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I have been learning about Ohm's law and testing the resistance across the plug of my household appliances and calculating the current.

For example, my kettle was 22 ohms (10.45 amperes) and is protected by a 13 A fuse.

This makes sense, and I'm okay with it, but then I tested the vacuum cleaner which had a resistance of 7.7 ohms which equates to 29.8 amperes which surely should blow the 13 A fuse, but it doesn't. I have now tested two different vacuum cleaners which have the same small resistance reading across the live and neutral.

Surely this would be a direct short, but it works fine so does the resistance change or what?

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    \$\begingroup\$ It's a big step taking ohms law in dc circuits and trying to apply it in ac circuits. How are you with complex numbers and reactances? \$\endgroup\$ – Andy aka Mar 24 '17 at 16:42
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    \$\begingroup\$ And also motor effects. \$\endgroup\$ – Trevor_G Mar 24 '17 at 16:44
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    \$\begingroup\$ Quick answer: Motors are not simply resistors, they also have inductive characteristics. \$\endgroup\$ – The Photon Mar 24 '17 at 16:44
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    \$\begingroup\$ The resistance only gives you information about the DC-current. Also keep in mind that this resistance might be current dependant. If you have ac circuits it depends on your impedance, thats where capacitors and inductance comes into the game. Edit: Important for you to note is that impedance is also frequency dependant \$\endgroup\$ – Felix Crazzolara Mar 24 '17 at 16:47
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    \$\begingroup\$ Good question by the way. The answer isn't always obvious until you've dealt with motors at an electrical level! \$\endgroup\$ – Cort Ammon Mar 24 '17 at 19:05
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The 7.7 ohms you measured is the winding resistance of the motor. But that is not the only factor that determines its operating current.

Your vacuum cleaner might draw close to the calculated 30A the instant power is applied, but as soon as the motor starts to rotate, it generates a voltage that is proportional to speed (called back emf) that opposes the applied voltage, decreasing the net voltage available to drive current through the windings. As the motor speed increases, the current (and therefore the torque produced by the motor) decreases, and the speed settles at the point where the torque produced by the motor matches the torque required to drive the load at that speed.

Fuses don't blow instantly. But if you were to lock the motor so it couldn't rotate, that fuse wouldn't last long.

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  • \$\begingroup\$ Sometimes the vacuum accidentally gets ahold of something that it shouldn't (like throw rugs) which prevents the brush (and presumably, from the sound, the motor) from spinning. Yet, I've never had this trip anything... \$\endgroup\$ – Michael Mar 24 '17 at 20:47
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    \$\begingroup\$ @Michael: I was referring to the main motor that drives the fan or impeller that creates the suction. Typically the brush is driven by a separate, smaller motor. Since the brush is reasonably expected to jam occasionally, that motor would be designed to tolerate that condition without drawing excessive current. \$\endgroup\$ – user28910 Mar 24 '17 at 21:00
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    \$\begingroup\$ Ah, and now I understand why my lights dim momentarily when I start a vacuum cleaner but they recover pretty much instantly. \$\endgroup\$ – fluffy Mar 25 '17 at 7:00
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    \$\begingroup\$ So I assume that the fact that my vacuum cleaner trips the breaker when I plug it into a particular socket just means that socket is already overloaded (distribution board)? \$\endgroup\$ – ArtOfCode Mar 26 '17 at 18:41
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    \$\begingroup\$ @MichaelKjörling which is why they always tell you not to plug your laser printer into a UPS! You can (and should) plug it into a surge protected outlet on the UPS, but not the powered side. I suppose, though, that if the UPS has enough capacity to run the PC, monitor(s), and everything else, and not give up when the laser fires up, there's no real reason not to (other than the screaming bloody murder). \$\endgroup\$ – FreeMan Mar 27 '17 at 13:43
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A vacuum cleaner isn't a resistor, and the line voltage from the outlet isn't DC (Direct Current). Ohm's law applies to resistors and DC. Ohm's law doesn't directly apply to your motor connected to an AC (Alternating Current) source.

For motors, you need to look into the rules for alternating current and inductors. They are far more applicable to your case.

