Why does an ideal voltmeter have infinite resistance?

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Leo Liu
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Context:
Printed text paragraph explaining how an ammeter is converted to a voltmeter by adding a series resistor, ending with the sentence that an ideal voltmeter would have infinite resistance, which is highlighted.

What I don't understand is why the little current that flows through the converted voltmeter can affect the measurement when the resistance of the resistor added to the ammeter is not infinite. Can someone please explain this to me? Thanks.
 
Leo Liu said:
Context:
View attachment 296294
What I don't understand is why the little current that flows through the converted voltmeter can affect the measurement when the resistance of the resistor added to the ammeter is not infinite. Can someone please explain this to me? Thanks.
I think it is simply a convenience so that we can discuss a voltage measurement that doesn't change the network it is connected to. You certainly can have a voltmeter with finite shunt resistance, but then the voltage you measure might not be the same as the voltage between those circuit nodes before or after the meter is connected. This is because the current that would flow through the voltmeter would change the result.
 
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An analog voltmeter has an indicator needle that is driven by coils which have a finite resistance that is adjusted for in the calibration.
 
Oops, sorry, I didn't really answer your question. The short version is that whatever current flows through the voltmeter must also flow through the circuit network you are measuring. This will cause additional voltage drops and redistribute the various node voltages.

Consider this simple circuit, with the ammeter having zero resistance (not including R):

Hand-drawn circuit schematic: a 5V battery in series with R1 (4 kΩ) feeds a node where R2 (1 kΩ) goes down to the return wire, and a branch with R in series with an ammeter labelled A returns to the bottom wire.


The voltage across R2 will be 1V when R=∞, or when the voltmeter is disconnected. The current through R1, I1=1mA.
But if R=1KΩ, then that would be the same as if R2=500Ω, in which case the voltage across R2 is 0.556V with the current through R1 being 1.11mA, equally divided between R2 and the meter.
Then in the extreme case of R=0, the voltage is 0V, the current through R1 is 1.25mA and it all flows through the meter.
 
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DaveE said:
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Ah I see. Thanks a lot!
 
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