I Why does an ideal voltmeter have infinite resistance?

AI Thread Summary
An ideal voltmeter is designed to have infinite resistance to prevent any current from flowing through it, ensuring that the measurement does not alter the circuit it is connected to. When a voltmeter has finite resistance, the current that flows through it can affect the voltage readings by causing additional voltage drops and redistributing node voltages. This means that the voltage measured may differ from the actual voltage present in the circuit before the voltmeter was connected. The discussion highlights how even small currents can significantly impact measurements in circuits with finite resistance. Understanding this principle is crucial for accurate voltage measurements in electrical circuits.
Leo Liu
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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.
 
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Leo Liu said:
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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):

20220130_215728.jpg


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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