"Franc Zabkar" wrote in message
...
On Sun, 12 Apr 2009 20:08:37 -0700, isw put finger
to keyboard and composed:

In article ,
David Nebenzahl wrote:

I have an old galvanometer (nice Weston in a slant-front black crackle
case). Its scale reads 30-0-30. What units are these? Or are they just
relative values for comparison?

With a standard alkaline cell and starting with a *very high* resistance
(5 megohms, say), drop the resistance until you get a decent deflection.
Then measure the resistance and the voltage, and do the math.

Isaac

A quicker way to do it might be to use two DMMs. Set the first to
measure resistance and the second to measure current. Start with the
highest resistance scale. DMM1 will supply a test current to the other
two meters.

DMM1 (ohms) DMM2 (amps) Weston Galvanometer

-- I -- I
o---------o-- A --o---------o-- M --o--|
| |
current |
source |
| |
o--------------------------------------|
I --

My DMM outputs the following test currents:

diode - 0.93mA
200R - 167uA
2K - 133uA
20K - 42uA
200K - 5uA
2M - 0.5uA

When testing on the diode range, the OP could set DMM2 for the highest
burden, ie the 2M scale. In that way the voltage to the galvanometer
would be reduced by 200mV (?). Alternatively he could insert a series
resistor or potentiometer.

If only one DMM is available, then replace DMM1 with a battery and
resistor and use your DMM to measure the current, otherwise you will
need to account for the burden of the galvanometer when doing your
Ohm's Law calculations.

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.

Every galvanometer I've seen in the past 60 years read in microamps