In my physics class, we were shown the picture below to measure the current and potential difference through a component. We were told that the resistance of the ammeter is as low as possible, and the resistance of the voltmeter as high as possible so that the electrons go through the component instead of the voltmeter. We were also told that the voltmeter measures the change in energy the electrons have before and after going through the component.

This has confused me as if the resistance of the voltmeter is so high that hardly any electrons pass through it, how can it measure how much energy the electrons are carrying? The few that do go through the voltmeter have surely avoided the component altogether, so how could you get a reading for how much energy has been transferred by the component?

  • $\begingroup$ It measures the difference between terminals, not the current passing through it. $\endgroup$
    – FGSUZ
    Feb 11 '19 at 17:46
  • $\begingroup$ The voltmeter doesn't really directly measure "the change in energy the electrons have before and after going through the component". What it measures is the voltage difference across the component, which you can think of as effectively being a measurement of the electronic "pressure" across the component. And knowing the voltage difference across the component is not in itself enough information to determine how much power is being dissipated by the component. You also need to know the current through the component, which is why the ammeter is in the circuit. $\endgroup$
    – user93237
    Feb 11 '19 at 18:51
  • $\begingroup$ @SamuelWeir That seems more like an answer than a comment. $\endgroup$
    – rob
    Feb 12 '19 at 0:55

Old-fashioned 'analogue' voltmeters (with a pointer and scale) did require a small current through them in order to work. They consisted of a sensitive ammeter (current meter) in series with a large resistance. They measured potential difference indirectly by using the fact that the current was proportional to the pd across the resistor (Ohms's law). Roughly speaking, the more you paid for a voltmeter, the more sensitive its ammeter, so the resistance could be higher, and the meter would have less effect on the pd you were trying to measure!

Digital voltmeters work on a totally different principle – and one that's much harder to explain! In theory they require no current to pass through them. First a very useful analogy: if we compare an electric current to the rate of flow of water through a pipe from a high reservoir to a lower reservoir, then we can compare potential difference to the height difference between the water levels in the reservoirs. Height difference is proportional to difference in gravitational PE per unit mass; Potential difference is proportional to difference in electrical potential energy per unit charge.

Modern voltmeters produce their own internal potential difference that rises at a constant rate (think about a water level rising) from zero. At the instant when the internal pd starts to rise, an internal clock is started. There is an electronic device that detects when the internally generated pd is equal to the unknown pd that you've applied to the voltmeter (water levels equal!). The device then stops the clock. The longer the clock has been going, the higher the unknown voltage must have been. The digital display of the voltage is made to be proportional to this elapsed time. The process repeats over and over again, keeping track of any (not too fast) changes in the unknown voltage.

Hope that you've been able to follow some of this!

  • $\begingroup$ even that voltmeter uses a little current over the internal resistance over which the voltage is measured; it is impossible not to as any measurement requires some amount of energy from the source whose parameter is measured. $\endgroup$
    – hyportnex
    Feb 11 '19 at 23:28
  • $\begingroup$ Indeed, though my cheapo digital voltmeter claims to have an impedance greater than 10 M ohm. $\endgroup$ Feb 11 '19 at 23:59

A voltmeter requires very little current to measure voltage. Therefore it uses very little of the energy of the electrons. Voltage is the energy required to move charge between two points. When work is work is done on the electron say by a battery it gains energy (voltage rise). When the electron does work to overcome resistance it loses energy (voltage drop). The sum of the rises equals the drops for conservation of energy ( Kirchhoff voltage law) Hope this helps


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