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.

V_{1} = E_{1} + I_{1} (R_{1} + jX_{1})

E_{r} = 12 (R_{2} + jsX_{2})

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:

- The magnetic leakage and leakage reactance of rotor: Higher in an induction motor than a transformer
- Magnetizing current: higher in the induction motor than the transformer because of the sir gap in the magnetic circuit.
- Ratio of stator and rotor currents and turns per phase in the rotor: equal in the transformer and not in induction motor
- Efficiency: Efficiency is lower in the induction motor than transformer because of higher losses at the induction end.

Fig. 21.Vector diagram of induction motor.

**18.1. Rotor output **

Talking about the primary current I_{1}, it consists of two parts:

- I
_{0} - I
_{2}’

which is later transferred into the rotor where I_{0} is used to meet the copper and the iron is lost in the stator itself.

Talking about the primary voltage V_{1}, some of the same is absorbed at the primary stage with remaining E_{1} transferring into the rotor.

If the angle between E_{2} and I_{2}’ is Φ

Then

Rotor input/phase = E_{1}H_{1}’ cos Φ

Total rotor Input = 3E_{1}I_{2}’ cos Φ

The electrical input to the rotor wasted in the form of the heat

= 3I_{2}E_{r}cos Φ (or 3I_{2}^{2} R_{2})

I_{2}’ = KI_{2} or I_{2} = I_{2}’/ K

E_{r} = sE_{2} and E_{2} = KE_{1}

**E _{r} = sKE_{1}**

**18.2. Equivalent circuit of the motor**

When the motor is loaded, the rotor current I_{2} is given by:

The formula illustrates that the rotor circuit consisting of the fixed resistance R_{2} along with the variable reactance sX_{2} connected across E; = sE_{2} is equivalent to the rotor circuit with fixed reactance X_{2} which is connected in the series along with variable resistance R_{2}/S and supplied with constant voltage E_{2 }is sown in the figure.

In additon to the same, the resistance R_{2}/s can also be illustrated as:

The formula consists of the two points:

- R2 represents the rotor copper less as the rotor resistance
- The second part is R
_{2}(1/s – 1) which is knows as load resistance RL.

RL is equivalent to mechanical load on the motor. We can also say that mechanical load is represented by non-inductive resistance [R_{2} (1/s-1)]