**Phasor Diagram of an Induction Motor:**

Induction motor is analogous to the transformer. For the stator winding

V_{1} =- E_{1} + I_{1} (R_{1} + j X_{1} )

where V_{1} = Supply voltage per phase,

I_{1} = Stator current per phase,

R_{1} = Stator resistance per phase,

X_{1} = Stator leakage reactance per phase, and

E_{1} = Voltage induced per phase in the stator.

and also,

E_{1} = 4.44 φ f N_{1} kw_{1}

Where φ = Mutual flux per pole,

f = Stator frequency,

N_{1} = Stator winding turns per phase, and

kw_{1 }= Winding factor of stator.

We know that rotor winding is short circuited at the ends

V_{2} = 0 = E_{2 s }- I_{2} ( R_{2} + j X _{2 s} )

where I_{2} = Rotor current per phase,

R_{2} = Rotor resistance per phase, and

X_{2s} = Leakage reactance of rotor at slip frequency per phase.

X _{2s} = 2π f2 L2 ;

f_{2} = S f

X_{ 2} s = 2π S f_{2} L_{2}

X _{2} s = S X _{2}

where X_{2} = Leakage reactance of rotor as stand fill per phase, and

E_{2s} = Induced emf per phase in the rotor of induction motor.

E_{2 s} = 4.44 φ f_{2} N_{2} Kw_{2}

= S E_{2}

where E_{2} = Induced emf in rotor per phase at stand still,

N_{2} = Rotor winding turns per phase, and

Kw_{2} = Winding factor of rotor.

From the Eqs. (39) and (41), we have

E_{1}/ E_{2 }= N_{1} Kw_{1} /N_{2} Kw_{2} = N_{1}′ / N_{2}′ = a

where, N_{1}′ and N_{2}′ are the effective number of stator and rotor turns respectively and "a" is called the effective turns ratio of an induction motor.

| V_{1 }| ≈ | E_{1} |

Since the value of R_{1} and X_{1} are small.

The exciting current in the induction motor is specified by

I¯_{0} = I¯_{i } + ¯I_{m}

Where I_{m } is the per phase value of the magnetising component of the exciting current and I¯_{i} is the per phase value of the loss component of the exciting current.

The no load current in induction motor is more than that of transformer due to the air gap among the stator and rotor winding.

From the current and voltage equations discussed above, we may develop the phasor diagram.

The rotor power factor angle at any slip, S is

θ = tan ^{-1 }(S X _{2 }/ R_{2}

**Figure: Phasor Diagram of an Induction Motor for a > 1 and S < 1**

where

I_{2}′ = (N_{2}' /N_{1}′ )I_{2} = (1/a) I_{2}

and

I¯_{1} = I¯_{0} + I¯_{2}′