I'm a beginner in this field so please forgive me if I'm confusing with my question.

There is a component that I can't understand with Ohm's law which is a washing machine drain pump. Washing machine drain pumps from most manufacturers have similar specifications. Their winding resistance is usually between 10-20 Ω and it operates under 120 VAC.

From drain pump troubleshooting

Drain pump resistance

Drain pump specification

However the specifications written on the label are quite different. 120 VAC, 1.1 A, and 80 W.

Drain pump current draw

The actual current draw, 0.9 A, is close to the specification value which is 1.1 A.

I really don't understand that according to Ohm's law the resistance value calculated per the specification should be (R = U/I) 133.33 Ω where U is 120 V and I is 1.1 A.

But why is the winding giving me 14.8 Ω?

Shouldn't it draw 8.11 A as I = U / R = 120 V / 14.8 Ω = 8.11 A?

  • 8
    \$\begingroup\$ There is a thing called inductance \$\endgroup\$
    – PlasmaHH
    Mar 23, 2018 at 14:44
  • 7
    \$\begingroup\$ Ohm's law is perfectly fine, but for AC you have to take more into account than the DC resistance. AC has this thing called impedance, which you can't measure with an ohmmeter. \$\endgroup\$
    – JRE
    Mar 23, 2018 at 14:46
  • 2
    \$\begingroup\$ Ohm's Law "works" wherever it is applicable, but there are many situations to which Ohm's law does not apply. As the answers below tell you, describing the behavior of motors is one of those. In the DC realm, Ohm's Law only applies to resistors and conductors. For an AC circuit operating at a given, fixed frequency, there is a version of Ohm's law that works for inductors and capacitors as well, but instead of "resistance", we use the word "impedance" in that case, and you have to do the math using complex numbers. \$\endgroup\$ Mar 23, 2018 at 14:55
  • \$\begingroup\$ Induction Motors loaded impedance are usually 5~8x DCR which defines the average rms surge current ratio as well. here the apparent impedance is 8.1A/0.9A or 9x the DCR of the coil, so it is not being fully loaded. \$\endgroup\$ Mar 23, 2018 at 15:43
  • \$\begingroup\$ This is either an induction motor (should be of the shaded pole type but this feature was not visible on the pole pieces and I think is required for self starting and direction selection) or more likely a permanent magnet AC motor that can start in either direction. The impeller looks reversible so no information there. If the motor turns under power it is likely to be working. It may wear faster if not water lubricated. Take care with mains voltages. \$\endgroup\$
    – KalleMP
    Mar 23, 2018 at 23:20

7 Answers 7


Have you ever played around with an electric motor connected to something like a light bulb or another motor? If you spin the motor, the motor acts like a generator and spins the other motor or lights the light bulb. The same thing happens when the motor is spinning under electrical power, the motor will behave like a generator, looking something like this:


simulate this circuit – Schematic created using CircuitLab

Notice how although you see 12V across the motor, the motor resistance only sees 1V, making the current through the motor 100mA instead of 1.2A. This phenomenon is called Back-EMF, and is the reason why motors will draw a huge current on startup, but not much when running normally (when you turn on your vacuum the lights dim for an instant).

  • 8
    \$\begingroup\$ This. The coil will have inductance too, but this, not inductance, is the main reason for the low current draw. Stall the motor and it'll still have inductance but the current draw will be MUCH higher. (Actually, don't do that...) \$\endgroup\$
    – user16324
    Mar 23, 2018 at 14:52
  • \$\begingroup\$ +1 for mentioning back EMF. I failed to mention that in my answer \$\endgroup\$
    – DerStrom8
    Mar 23, 2018 at 15:21
  • 1
    \$\begingroup\$ This experiment will not translate directly to the AC type pump motor in question but makes some sense of what is happening. \$\endgroup\$
    – KalleMP
    Mar 23, 2018 at 23:06

You're missing the reactance, which is the AC resistance (EDIT: And back-EMF - see comments). When you measure the resistance with a meter you're only measuring DC resistance and you're missing a significant part of the system.

Reactance comes from either capacitance, inductance, or a combination of the two. In the case of a motor most of the reactance will be inductive due to the inductor-like nature of the windings.

When using Ohm's Law in AC systems you use impedance instead of just resistance. Impedance, usually denoted Z, is a combination of the DC resistance and the AC reactance.

