Wound motor operates in the same manner as squirrel-cage motor when the external controller is set at zero resistance as the rotor will be short-circuited. Imagine now that the torque is carrying a given load because of the rotating of rotor at a constant speed. The rotor resistance will be doubled per phase as the controller is moved.
Reactions
Following are some reaction that will be observed for the mentioned example.
The current flowing in the rotor will switch to half of the previous value neglecting the transient effect
Rotor resistance along with the reactance increases in the same portion even on the condition when rotor factor remains same. The induction motor will develop the same value of torque which is rotating at a lower speed but carrying the same current as before. It happens because of the slid before the increase in the rotor circuit resistance.
So if you place more load on the induction motor, I will always deliver higher slip and rotor developing the required torque in the complete process. Talking about the situation when pullout torque will be attained” it will happen when slip will be twice as great as the shortened rotor. This is because of rotor resistance which is twice the value of the rotor alone. The value of the pull out torque, thus, will be same as of short circuited rotor.
Fig. 18 shows the torque-slip characteristic of the shorted rotor, together with that for a total resistance of, Rt = 2R.
It will not be worth if the torque is developed at the standstill position which is higher along with increase in rotor resistance. Why, if you ask? The fact is that the rotor frequency is equal to the line frequency which results into maximum rotor resistance.
In order to conclude the best results, the additional resistance in the rotor circuit will result into improved power factor in the circuit when the rotor current is reducing. The develop torque will be greater with the added resistance because of the standstill rotor resistance. The standstill rotor resistance is greater than its town resistance; it helps in increasing the power factor as compared to decrease in the rotor current resulting into great torque.
Examples:
Example 1. A 3-phase, 4-pole, 60-Hz induction motor is running at 1280 r.p.m. Determine the slip speed and slip.
Solution. Number of poles, p = 4
Frequency, f = 60 Hz
Actual speed of rotor, N = 1280 r.p.m.
Slip speed = ?
Example 2:
A 3-phase induction motor has 2 poles and is connected to 500V, 60Hz supply. Calculate the actual rotor speed and rotor frequency when the slip is 5%.
Solution:
Given, p = 2;
S = 5% = 0.05; f = 60Hz
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