start with the simpler series resonance (X = Xl + Xc)
case in which the measured reactance X is the end to end reactance.

And please, do not try to invent your own formulas such as
L = X / ( 2 pi f) when the components of X are unknown. That formula
is only valid in a special case when X is known to be a pure inductive
reactance Xl.

Paul OH3LWR