  • 1
    \$\begingroup\$ ...and which, in the case of a motor, depends on the speed and the torque. \$\endgroup\$ Mar 23, 2018 at 14:46
  • \$\begingroup\$ @jameslarge Yes indeed. If you jam the motor shaft then the reactance will drop significantly and you'll draw a heck of a lot more current. \$\endgroup\$
    – DerStrom8
    Mar 23, 2018 at 14:47
  • \$\begingroup\$ This same thing happens for DC motors, it's not only reactance \$\endgroup\$
    – C_Elegans
    Mar 23, 2018 at 14:50
  • 1
    \$\begingroup\$ Motors also have a back emf which is speed dependent. \$\endgroup\$
    – vini_i
    Mar 23, 2018 at 15:01
  • \$\begingroup\$ @C_Elegans Sure, but the current increases for a different reason. In a DC motor the "active" coil is determined by which connections are being made on the commutator at what time, and this constantly changes as the motor is running. Constantly changing which coil is connected limits the amount of time each coil is connected, so the average current remains low. If you stall a DC motor then only one coil is connected for as long as the shaft is stalled, and only the DC resistance limits the current which then increases drastically. \$\endgroup\$
    – DerStrom8
    Mar 23, 2018 at 15:02

Aside from the excellent answers on the differences with AC motors, the thing you need to understand is that what they were wanting from having you check the DC resistance would be to see if it was too LOW, which would indicate it was shorted out, or WAY TOO HIGH, as in an Open Circuit because of a broken conductor. Anything in between just meant it was NOT one of those obvious forms of failure.


The DC resistance of the winding conforms to Ohm's law perfectly, and if you actually and directly (without eg a commutator) fed that winding 120V DC, it would perfectly dissipate 80 watts of heat and perfectly go up in smoke, perfectly in accordance with Ohm's law.

The actual power draw is dominated by inductance - any power dissipated in the DC winding resistance is actually LOST, all it does is heat up the motor (there is a magnetic field built, but you would get just the same field from a lower voltage if the winding resistance was lower).

The inductance of the windings alters with motor load (energy conservation law has something do with it) - an idling motor (if the motor design is safe to idle - some are not!) might actually draw even LESS current than the nameplate says, while a heavily overloaded motor (say if you pumped molasses with that pump) will get closer to the above scenario - very little inductance will be in effect, and DC losses will dominate and eventually overheat the motor.


Ohm's law is not a fundamental law of nature.

It's just a law that some very specific components observe; we call those resistors.

Now, it so happens that quite a bunch of components that aren't specifically designed as resistors still behave as if they were resistors – but only under specific circumstances. In particular, simple homogeneous metal parts obey a local Ohm's law. That includes also the wire with which the coils of an electric motor are wound, which is the reason you can some sort of reading when using an Ohmmeter with the motor.

Nevertheless, the motor as a whole does not actually obey Ohm's law, because the wire is electromagnetically coupled to other stuff: in operation, there's a constantly changing magnetic field inside the motor, and such a field induces voltages in the coils. It is these voltages that dominate the electrical behaviour of the motor in any real use situation, not the voltage from Ohmic resistance.

Only if you let a small DC current flow through the coils, nothing actually moves in the motor, the magnetic field is everywhere constant, and since induction only depends on the time-variation of the magnetic field, you then get a much voltage reading that corresponds to the ohmic resistance of the wire alone. That's why your Ohmmeter shows such a small value.


There are two effects at work here: one is that the DC resistance may be \$15\Omega\$, but that does not tell you the impedance of the coil.

At 1.1A, 120V, and 80W, the \$\cos\phi\$ is 0.6 which does not actually correspond to \$15\Omega\$, so what gives? It turns out that the DC resistance as measured corresponds to a blocked motor. If the motor is allowed to turn (and to pump water), additional countervoltage is induced significantly lowering the amount of current that is actually flowing.

So you have both the difference between DC and AC impedances here, and the difference between blocked and rotating (though loaded) motor.

  • \$\begingroup\$ The DC resistance does not know the state of the fotor in an induction motor. The motor is not and will not try to turn. All you see if the copper resistance. When AC driven the rotor speed will affect the current, highest when stationary. We exclude the possibility of generating, though I don't think a single phase AC motor is easy to use for generation. \$\endgroup\$
    – KalleMP
    Mar 23, 2018 at 23:16

The manufacturer is stating the coil resistance so that you as the technician can determine the "health" of the motor winding/s. Each winding should be the same as the others (if 3 phase) and the same as the manufacturers specification. This as well as an insulation resistance test between each phase and earth and between phases should form part of any motor inspection regime to determine the servicability of the motor windings.


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