Fig. 16 shows a wound-rotor induction motor with controller (rheostat).

If the external controller is set at zero resistance, the motor operates almost exactly the same as the squirrel-cage motor, since the rotor is short-circuited. Suppose now the rotor is rotating at a constant speed and developing torque to carry a given load. The controller is moved so that the rotor resistance per phase is doubled per phase. The following reactions will be observed:

- The current in the rotor falls to half of its previous value (neglecting a transient effect), since the rotor speed, and hence the rotor voltage, cannot change instantaneously.
- The developed torque decreases due to reduced rotor current.
- The load on the motor has not changed, and with the external applied, or load, torque now greater than the developed torque, the motor slows down.
- The slip increases (increasing the rate at which the stator flux cuts the rotor conductors).
- The rotor induced voltage increases.
- The rotor current now increases until it is sufficient to develop enough torque to again carry the load at a constant speed.

The induction motor now rotates at a lower speed, but with the same current as before, and develops the same torque as it slid before the rotor circuit resistance was increased. The frequency in the rotor is greater, as is the rotor resistance. Rotor resistance and reactance are both increased in same proportion, however, and the rotor power factor therefore remains the same. If more load is now placed on the shaft, the rotor will develop the required torque, but always at a lower speed i.e. a higher slip. Eventually pull-out torque is attained. The value of pull-out torque will be the same as with the short-circuited rotor, but it will occur at a slip which is twice as great as that with the shorted rotor, since the rotor circuit resistance is twice that of the rotor alone.

Fig. 18 shows the torque-slip characteristic of the shorted rotor, together with that for a total resistance of, R_{t} = 2R.

It is worthnoting that the torque developed at standstill is higher with increased rotor resistance. This is explained by the fact that at standstill the rotor frequency is equal to the line frequency and rotor resistance is maximum. The addition of resistance in the rotor circuit obviously improves the power factor in that circuit while reducing the rotor current. Since standstill rotor resistance is much greater than its resistance, the power factor increases more rapidly than the decrease of rotor current, and hence the developed torque is greater with added resistance.

**SLIP, STARTING TORQUE, MAXIMUM TORQUE**

**Example ****1***. A **3-phase, 4-pole, 50-Hz induction motor is running at 1440 r.p.m. Determine the slip speed and slip. *

**Solution**. Number of poles, p = 4

Frequency, f = 50 Hz

Actual speed of rotor, N = 1440 r.p.m.

**Slip speed ****= ****? **

**Example 7.** *A 3-phase induction motor runs at almost 1000 rpm at no-load and 950 r.p.m. at full-load when supplied with power from **a 50 Hz 3-phase line.*

*(i) **How many poles has the motor? *

*(ii) **What is the percentage slip at full load**? *

*(iii) **What is the corresponding frequency of rotor voltages?*

*(iv) **What is the corresponding speed of the rotor field with respect to rotor? *

*(v) **What is the corresponding speed of the rotor with respect to the stator? *

*(vi) **What **is the corresponding speed of the rotor field with respect to the stator field? *

*(vii) **What **is the rotor frequency at the slip of 10 percent **? *

**(AMIE, 1997) **

**Solution.** Given:

N_{0} = 1000 r.p.m.,

N_{f} = 950 r.p.m.,

f = 50 Hz

Since no-load speed of motor is almost 1000 r.p.m., hence synchronous speed near to 1000 r.p.m. is 1000 r.p.m.

**Example 9***.A 50 Hz, 440 V, 3-phase, 4-pole induction motor develops half the rated torque at 1490 r.p.m. With the applied voltage magnitude remaining at the rated value, what should be its frequency if the motor has to develop the same torque at 1600 r.p.m. **? **Neglect stator and rotor winding resistances, leakage reactances and iron losses.*

**(GATE, 1995) **

**Solution**. Given:

f = 50 Hz;

p = 4;

N = 1490 r.p.m. ;

New speed, N_{n} = 1600 r.p.m.

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