It will not be wrong to say that induction motor is an essential transformer where the stator forms the primary and rotor forms the short circuited rotating secondary. How is it possible?
It is possible because of the transfer of the stator to the rotor. This takes place in the induction motor with the help of flux which links the two on mutual basis.
Fig. 20.Induction motor as transformer.
The vector diagram (Fig. 21) is similar to that of a transformer.
V1 = E1 + I1 (R1 + jX1)
Er = 12 (R2 + jsX2)
Though an induction motor might act as a transformer, but still are different from each other. Following are few points illustrating the difference between the two:
Fig. 21.Vector diagram of induction motor.
18.1. Rotor output
Talking about the primary current I1, it consists of two parts:
which is later transferred into the rotor where I0 is used to meet the copper and the iron is lost in the stator itself.
Talking about the primary voltage V1, some of the same is absorbed at the primary stage with remaining E1 transferring into the rotor.
If the angle between E2 and I2’ is Φ
Rotor input/phase = E1H1’ cos Φ
Total rotor Input = 3E1I2’ cos Φ
The electrical input to the rotor wasted in the form of the heat
= 3I2Ercos Φ (or 3I22 R2)
I2’ = KI2 or I2 = I2’/ K
Er = sE2 and E2 = KE1
Er = sKE1
18.2. Equivalent circuit of the motor
When the motor is loaded, the rotor current I2 is given by:
The formula illustrates that the rotor circuit consisting of the fixed resistance R2 along with the variable reactance sX2 connected across E; = sE2 is equivalent to the rotor circuit with fixed reactance X2 which is connected in the series along with variable resistance R2/S and supplied with constant voltage E2 is sown in the figure.
In additon to the same, the resistance R2/s can also be illustrated as:
The formula consists of the two points:
RL is equivalent to mechanical load on the motor. We can also say that mechanical load is represented by non-inductive resistance [R2 (1/s-1)]
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