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    \$\begingroup\$ Ohm's Law also works with purely resistive (and incandescent) loads on AC. That's why AC is RMS: when they switched from 110VDC to 110VAC, they chose the voltage that would make heaters and bulbs run right. Not motors obviously. Can't put AC on a DC motor. \$\endgroup\$ – Harper Mar 24 '17 at 17:47
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    \$\begingroup\$ @Harper Ohm's Law always holds true, but you need to use impedance when talking about AC, not just resistance. \$\endgroup\$ – DerStrom8 Mar 24 '17 at 18:12
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    \$\begingroup\$ Again, the rules for inductors and capacitorsare relevant but unimportant in a motor - back EMF ( the motor acting as a generator and cancelling out most of the applied voltage) is what matters here. \$\endgroup\$ – Brian Drummond Mar 24 '17 at 18:26
  • \$\begingroup\$ > back EMF ... Bingo! \$\endgroup\$ – dannyf Mar 24 '17 at 18:54
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    \$\begingroup\$ @DerStrom8 If you take Ohm's Law as the definition of resistance, then it is always true (by definition), but still useless for some devices that have a constantly changing resistance. \$\endgroup\$ – immibis Mar 25 '17 at 8:56
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"Resistance" is for DC circuits. While resistance still plays a role in AC, there is also another characteristic for AC circuits called "Reactance", which is effectively just resistance to alternating current. "Reactance" is provided by inductance and capacitance and changes with frequency, per the following formulas:

$$X_L = 2\pi f L$$ $$X_C = \frac{1}{2\pi f C}$$

where \$X_L\$ is inductive reactance (in ohms), \$X_C\$ is capacitive reactance (in ohms), \$f\$ is frequency (in Hertz), \$L\$ is inductance (in Henrys) and \$C\$ is capacitance (in Farads).

Together, resistance and reactance (whether inductive or capacitive) become a complex number of the form

$$Z = R \pm jX$$

where \$R\$ is the resistance, \$j\$ is an imaginary number (\$\sqrt{-1}\$), and \$X\$ is the reactance. The resulting complex number is called "impedance", denoted by the letter \$Z\$, which affects current draw of your device. You can use \$Z\$ in place of \$R\$ anywhere in Ohm's law and it will work, but you must do the math properly with the complex numbers. It is a bit more difficult, however, because there is a lot more to a motor than just inductance, for example. The windings themselves have capacitance and resistance, so it may be difficult to find all of the necessary variables in order to accurately calculate current.

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    \$\begingroup\$ Reactance isn't that important in a motor, back EMF is. \$\endgroup\$ – Brian Drummond Mar 24 '17 at 18:21
  • \$\begingroup\$ back EMF is certainly important, but if you're trying to determine the current draw then you can't ignore reactance. \$\endgroup\$ – DerStrom8 Mar 24 '17 at 18:52
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"So does the resistance change or what?"

The short answer is yes...

Long answer is a lot more complex, but I won't confuse you with the details.

In a nut-shell, your vacuum cleaner has magnetic coils in it. Coils and especially motors are complex loads, not simply resistive like your kettle. These loads are especially sensitive to AC power. That produces an "effective resistance" which is MUCH, MUCH larger than the DC resistance you measure with your multi-meter.

And yes, you didn't ask yet, but when you first switch it on, the initial current surge can be LARGE.

However, the effective resistance very rapidly increases as the motor starts. The appliance is designed so that surge is very short, short enough that the fuse does not have time to heat up and melt.

Though in some countries, like most of North America, you may notice the lights on the same circuit dim briefly when you turn on the "hoover".

Stalling the motor CAN however, create some beefy currents. When you catch the edge of that carpet with the vacuum and the motor starts whining... turn it off.

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  • \$\begingroup\$ When you catch it with the rotor?! \$\endgroup\$ – Gregory Kornblum Mar 24 '17 at 18:31
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Motors create a voltage opposed to the source, Back EMF. So Ohm's law works, but it's not just resistance and source voltage in the equation.

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Why does Ohm's law not hold for vacuum cleaners?

It doesn't work for essentially the same reason Newton's laws don't work for resistors (if you apply a force \$F\$ to a resistor of mass \$m\$ which happens to be soldered into a circuit, the resistor will not accelerate with \$a = F/m\$, as the solder joints will hold it back). Or, as an even more absurd example, for the same reason Asimov's laws of robotics don't work for celestial bodies.

All laws, certainly all physical laws, only work for a particular, well-defined setting. Ohm's law (in its simplest form, which is what a multimeter assumes) works for idealised resistors. It so happens that a water kettle behaves pretty much like an idealised resistor, and obviously the resistors you use electronic circuits do as well. But a priori, there's absolutely no reason to think a given, unknown component should obey Ohm's law, like there's no reason to assume that Kepler's laws of planetary motion should hold for your water kettle.

Only in a few cases, one finds out that a law which works for some physical object A turns out to also work for a completely different object B. Those incidences are the really exciting moments in physics, like when Einstein proposed that Lorentz invariance, which was first only known as a property of Maxwell's laws of electrodynamics, also holds for massive bodies. That this unwarranted prediction turned out to be true is what makes relativity theory a proper physical theory, as opposed to just some law – like Ohm's law, which is just a description of what, well, resistors do.


