# Slip-ring Induction Motors-Starting of

20.2. Slip-ring Induction Motors-Starting of

(i) Rotor rheostat. The slip-ring induction motors are practically always started with full line voltage applied across the stator terminals. The value of starting current is adjusted by introducing a variable resistance in the rotor circuit. The controlling resistance is in the form of rheostat connected in star (Fig. 33), the resistance being gradually cut out of the rotor circuit as the motor gathers speed. By increasing the rotor resistance, not only is the rotor (and hence stator) current reduced at starting but at the same time torque is also increased due to improvement in power factor.

The rheostat is either of stud or contractor type and may be hand operated or automatic.

As discussed earlier, the introduction of additional external resistance in the rotor circuit enables slip-ring motor to develop a high starting torque with reasonably moderate starting current. Hence such motors can be started under load. When the motor runs under normal conditions the rings are short-circuited and brushes lifted from them . Fig. 33. Starting of slip-ring induction motor.

Example 20. The short-circuit current of a small 3-phase induction motor is 3 times the full-load currents. Determine the starting torque as a percentage of full-load torque if full-load slip is 3 per cent.

Solution. Short-circuit current, Ist = 3If

Full-load slip, sf = 3% or 0.03 Hence, starting torque is 27% of full-load torque. (Ans.)

Example 21. A 3-phase, squirrel-cage induction motor has a short-circuit current equal to 4 times the full-load current. Find the starting torque as a percentage of full-load torque if the motor is started by:

(i)                           direct switching to the supply;

(ii)                        a star delta starter;

(iii)                      an auto-transformer; and

(iv)                      a resistance in the stator circuit.

The stator current in (iii) and (iv) is limited to 2 times the full-load current and the full-load slip is 3%. .

Solution. Short-circuit current I­ = 4If

Full-load slip, sf=3% or 0.03

(i) Starting torque with direct switch, Example 22. A 400 V, 50-Hz, induction motor, when started directly from the mains takes 4 times the full-load current and the torque produced is twice the full-load torque. Determine:

(i)      The motor current, the line current and the starting torque when started by means of an auto- transformer of ratio 2.5: 1.

(ii)      The voltage to be applied and the motor current if the full-load torque is to be obtained at starting.

Solution. Supply voltage = 400 V

Frequency = 50 Hz   