From: Joe Foley (redmenaced@yahoo.com)
Date: Thu Apr 03 2003 - 16:22:43 PST
Well, yeah,
I think what you are trying to tell us is the
IMPEDANCE of the capacitor since you have frequency in
that equation.
You can always remove subject capacitor from the
circuit.
Otherwise you would use Kirchoff's Law and put the
test voltage on each side of the capacitor, it CAN'T
go anywhere else! It has to go through the capacitor.
But, if you still want a test while the thing is in
the circuit you could get a ZM-11 which will put a
10.79 MHz signal through the capacitor and will let
you vary the voltage according to what the circuit and
cap like.
Joe
--- Patrick Jankowiak <eccm@swbell.net> wrote:
> Well I guess what I meant was the kind of
> capacitance meter than
> derives the capacitance by passing a small AC signal
> through it and
> measuring its impedance (AC resistance). The
> capacitance is given by:
> C[farads] = 1 / ( 2 * pi * F[Hz] * X [ohms] )
>
> Since the system contains diodes and other voltage
> sensitive devices,
> DC bias should be applied from a low resistance,
> high impedance source
> so that on the negative excursion of the tester's AC
> waveform, any
> diodes (such as in the alternator) do not become
> forward biased.
>
> Because the system is a 24VDC system, the tester
> must not output more
> than 24VAC peak to peak, and a DC bias of half of
> the peak to peak
> testing voltage would be in order. This insures the
> voltage from the
> tester will not exceed 24VDC and will not go
> negative.
>
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