Equivalent Circuit of Three Phase Induction Motor

EQUIVALENT CIRCUIT

Similar to a transformer, an induction motor can be represented by an equivalent electrical circuit. Equivalent circuit enables the performance characteristics of the induction machine to be evaluated for steady state condition by ordinary processes of AC network solution.

In the transformer the load on the secondary is electrical, whereas in the induction motor, the load is mechanical. However, in an equivalent circuit of the induction motor, the mechanical load can be replaced by a pure resistance. The equivalent circuit of an induction motor is drawn for only one phase.

As in the case of a transformer, in this case also, the secondary values may be transferred to the primary and vice versa.

Derivation of Equivalent circuit

1. Stator Side

In the stator side V₁ = E₁+I₁(R₁+jX₁) where V is the supply voltage and E₁, is the emf induced in the stator winding as shown in Fig. 5.14 (a).

We know that I_μ lags behind V₁ by 90° and so it is presumed to flow through an assumed inductive reactance X₀ ohms. Similarly the loss component of the no‒load current, I_w is assumed to flow through a fictitious resistance R₀.

Hence the above components of current may be represented as in Fig. 5.14 (b). The Sum of these 2 currents is I₀ known as no‒load current.

2. Rotor Side

When the motor is loaded (under running conditions) I₂ is given by

Rotor current, I₂ = E_r / Z_r = rotor emf/ϕ under running conditions / rotor impedance/ϕ under running conditions

From the above relation, it appears that the rotor circuit consists of a fixed resistance, R₂ and a variable reactance sX₂ (proportional to slip) connected across E_r = sE₂ as shown in Fig. 5.15 (a). This can also be stated that the rotor circuit is having fixed reactance X₂ connected in series with a variable resistance R₀/s (inversely proportional to slip) and supplied with constant voltage E₂ as shown in Fig. 5.15 (b).

Now the resistance, R₂/s = (R₂+ R₂/s) − R₂

 R₂/s = R₂+ R₂ [(1/s) ‒ 1]

It consists of 2 parts,

• The first part R₂ is the rotor resistance itself and represents the copper loss.

• The second part is R₂[(1/s) ‒ 1]. This is known as the load resistance R_L and is the electrical equivalent of the mechanical load on the motor. The equivalent circuit of the rotor along with the load resistance R_L may be drawn as in Fig. 5.15 (c).

As in the case of transformer, in this case also the secondary values may be transferred to primary and vice versa. When shifting impedance / resistance from secondary to primary, it should be divided by K², whereas current should be multiplied by K where K is the transformation ratio.

The approximate equivalent circuit is obtained by shifting the exciting circuit to the left. The calculations with this modified circuit is also simplified.