The equivalent circuit for a polyphase induction motor is shown in Fig. 24, where

V_{1} = applied voltage per phase

R_{1} = stator resistance/phase

R_{2} = rotor resistance/phase

X_{1} = stator leakage reactance/phase

X_{2} = rotor standstill leakage reactance/phase

K = turn-ratio of secondary to primary

R_{0} = no-load resistance/phase

X_{0} = 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,

R_{2} = 0.09 Ω/phase

Gross power output, P_{g} = 8.5 kW or 8500 W

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