[R-390] Meters

[Roy Morgan]

At 11:16 AM 5/8/03 -0500, Scott, Barry (Clyde B) wrote:
>That's what I was thinking.
>
>Assume a 0.1mA meter (100uA) with internal resistance of 100 ohms.  The
>parallel resistor should be 1/10 the internal resistance so put a 10-ohm
>across it.  The new "internal" resistance will now be 0.11 ohms.

Barry,

I think the parallel resistor would be one ninth of the 100, actually, but 
close enough.  (So that nine tenths of the one ma goes through the resistor 
and one tenth or 100uA goes through the meter.)

The equivalent resistance of the 100 ohm meter and a 10 ohm resistor in 
parallel will be
(R1 x R2) / (R1 + R2) or 9.09 ohms.  Then you'd need 17 - 9.09 = 7.91 ohms 
in series with that combination.

Note that Mr Li reported measured values about twice the expected 17 ohms.

No doubt a clever person could develop a curve or family of curves to tell 
you the values you'd need to make this work with meters of various internal 
resistances and full scale currents. I would expect a parabola or some 
thing like that.  And, regions of "it won't work" values could also be plotted.

Roy

- Roy Morgan, K1LKY since 1959 - Keep 'em Glowing!
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