# Why is the resistance of an ideal voltmeter infinite?

Consider an analog voltmeter. I understand that the internal resistance of the voltmeter should be high enough to not lower the actual voltage across the load connected to the voltmeter, but, at the same time, if the resistance was infinite, there will be no current flowing through the voltmeter which means no magnetic field will be produced and the pointer wouldn't deflect.

• Because an ideal voltmeter does not draw any currrent.
– Chu
Jan 10, 2018 at 9:19
• Because if you had an ideal voltmeter you could make a measurement without affecting the circuit being measured, as it is you may need to account for the meter resistance when working on some circuits. Jan 10, 2018 at 15:31
• The word 'ideal' usually means something that cannot exist in practice but which would make all the involved mathematics simple. An ideal voltage meter is also spherical, has zero mass and homogenous density, and its display always points at the viewer. Jan 10, 2018 at 17:23
• Related: there are no ideal resistors. Every resistor you ever buy will not perfectly obey ohm's law. Also: no ideal capacitors, inductors, voltage sources, current sources, or switches. Jan 10, 2018 at 19:07
• @ilkkachu so what I should understand by the word 'ideal' in the context of this analog voltmeter is that (ideally) no current is supposed to flow through the voltmeter and, consequently, there's no voltage drop across the load, but in real life current has to flow into the voltmeter so there must be a voltage drop; however, designers try their best to minimize the amount of current flowing through the voltmeter by choosing the appropriate resistance. Am I correct? Jan 11, 2018 at 7:12

Your reasoning is correct if you take “voltmeter” to mean galvanometer (a device which deflects a needle using a coil in a magnetic field). But “voltmeter” just means a device which measures voltage, and we can imagine taking that measurement without disturbing the circuit.

The whole idea of an ideal instrument of any sort is to ignore the limitations of a physical implementation of the concept; there does not have to be a possible way to do it.

That said, there are ways to measure DC voltage with a nearly infinite input impedance.

• The electrostatic voltmeter or mechanical electrometer. An electrostatic voltmeter has a needle and pivot like a galvanometer, but instead of a coil, has a pair of shaped plates, much like a variable capacitor, which are electrostatically attracted by the voltage across them.

The electrostatic voltmeter electrically resembles a capacitor, so its input impedance is infinite at DC (if you ignore leakage across insulators). If you remove it from a circuit, the needle will keep its reading unless you short the meter terminals!

In this type of meter the force is very small compared to a magnetic type and so it only works well for voltages in the tens to thousands of volts, and takes a long time to settle. Compared to a galvanometer, it is less practical to be adapted to measure current or have multiple voltage ranges.

• A null voltmeter or (obsoletely) potentiometer uses the bridge principle to measure voltage. In manual operation, an adjustable reference voltage is connected through a galvanometer to the circuit being measured, then the reference voltage is adjusted until the galvanometer is not deflected in either direction ("nulled"). The same principle can be implemented electronically without manual controls.

This type of voltmeter will either draw or supply (!) power from/to the circuit under test, until the nulling is done; the effective input impedance depends on how precise the null is.

(One of the original applications of these devices was measuring the voltage generated by thermocouples.)

An ideal voltmeter isn't something you can go out and buy. It's a conceptual idea.

If you had an ideal voltmeter, you could connect it to any circuit and measure the voltage, without having any effect whatsoever on the circuit you're measuring.

A moving coil voltmeter can't be ideal since it requires some current to operate the coil.

An ideal voltmeter would not draw any current at all.

A good voltmeter would draw as little as possible.

The less current a voltmeter draws, the less error it generates when drawing current from a real circuit.

In the bad old days when I started electronics, a reasonable meter would take 20uA for FSD (full scale deflection), also rendered as '50k/volt', so you could work out the series resistor needed to get any voltage range.

These days, amplifiers in ordinary meters take much less current than that, my present £10 DVM will read 200mV with 10Meg input impedance, that's 20nA for FSD, and specialist low input current (electrometer) amplifiers can be used to draw orders of magnitude less than that.

• "a reasonable meter would take 20uA for FSD..." This is what's confusing me; if FSD is achieved when a current of 20uA is flowing, this means that current must flow for the meter to work; hence there will be no reading if no current was flowing... Am I missing something? Jan 10, 2018 at 6:10
• if no current is flowing, there is a reading, and the reading is correct at 0! If there's zero current flowing through the voltmeter circuit, then there's zero voltage at the measurement point. So on the 20v range, I'd have a series resistor of 1Meg. With 20v, 20uA flows, and I get FSD. With 2v, only 2uA flows, and I'd get 10% of FSD, which is a reading of 2v. Jan 10, 2018 at 6:41
• @DigiNinGravy that meter is not ideal, obviously. Jan 10, 2018 at 23:18

Current has to go through the coil of an analog voltmeter to get the needle to move (against the torque of hairspring), however there is no requirement that the current come from the voltage being measured.

Devices like vacuum tubes, JFETs, and MOSFETs draw almost no current on their grid or gate and can control usable amounts of current. Most of the current from the voltage being measured in meters that use such devices is typically the input divider, and it is usually in the M$\Omega$. Some meters have the ability to switch the divider out, yielding input resistance in the G$\Omega$. Special meters designed to have the highest possible input resistance are in the hundreds of T$\Omega$. Keithley, for example, claims "reliable measurements down to 10aA" (that represents about 63 electrons per second).

The 10A range on your multimeter is eighteen orders of magnitude greater full-scale than that sensitivity. Roughly the ratio between the size of a hydrogen atom and the circumference of the earth.