I am currently performing an experiment in which I measure the Current and Voltage of a motor that experiences changes in load. I am using a 3V (3V battery) supply however the voltmeter's (connected to the terminals of the motor) reading changes from 3V at no load down to 1V at max load. I understand how the current changes since less back-emf is produced at lower speeds. I also understand how using V = IR could result in that change of Volts. But wouldn't that require constant resistance and that is not the case with the values I have. For example, I tried calculating R using the equation and each time I got a different answer, Which itself increased in respect to the load. Is this a case of simple experimental error or am I missing something rudimentary? And if the voltage is changing is calculating power with P = I*3(V) all through the experiment incorrect?

  • \$\begingroup\$ "I tried calculating R using the equation" - which equation did you use, and what were the results (volts, amps, rpm)? \$\endgroup\$ Commented Mar 10, 2021 at 2:23

1 Answer 1


A motor isn't just a resistor. It's not just a resistor and BEMF source either. You cannot use V=IR to try and calculate a motor winding resistance while it is spinning because a bunch of other variables are also present.

At the absolute minimum, your model should also include an inductance:


simulate this circuit – Schematic created using CircuitLab


simulate this circuit

You can then measure the current and voltage at the motor terminals at various supply voltages and RPM/loads to figure out what the kV (BEMF voltage per RPM), inductance, and winding resistance are.

But you will need to be able to measure the RPM for this since the impedance drop presented by the inductance varies with frequency which varies with RPM (and so too does the BEMF voltage).

But if you have a smart meter to measure things that is able to separate out the reactive voltage drop due to the inductance from the resistive voltage drop, it will still measure a higher resistance while running, and the faster it runs the higher this resistance will measure.

The rotating magnetic field induces eddy currents to circulate with the iron core itself. These only appear while the motor is running and increase with frequency (RPM). They manifest as a voltage drop and with all the characteristics I listed, at first glance you would think they appear to be just like the voltage drop due to the inductance. But they are not. They have one critical difference: They are still resistive losses that permanently remove energy from the system unlike the inductance which absorbs energy only to release it later in the future. So there is actually a fixed resistor in the motor due to winding resistance but eddy currents in the core also produces the equivalent of a second resistor which is frequency dependent.

If this is for a science fair you can try to weed out those parameters and if you do it well, you should be able roughly predict things how it will behave.

  • \$\begingroup\$ I am simply playing around with motors and I wish to understand it better. That said your answer actually clears a lot of stuff up thank you. However, It wouldn't be possible for me to simply use the cells voltage and the ammeter reading to calculate the power into the motor right? Is there a way to keep the voltage constant? \$\endgroup\$ Commented Mar 10, 2021 at 2:43
  • 1
    \$\begingroup\$ @AnderMaxwell Well that would give you the input power. But input power is not the same as output power which is considerably more difficult to obtain. There are power supplies where you can dial in a voltage or dial in a current and it will provide either a constant voltage OR a constant current. \$\endgroup\$
    – DKNguyen
    Commented Mar 10, 2021 at 2:59
  • \$\begingroup\$ Don't forget the mechanical commutator. The configuration of the commutator in a DC motor allows more than one winding to be connected at certain rotation angles. \$\endgroup\$ Commented Mar 10, 2021 at 5:27
  • \$\begingroup\$ @ScienceGeyser I don't know of a simple way to convey that in a schematic. The assumption above is that even though you're feeding the motor DC the inductance is being subject to a frequency based on the RPM via the commutator and not the actual 0Hz DC of the supply. \$\endgroup\$
    – DKNguyen
    Commented Mar 10, 2021 at 5:31
  • \$\begingroup\$ It may be better to model as an AC circuit since all motors inherently are AC operated. In this case you would replace the DC supply with an AC supply that represents the commutation ring on the armature. \$\endgroup\$ Commented Mar 10, 2021 at 5:53

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