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I noticed that in deriving the equivalent circuit for an induction motor in the rotor component, that a "variable" resistance was introduced. This really confused me as I didn't understand how power was equivalent when they divided everything by slip in the derivation. Is there a back EMF being induced that forms the $$\frac{(1-s)}{s}R$$ portion of the circuit or something along those lines. Is this like an "induced resistance" or a voltage drop depending on slip that would form the component being converted to mechanical power?

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There is back EMF in an induction motor, but that concept is not used in the equivalent circuit of an induction motor as it is in a DC motor. The universally accepted equivalent circuit represents the motor as a transformer and the conversion of electrical to mechanical energy as a resistor with a value that varies with slip. The concept of back EMF is rarely used in discussing induction motors.

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Yes. The armature induces current in the rotor at \$s\$ times the line frequency. That current depends on slip, and the amount of coupling depends on slip, which is where the above relationship comes in.

(The rotor current sets up a magnetic field which induces back-emf in the armature. There's some lag, which is why induction machines look inductive.)

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  • \$\begingroup\$ Is this the reason for the drop in voltage other than just the resistor? Also is this back EMF*I = the developed power? \$\endgroup\$ – Colin Hicks Nov 6 '18 at 23:28
  • \$\begingroup\$ Hopefully clarified. What drop in voltage are you referring to? The back EMF is not directly related to the developed power, but when the motor is at a reasonable operating point the developed power increases as slip increases and back EMF decreases. \$\endgroup\$ – TimWescott Nov 7 '18 at 1:25

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