The equivalent circuit for a polyphase induction motor is shown in Fig. 24, where
V1 = applied voltage per phase
R1 = stator resistance/phase
R2 = rotor resistance/phase
X1 = stator leakage reactance/phase
X2 = rotor standstill leakage reactance/phase
K = turn-ratio of secondary to primary
R0 = no-load resistance/phase
X0 = no-load reactance/phase
Fig. 24. Equivalent circuit of an induction motor.
As shown in Fig. 25 the exciting circuit may be transferred to the left, because inaccuracy involved is negligible but the circuit and hence the calculations are very much simplified. This is known as the approximate equivalent circuit of the induction motor.
Maximum Power Output. Refer Fig. 25.
EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR
Example 16. In a 3-phase induction motor maximum torque occurs at a slip of 10 per cent. The equivalent secondary resistance of the motor is 0.09 Ω/phase. If the gross power output is 8.5 II W. Calculate.-
(i) Equivalent load resistance,
(ii) Equivalent load voltage, and
(iii)Current at this slip.
Solution. Slip at maximum torque, s = 10 per cent
The equivalent secondary resistance of the motor,
R2 = 0.09 Ω/phase
Gross power output, Pg = 8.5 kW or 8500 W