Let's say I've got a circuit with a two-pole Thévenin equivalent that has DC voltage difference U and out-resistance Ro. What does the voltmeter do? I suppose it has an internal, very large resistance (Rv) that is connected in parallel with Ro. But what is measured across Rv, and how?
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The Thevenin model is a series resistance, so the meter would be connected in series with this. It's the Norton model that has a shunt resistance the meter would be connected in parallel with. – Chris Stratton Dec 11 '11 at 01:54
3 Answers
Voltmeters measure the voltage they see after loading it with their "input resistance". For meters with electronic circuitry, this is usually in the 10 MΩ range. Therefore, the Ro impedance of the source voltage and the voltmeter input impedance form a voltage divider, and the voltmeter tells you the voltage out of that divider.
As long as the source voltage impedance is substantially less than the voltmeter's input impedance, the voltmeter will read the voltage accurately. This is why higher input resistance is better for voltmeters. In practise, most of the time loading something with 10 MΩ won't change its voltage to the point you care. Still, you have to be aware of the limitations of your tools to know when they won't give you accurate readings.
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Thanks, that makes it clearer! How is the actual measurement of voltage done in most voltmeters? I suppose they don't count electrons and divide by the resistance? – Astrid Dec 10 '11 at 21:47
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@Astrid: In modern electronic meters, the voltage is suitably amplified, converted to a number by a A/D converter, that converted to decimal in a microcontroller, then that displayed to the user. Old meters applied the voltage directly to the coil of a meter movement. These had substantially lower input resistance even though the coil was wound with as many turns of thin wire as possible. – Olin Lathrop Dec 11 '11 at 15:17
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BTW still there was an old voltmeter with very high (theoretically, infinite) input resistance using a potentiometer as a "measuring instrument". I agree with @starblue that conventional old movement voltmeters actually measure the current. There the ballast resistor acts as a voltage-to-current converter. – Circuit fantasist Jan 18 '15 at 14:03
There are a variety of ways of making the actual measurement.
One of the more traditional is a moving-coil micro-ammeter, consisting of a winding on an iron core that rotates between the poles of a permanent magnet, working against the force of a spring. The meter circuitry and selector connects various resistors to scale the input voltage to a tiny current. A crude, related instrument can be made by winding a few turns of wire across a plastic-body magnetic compass, with the introduced field forming a vector sum with the earth's field and resulting in a new pointer angle.
The next major development consisted of using a high impedance vacuum tube amplifier between the circuit under test and the meter movement, producing the Vacum Tube Volt Meter or VTVM. Later the tube was replaced with a field effect transistor.
The third major development would be replacing the meter movement with an analog to digital converter. This usually consists of a comparator which compares the unknown input against known reference voltages; either sequentially as the reference voltage is changed by a digital to analog converter under a search algorithm, or by charging a capacitor while measuring time, or for high speed applications bu using a number of of comparators and voltage sources acting in parallel to produce a faster answer (though that form is more likely to be found in a high speed instrument such as an oscilloscope than a typical digital multi meter).
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A volt meter calculate the potential deference between the hot wire and the neutral Because current flows form hot to neutral 0-(-115)= 115 and 115 -0=115 alternating current neutral stays at zero where as the hot goes from a negative hot to a positive hot.
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This is not an answer to this question. OP is asking about the working of voltmeter. – nidhin Jan 18 '15 at 11:56