I am a bit confused about this idea. When ONLY a voltmeter is connected to a cell, there is no current flow since the voltmeter has infinite resistance. So, the voltmeter gives the reading of the actual emf since there is no voltage across the cell's internal resistance (as I=0).But if there is no current flow at all, it must mean that there was no energy transfer in the circuit at all. Then how does the voltmeter show a reading of the emf?

  • $\begingroup$ Zeroes and infinities can cause this sort of apparent paradoxes. Here's another one: if a wire is assumed perfectly conducting, how can current flow when there is no voltage difference between its extremes? As with your voltmeter doubt, it can be resolved by considering finite quantities: the conductivity of the wire is not infinite but very high; the voltmeter's internal resistance is not infinite but very high (as explained in Farcher's answer) $\endgroup$
    – Peltio
    Commented Apr 8, 2023 at 21:53

4 Answers 4


When ONLY a voltmeter is connected to a cell, there is no current flow...

That is not strictly true. Even if a voltmeter truly did have infinite resistance, then there still must be a transient current flow when the volt meter is connected to the battery. There must be a transient current flow any time the voltage reading changes. An ideal voltmeter would measure the electric charge between two isolated conductors somewhere within itself, and in order for the amount of charge to change, electrons must enter or leave at least one of the conductors.

Once the circuit reached a stable state (microseconds or less) then the current could completely stop flowing, and the meter could display a steady reading.

There are several ways to measure voltage. Old-school voltmeters were built around a D'Arsonval galvonometer that actually measured the current through a high-value resistor. The scale behind the pointer of the galvo was calibrated to directly read out the number of volts needed to push the measured current through the resistor. This type of "volt meter" requires a steady state current in order to give a steady, non-zero reading.

Relatively high voltages can be measured by an electroscope, which measures the force between two isolated conductors. In principle, no current need flow at all in order for the electroscope to give a steady reading. But, I am not aware of any practical, precision volt meters based on electroscope technology. The only electroscopes that I have ever heard of were built for lab/lecture demonstrations. (I.e., They were built to show off the principle, rather than to put it to practical use.)

Small voltages can be measured by applying the voltage to the gate of an insulated gate field effect transistor (a.k.a., "igFET" or "MOSFET".) The strength of the electric field surrounding the gate electrode modulates a current drawn from a separate power supply, and the gate voltage can be inferred from that. In the steady state , in principle, no current would be drawn from the circuit under test, but in practice, there will be a miniscule leakage current.

Modern, practical volt meters must be able to measure a wide range of voltages. I don't know much about how such instruments work in practice, but most of them do allow some small amount of current to flow. The spec sheets will tell you the input impedance of the instrument, which typically is in the tens of megohms, but for some, specialized instruments, can be tens of gigohms.


When ONLY a voltmeter is connected to a cell, there is no current flow since the voltmeter has infinite resistance.

In the real world voltmeters will have a resistance which could be very large (but not infinite), compared to the the internal resistance of the cell, and a current will flow.
With a very large resistance the current will be very small and that will mean that the potential drop across any internal resistance will be very small compared to the emf of the cell.

  • 1
    $\begingroup$ That an ideal voltmeter needs to have a resistance is a misconception. What it does need and does have is a capacitance. A typical example of such an "ideal voltmeter" would be an electrometer. $\endgroup$ Commented Apr 9, 2023 at 0:37

There are different types of voltmeters, using different effects to measure or indicate voltage values.

Take an electroscope (a), where two metallic blades hang electrically connected. Charge it (with high voltage) and they‘ll separate. Discharge a little and they‘ll come closer again. So angle is an indication about voltage, while almost no current flows in the beginning, and zero current after a while.

Take an oscilloscope (b), where a cathode emits an electron beam towards a screen. The tubes in the input have practically infinite resistance. Yet it can supply a voltage to two plates around the beam, deflecting it. Again, angle indicates voltage.

Old voltage meters used coils and a magnetized needle (c) to indicate voltage … they NEED to draw some current, giving you a wrong reading of the sources voltage.

Modern voltage meters provide an operational amplifier (d) as input. Its important characteristic is almost infinite amplification. With a simple resistor network you can make it having almost infinite input resistance, while even an old fashioned voltmeter draws enough current from the OpAmps output, i.e. from the OpAmps supply voltage … and not from the input. This is a form of decoupling.

In electronics we deal with ideal components (e), like a voltmeter, which draws no current while measuring the potential difference between two points, e.g. between two poles of a battery.

some voltmeter types




\begin{figure}% save images in same directory, names as given
	\subfigure[Electroscope: based on repelling charges\\]{\includegraphics[height=5cm]{electroscope}}
	\subfigure[Oscilloscope: repelling charges]{\includegraphics[height=5cm]{scope}}
	\subfigure[Coil-based: repelling poles]{\includegraphics[height=5cm]{coil}}
	\subfigure[OpAmp: $R_{i+} \approx \infty$, meter supplied by $2 \times 12$ V. Because $v \approx \infty$, the OpAmps input difference $\Delta V = V_+ - V_- \approx 0$ V, which "mirrors" the voltage on the red line to the negative input.]{\includegraphics[height=5.0cm]{opamp}}
	\subfigure[Ideal(ized) electric component \textit{voltmeter}]{\includegraphics[height=3cm]{ideal}}
	\caption{Various voltmeters, using different effects. Histroically, electroscope-based voltmeters were the first ($R_i \approx \infty$), followed by coil-based instruments ($R_i \approx 0$ Ohms, needing some series resistor to indicate voltages).}

  • $\begingroup$ Can you link to a source for your excellent figure? $\endgroup$
    – rob
    Commented Apr 9, 2023 at 11:00
  • $\begingroup$ Sure + done, please see above. $\endgroup$
    – MS-SPO
    Commented Apr 9, 2023 at 11:15
  • $\begingroup$ Oh, if the figure itself is your own creation, we don’t need the source code. I just thought it might have come from some paper, since most of our users would have included each image in the MarkDown rather than using LaTeX for compositing. I appreciate the links to the individual images. $\endgroup$
    – rob
    Commented Apr 9, 2023 at 22:53
  • $\begingroup$ No problem. I use this approach whenever I want things to be more compact than MarkDown offers. $\endgroup$
    – MS-SPO
    Commented Apr 10, 2023 at 3:37

The first thing I would ask is why you expect there to be an energy transfer to measure voltage. Take the often-used analogy of water flowing in a circuit of pipes. In this analogy, water pressure corresponds to voltage, flow rate (flow of water volume per unit time) corresponds to current, and water might be driven by a pump, corresponding to a battery. We can measure the pressure difference between two points in the circuit by two gauges. The difference in the pressure readings corresponds to the potential difference between these points, i.e. the voltmeter reading.

Note that at steady state (after the circuit has settled so nothing is changing with time), there needs to be no energy transfer to a pressure gauge to get a reading. As mentioned in Solomon Slow's excellent answer, you can likewise have a voltmeter that doesn't require energy transfer at steady state (it can have infinite input resistance), although even then it will still have non-zero input capacitance, so there will be some energy transfer when you first connect the voltmeter.


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