Induction motor currents

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cnh1995
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Suppose a 3-phase IM is operating on a 50Hz power supply. When the motor is loaded, more current is drawn from the stator because of the demagnetizing effect of the rotor current. But the rotor current has a lower frequency i.e. slip*stator frequency. Still the reflected current in the stator is of 50Hz frequency. What's the physics behind this?
 

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cnh1995
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I understand the operation of the motor. In transformers, when a load is connected to secondary, current flows in the secondary. This current tends to demagnetize the core and hence, primary current is increased to nullify the demagnetizing secondary ampere-turns, which gives the expression NpIp=NsIs. But all the currents in transformer have same frequency, say 50Hz. In IM, if rotor current has lower frequency than the magnetizing stator current, how does the stator current due to loading has the same 50Hz frequency?
 
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if rotor current has lower frequency than the magnetizing stator current, how does the stator current due to loading has the same 50Hz frequency?
Reread that paragraph in the wiki article again, carefully parsing each word. It is like a transformer yes, but a rotating transformer. Focus first on what happens when slip is zero and rotor currents are zero. A regular transformer with shorted secondary does not have zero current.

If you google more, you can find a time domain analysis or a detailed animation that shows one rotation.
 
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Averagesupernova
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The stator current frequency will always be the frequency of the source. Why would you expect any different?
 
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cnh1995
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The stator current frequency will always be the frequency of the source. Why would you expect any different?
Say the motor is operating on 50Hz. The rotor current demagnetizes the air gap and its frequency is less than the stator current frequency. So, demagnetizing mmf has less frequency than the magnetizing mmf. To cancel out the de-magnetizing mmf, the additional stator current drawn has also a frequency of 50Hz. That's why I was wondering how stator current of 50Hz cancels out the demagnetizing mmf of a lower frequency.
 
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Let's say it is about 3 phases induction motor and for easy explanation let's take a wound rotor. The stator windings dispersion along the air gap and the symmetrical delay between supply voltages [and current] produce a virtual ring of moving magnetic poles.
The velocity of moving poles will be then rpmsynchron=frequency*60/p.
This "ring" actually does not move.
Let's take now the wound rotor.The magnetic process is identical. The virtual velocity is rpmvirtual=slip*frequency*60/p [ since the rotor current frequency is slip*frequency, and the rotor
no. of pole pairs is the same] .But there is also an actual velocity of the rotor rpmactual=
(1-slip)*rpmsynchron
The total velocity of the virtual rotor ring of poles is then:
rpmvirtual+rpmactual=rpmsynchron.
That means both magnetic fields rotate synchronously.
 
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cnh1995
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Let's say it is about 3 phases induction motor and for easy explanation let's take a wound rotor. The stator windings dispersion along the air gap and the symmetrical delay between supply voltages [and current] produce a virtual ring of moving magnetic poles.
The velocity of moving poles will be then rpmsynchron=frequency*60/p.
This "ring" actually does not move.
Let's take now the wound rotor.The magnetic process is identical. The virtual velocity is rpmvirtual=slip*frequency*60/p [ since the rotor current frequency is slip*frequency, and the rotor
no. of pole pairs is the same] .But there is also an actual velocity of the rotor rpmactual=
(1-slip)*rpmsynchron
The total velocity of the virtual rotor ring of poles is then:
rpmvirtual+rpmactual=rpmsynchron.
That means both magnetic fields rotate synchronously.
Classic!:smile: This is the explanation I was looking for! The expressions were right in front of my eyes but I wasn't able to see how both the fields rotate synchronously! Thank you very much @Babadag!
 

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