Equivalent Circuit of Induction Motor : Part 1

We have already seen that the induction motor can be treated as generalized transformer. Transformer works on the principle of electromagnetic induction. The induction motor also works on the same principle. The energy transfer from stator to rotor of the induction motor takes place entirely with the help of a flux mutually linking the two. Thus stator acts as a primary while the rotor acts as a rotating secondary when induction motor is treated as a tranformer.

If E₁ = Induced voltage in stator per phase

E₂= Rotor induced e.m.f. per phase on standstill

k = Rotor turns / Stator turns

then k = E₂/ E₁

Thus if V₁ is the supply voltage per phase to stator, it produces the flux which links with both stator and rotor. Due to self induction E₁, is the induced e.m.f. in stator per phase while E₂ is the induced e.m.f. in rotor due to mutual induction, at standstill. In running condition the induced e.m.f. in rotor becomes E_2rwhich is s E₂.

Now E_2r= Rotor induced e.m.f. in running condition per phase

R₂= Rotor resistance per phase

X_2r= Rotor reactance per phase in running condition

R₁= Stator resistance per phase

X₁= Stator reactance per phase

So induction motor can be represented as a transformer as shown in the Fig. 1.

Fig. 1 Induction motor as a transformer

When induction motor is on no load, it draws a current from the supply to produce the flux in air gap and to supply iron losses.

1. I_c= Active component which supplies no load losses

2. I_m= Magnetizing component which sets up flux in core and air gap

These two currents give us the elements of an exciting branch as,

R_o= Representing no load losses = V₁/I_c

and X_o= Representing flux set up = V₁/I_m

Thus, Ī_o = Ī_c + Ī_m

The equivalent circuit of induction motor thus can be represented as shown in the Fig. 2.

Fig. 2 Basic equivalent circuit

The stator and rotor sides are shown separated by an air gap.

I_2r= Rotor current in running condition
= E_2r/Z_2r = (s E₂)/√(R₂² +(s X₂)²)

It is important to note that as load on the motor changes, the motor speed changes. Thus slip changes. As slip changes the reactance X_2r changes. Hence X_2r= sX₂is shown variable.

Representing of rotor impedance :

It is shown that, I_2r= (sE₂)/√(R₂² +(s X₂)²) = E₂/√((R₂/s)² + X₂²)

So it can be assumed that equivalent rotor circuit in the running condition has fixed reactance X₂, fixed voltage E₂but a variable resistance R₂/s, as indicated in the above equation.

Now R₂/s = R₂+ (R₂/s) - R₂

So the variable rotor resistance R₂/s has two parts.

1. Rotor resistance R₂ itself which represents copper loss.

2. R₂(1 - s)/s which represents load resistance R_L. So it is electrical equivalent of mechanical load on the motor.

Key Point : Thus the mechanical load on the motor is represented by the pure resistance of value R₂(1 -s)/s.

So rotor equivalent circuit can be shown as,

Fig. 3 Rotor equivalent circuit

Now let us obtain equivalent circuit referred to stator side.

Equivalent circuit referred to stator :

Transfer all the rotor parameters to stator,

k = E₂/E₁ = Transformation ratio

E₂' = E₂/ k

The rotor current has its reflected component on the stator side which is I_2r'.

I_2r' = k I_2r = (k s E₂ )/√(R₂² +(s X₂)²)

X₂' = X₂/K² = Reflected rotor reactance

R₂' = R₂/K² = Reflected rotor resistance

R_L' = R_L/K² = (R₂/K²)(1-s / s)

= R₂' (1-s / s)

Thus R_L' is reflected mechanical load on stator.

So equivalent circuit referred to stator can be shown as in the Fig. 4

Fig. 4 Equivalent circuit referred to stator

The resistance R₂' (1 -s)/ s = R_L' is fictitious resistance representing the mechanical load on the motor.

1.1 Approximate Equivalent Circuit

Similar to the transformer the equivalent circuit can be modified by shifting the exciting current (R_o and X_o) purely across the supply, to the left of R₁ and X₁. Due to this, we are neglecting the drop across R₁ and X₁due to I_o, which is very small. Hence the circuit is called approximate equivalent circuit. The circuit is shown in the Fig.5.

Fig. 5 Approximate equivalent circuit

       Now the resistance R₁ and R₂' while reactance X₁ and X₂' can be combined. So we get,
                 R_1e = Equivalent resistance referred to stator = R₁ + R₂'
                   X_1e = Equivalent reactance referred to stator = X₁ + X₂'
                   R_1e = R₁+ (R₂/K²)
and              X_1e = X₁ + (X₂/K²)
While Ī₁ = Ī_o + Ī_2r'               .........phasor diagram
and               Ī_o = Ī_c + Ī_m
       Thus the equivalent circuit can be shown in the Fig.6.

Fig. 6

Power Equations from Equivalent Circuit

With reference to approximate equivalent circuit shown in the Fig. 6, we can write various power equations as,

                   P_in = input power = 3 V₁I₁cos Φ
       where    V₁ = Stator voltage per phase
                   I₁= Current drawn by stator per phase
                 cos Φ = Power factor of stator
       Stator core loss = I_m²R_o
       Stator copper loss = 3 I₁² R_o
where             R₁= Stator resistance per phase
                      P₂ = Rotor input = (3 I_2r'²R₂')/s
                      P_c = Rotor copper loss = 3 I_2r'²R₂'
Thus               P_c = s P₂
                      P_m = Gross mechanical power developed

T = Torque developed

where N = Speed of motor
But N = N_s (1-s) =, so substituting in above

and                 I_2r' = V₁/(( R_1e + R_L') + j X_1e )
where             R_L' = R₂' (1-s)/s
                     I_2r' = V₁/√(( R_1e + R_L')² + X_1e² )
Key Point : Remember that in all the above formula all the values per phase values.

See part 2

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