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Let's say I take my multimeter and measure the voltage across the poles on my battery, I get some reading X volts. I then put a resistor between one terminal and the adjacent probe, this resistor has resistance Y ohms.

First:

schematic

simulate this circuit – Schematic created using CircuitLab

Then:

schematic

simulate this circuit

What will the new voltage reading be?

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An ideal voltmeter (which does not exist in real life) has infinite resistance and no current flows when a measurement is made. In that case there would be no current flow in the series resistor, so no voltage drop, so the metter would still read full battery voltage.


A real voltmeter has finite resistance. A cheap meter may have a 1 megohm resistance, maybe even 100 kohm - but there is no good reason for a digital meter to be under 1 megohm. When connected current will flow in the meter. This will have minimal effect on the battery voltage, but the current will cause voltage drop in the series resistor and the meter will read low by this amount. As the same current flows in meter and in the series R and as V= I x R , the voltages across resistor and meter will be in proportion to their resistances.
Say Rtotal = Rmeter + Rseries.
All the voltage will drop across Rtotal (of course).
So the proportion across the resistor will be Vbattery x Rseries/Rotal.
And the proportion across the meter will be Vbattery x Rmeter/Rotal.

If Rmeter = 1,000,000 ohms and Rseries = 10,000 ohms then
Rmeter/Rtotal = Rmeter / (Rseries + Rmeter)
= 1,000,000 / (1,000 + 1,000,000) = 1,000,000/1,001,000 = 0.999000999 ~= 0.9990 of original
ie for small Rseries compared to Rmeter the drop in voltage will be proportional to the series resistance compared to the meter resistance.
So a 100 k resistor with a 1 megOhm meter will cause about a 10% drop. A 20k resistor will cause a ~= 2% drop etc.

So, in most cases, things like contact resistance and wiring resistance will have minimal effects on voltage measurements as long as there are no other currents flowing to affect the reading.

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  • \$\begingroup\$ that's what I mean \$\endgroup\$ – Lone Kaffekande May 16 '12 at 18:00
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X volts. The resistor doesn't have any current through it since there is no return path (assuming you have a meter with high resistance, which good ones do).

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    \$\begingroup\$ I don't think that that is what he meant. I think he means that the resistor is added in series with the meter. But, I may be wrong. \$\endgroup\$ – Russell McMahon May 16 '12 at 18:17
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The voltage will be X * (Rm / (R + Rm)) where Rm is the internal resistance of your meter. Usually this is a large resistance like 10 megohms. So if R << Rm, then Rm/(R + Rm) ~= 1, and so the voltage reading is still X. But if your meter's internal resistance is 10 megohms, and you use a 10 megohm resistor in series, you will see a reading of X/2.

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