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[SOLVED] Voltmeter Internal Resistance

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Jan 4, 2013
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Based on Boylestad's Introductory Circuit Analysis, a voltmeter places an additional resistance across the element of interest. As I understand this is voltmeter internal resistance, which should be very high. However, in another book, Nilsson & Riedel Electric Circuits, it shows that a voltmeter internal resistance is connected in series with the voltmeter itself. I'm lost here, which configuration is correct?


In essence all old fashioned voltmeters were actually current meters. A good one used a 50 microamp movement, so needed an external resistance of 20,000 ohms per volt measured, hence a voltmeter (meter calibrated in volts) would have a high resistance in series with it, although normally internally. Modern DVMs or old fashioned VVTMs, normally have a constant impedance on their voltage range of 10 M ohms, so can be disregarded for most measurement.

So far as the other book and resistance shown in series is concerned, this is in fact what is shown in post #2 by Chuckey.
The voltmeter conventionally is a current meter (sensitive galvanometer) connected in series with a high value resistance. So many times a voltmeter is shown as a meter with a series resistor if the internal resistance of the voltmeter is to be included in the circuit analysis ( if it contributes substantial load).

An ideal voltmeter has infinite resistance. It does not take any current from the circuit. A real voltmeter will take current from the circuit under measurement. So to model a real voltmeter you can use an ideal voltmeter with a resistor in parallel.
A resistor in series with an ideal voltmeter does not make any sense. Because the ideal voltmeter is an open circuit.

An ideal ammeter has zero resistance. To model a real ammeter you place an ideal ammeter in series with a resistance.

In practice as chuckey said, a voltmeter can be made with a resistance in series with an ammeter. Probably that is what you saw at the 2nd book.

So when asked about the reading of voltmeter of a given internal resistance without specifying what type of voltmeter (could be digital or an old analog), we add the internal resistance in parallel to the element/points/branch across which we want to find the voltage, and then find the voltage across this internal resistor?

That is right. Suppose that you want to measure the voltage of a 12v battery which has a 100k resistor in series.
An ideal voltmeter will measure exactly 12v.
If you place a voltmeter with 20000 ohm per volt sensitivity in the 15v scale it will have 20K x 15v = 300k internal resistance.
With that voltmeter you will measure V= 12v x 300K /( 100K + 300K) = 9 volts . Because the internal resistance of the voltmeter forms a voltage divider.
In the same way. A modern DVM with 10M internal resistance will measure V= 12 x 10M/(10M+100K)= 11.88v

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