RE: strange vttc

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Randy & Lori" <rburney6-at-comcast-dot-net> 

Matt, Thanks.
I have never measured the drop of a 10M resistor in a string, and never
considered that my meter would play a role in increasing current and
throwing off a reading.  I used 10M as an example because everyone
(almost) has them, and current from the transformer would be limited.  I
teach and prove my method all the time, but with smaller resistors and
voltages, and the meter influence has never come into play.  I learn
something new every day on this list.  Thanks

Randy

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Wednesday, March 31, 2004 7:45 PM
To: tesla-at-pupman-dot-com
Subject: Re: strange vttc

Original poster: Mddeming-at-aol-dot-com

In a message dated 3/31/04 10:03:39 AM Eastern Standard Time,
tesla-at-pupman-dot-com writes:
Instead of a HV probe, couldn't one just wire several resistors in
series, read the drop across one of them and just apply Ohm's/
Kirchhoff's law? Maybe those 10M resistors used for draining the primary
caps? Three resistors will give 1/3 of the applied voltage across any
single resistor; 4 resistors-1/4.  If you think the 4000V label might be
correct, go with 5 resistors, a 1000V meter could do that one.  I don't
know why I haven't done this myself, I have been putting 120VAC into the
HV windings and measuring the output of the low voltage side, and figure
my ratios.  I'm not sure how accurate this is, but it seems to be pretty
close.

Hi Randy,
Putting a voltmeter across one of the resistors will give good readings
IF
and ONLY IF the internal resistance of the meter >> resistance of the
resistor. Let's say you have 5 10Meg resistors in series across  a 5 kV
source. The voltage drop across each will be 1 kV and the current will
be
100uA. If you place a meter with an internal resistance of 10Meg across
one
of the resistors, the effective resistance of the string is now 45Meg
and
the current through the string and meter combination is 111uA, half
through
the resistor. The drop across the resistor while being measured is thus
only 555V. Error >44%.  With a 100Meg input meter,
the effective resistance of the string is now 49.1Meg, the total current
is
~102uA of which 90% flows through the resistor making the voltage drop
~918V or about 8% error. Meters with an input Z of about a Gig are sort
of
pricey.
Possible solutions:
1) Always calculate the change caused by the meter
2) Many of the old Tripplet analog multimeters have a 5000VAC or 6000VAC

scale and cost ~$20-$30 on eBay. This eliminates the need for the
voltage
divider.

Hope this helps,
Matt D.