Well, on a level Newton's laws do of course work for resistors: if you apply a force to a that resistor, it will very briefly accelerate until the solder joints apply a counter-force holding it back. All forces together, Newton's law is then again fulfilled. Similarly, even a vacuum cleaner may actually in a generalised sense fulfill Ohm's law, if you consider the motor's inductances as extra (imaginary) impedances/reactances. Those are just not visible to your multimeter, much like the solder joints holding your resistor down aren't visible to the guy who weighed it before you included it in the circuit.

Even that is not completely true though: in fact resistance depends on temperature, which is also influenced by the current; and there are more tricky effects like Johnson noise. In a sufficiently pedantic sense, resistors do thus not obey Ohm's law!

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  • \$\begingroup\$ And sadly, non-Newtonian resistors are rather expensive. :) \$\endgroup\$ – Wossname Mar 27 '17 at 7:35
  • \$\begingroup\$ Ohms law works for everything. It only applies to the resistive part. Everything has such. It's VERY large in some cases. Very. And when the resistance changes Ohms law still applies but the result varies as the resistance changes. \$\endgroup\$ – Russell McMahon Mar 27 '17 at 8:45
  • \$\begingroup\$ @RussellMcMahon no, it doesn't work for everything. Sure, you can for anything measure some voltage / current pairs, perform a simple linear fit and call the linear coefficient “the resistance”. That's essentially what a decent multimeter does. But generally, the result is not well-defined, it will depend substantially on how you choose the boundary conditions of the measurement; only for linear systems can you actually specify a procedure that always gives consistent results corresponding to the resistive part. \$\endgroup\$ – leftaroundabout Mar 27 '17 at 11:59
  • \$\begingroup\$ ...For nonlinear components, the best you can do is consider the “local Ohm's law” \$\frac{∂U}{∂I} = R\$ in the low-frequency limit, but nontrivial circuits have no real equilibrium state so you will never get consistent results. \$\endgroup\$ – leftaroundabout Mar 27 '17 at 11:59
  • \$\begingroup\$ @leftaroundabout - Please to see my answer written a while ago. I still state that it applies to everything BUT I agree with what you say for practical purposes. My point (see answer) is that it works essentially by definition for what it applies to, even though this may not be overly useful. From my "tower bridge resistance" example: " ... It's probably immensely large, varies continually and is not an overly useful measure of anything. ... When the resistance of an object changes Ohms law still applies but the result varies as the resistance changes." as you say \$\endgroup\$ – Russell McMahon Mar 27 '17 at 12:49
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Ohm's law can be considered either to be an exact relationship when dealing with ideal resistors, or an approximation when dealing with non ideal resistors or part of an overall set of 'laws" when dealing with resistors plus "something else" or with resistors which are significantly affected by their environment in some way.

Ohm's law always applies to the things it's meant to apply to -
ie to pure invariant resistors.

If it doesn't work for a 'thing' then the thing is not a pure invariant resistor.
It may be

  • a resistor plus inductance, or
  • a resistor affected by voltage applied or
  • current or
  • electric field or
  • thermal effects or
  • ... .

In the case of a vacuum cleaner motor you "see" a field inductor, plus a rotor inductor, plus the resistance of both those plus some wiring resistance. The applied AC tends to be more affected by the inductance than by the resistance.

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The following apparently stupid and pedantic statements (which may in fact be stupid and pedantic :-) ), still happen to explain the overall real-world situation well:

  • Ohms law works for everything.
  • It only applies to the resistive part.
  • Everything has a resistive part.
  • It's VERY large in some cases. eg Tower Bridge in London England has a resistance which can be measured from two chosen points at either end. It's probably immensely large, varies continually and is not an overly useful measure of anything.

  • When the resistance of an object changes Ohms law still applies but the result varies as the resistance changes.

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  • \$\begingroup\$ "Everything has a resistive part" Though some people work very hard to find ways to make materials that, um, resist this statement... \$\endgroup\$ – a CVn Mar 27 '17 at 8:53
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A motor has coils and hence posses inductance. Inductance always tries to oppose the cause producing it by back emf. Motor also has the ability to rotate. Hence the motor turns in a direction which opposes the change in magnetic field or currect due to ever changing alternating current.

Hence the AC current is obstructed both by the back emf and the turning of motor. Thus although resistance is small the obstruction to current flow is high. This is the reason that the current drawn is very high when the motor is jammed and while it is starting (inistially at rest hence no rotation to block changing current).

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