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During the 19's, Schrödinger state that it is impossible to make a measurement without disturbing a system. To illustrate his statement, he made the famous experiment of the cat and the box. Poor cat btw..

This is also true in electronic nowadays. A multimeter is use in parallel of a load for a voltage measurement. This "add on" will change the equivalent resistor value and then change the "true" voltage value. This is also true for current or other kind of measurement.

How is it possible to compensate this lack of precision? I don't need to be that accurate, but I ask that question by pure curiosity

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    \$\begingroup\$ Pickup a used 5.5 or 6.5 digit multimeter and you will get a >1GOhm input resistance for low voltage work, also some input bias current in 100 pA range \$\endgroup\$
    – sstobbe
    Commented Jun 10, 2017 at 14:20
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    \$\begingroup\$ Just to point this out: Schrödinger never did, nor intended to do, such an experiment. It was just a thought experiment. en.m.wikipedia.org/wiki/Thought_experiment \$\endgroup\$
    – Timo
    Commented Jun 10, 2017 at 14:44
  • \$\begingroup\$ Just to point out, Schrodinger was MOCKING the idea of superposition. en.wikipedia.org/wiki/Schrödinger%27s_cat \$\endgroup\$
    – Passerby
    Commented Jun 10, 2017 at 21:37
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    \$\begingroup\$ Technically, the Schrodinger cat experiment regards the superposition of quantum states and the probabilistic nature of quantum phenomena. The Heisenberg Principle is what is usually brought up when talking about measurements. But even that principle does not state that it is impossible to make exact measurements. It only states that it is impossible to make simultaneous arbitrarily accurate measurements of conjugate variables (position and momentum, time and energy/frequency). Stationary states have well defined values of energy, for example. \$\endgroup\$
    – Sredni Vashtar
    Commented Jun 10, 2017 at 21:38
  • \$\begingroup\$ I want to create a superhero whose origin story is someone actually trying Schrodinger's experiment. \$\endgroup\$
    – Harper - Reinstate Monica
    Commented Jun 10, 2017 at 21:49

4 Answers 4

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If circuit has a impedance of 10 ohm, and the multimeter is 10Mohm, then the change is 0.0001%.
With the impedance of the circuit I mean the load, the power source, and everything combined. Together they determine the impedance of the circuit.

Sometimes the difference can be calculated, but not every circuit has a linear transfer. Sometimes is makes a big difference, for example with high voltage and low current applications and with current measurement. So yes, you have to be aware of it.

For example a Geiger counter with a Geiger tube that needs 400V with a very low current. Suppose the voltage measured with a multimeter is 400V, but after the multimeter is removed, the voltage may raise to 450V without you knowing it. One solution is to have a voltage divider with two resistors always connected to the high voltage. Once calibrated, the low voltage of the voltage divider can be used to calculate the actual voltage.

To do a current measurement the right way is a difficult subject on its own. For example, the wires themself could have influence. Do you know about 4-wire shunts ? Measuring high frequency current is very hard. And so on.

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Schrödinger made an statement about a quantum mechanical system. So I disagree your conclusion "This is also true in electronic nowadays." because in common electronic systems QM effects are not observed directly; i.e. other effects are more dominant (by several orders of magnitude) than the QM effects you are referring to.

So normally there is no need "to compensate this lack of precision" because it doesn't matter at all.

E.g. normally no QM effects prevent you from measuring voltage arbitrarily accurately without affecting current (many other effects you will need to overcome; but they have nothing to do with Schrödinger's statement).

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As you said, multimeter has an internal resistor to make different kind of measurement. That value can be found in the documentation. If you know that value, you can then make the calculation and find the impact of the multimeter on the circuit.

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No measurement can be made without disturbing the source. That is not only true for electrical measurements but for all kind of measurements. Nowadays we use instruments that consume far less energy then in the past therefore the influence on the source can be neglected in most of the occasions.

In general it can be stated that no measurement is possible without transfer of energy and therefore the source always becomes disturbed.

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  • \$\begingroup\$ That appears to be a hard claim to defend. Suppose I add a dab of reflective paint to an illuminated flywheel. Do you claim that using a photodetector to measure the wheel's speed in some way influences the wheel's speed? Do you claim that our measurement of the dimensions of astronomical objects in some way influences those dimensions? \$\endgroup\$
    – Magoo
    Commented Jun 11, 2017 at 1:29
  • \$\begingroup\$ @Magoo Now 1) the wheel is slightly unbalanced because of the extra paint and 2) you have to shine a light on the wheel which transfers a slight amount of momentum between the photons and the wheel. \$\endgroup\$
    – user20574
    Commented Jun 11, 2017 at 12:06
  • \$\begingroup\$ @Magoo. It is straigth forward. If we require energy to measure (light is a form of energy) then the loss of energy is influencing the object. Sometimes we can make use of energy that is already radiated from a source. In such a case we use small part of the already radiated energy for our measurements. \$\endgroup\$
    – Decapod
    Commented Jun 11, 2017 at 12:17

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