# Objective Questions on Induction Motor | 6

1. A 4 pole, 50 Hz, 3 phase induction motor runs at a speed of 1470 rpm the frequency of rotor electric current is ___________ Hz.

Frequency of rotor current, f2 = s × f = 0.02 × 50 = 1 Hz.

2. What are the starters used for starting of cage motors?

For starting of cage motors 3 starters are used - direct on line ( DOL ) starter, star - delta starter and auto transformer starter.

3. What is the relation between rotor copper loss and full load slip?

Rotor Cu loss / rotor output = s / ( 1 - s ).
Where s = full load slip.

4. 6 pole, 50 Hz 3 phase IM running at full load develops torque of 150 Nm at a rotor frequency of 5.0 Hz. Calculate the slip?

Slip S = rotor frequency / supply frequency = 5.0 / 50 = 0.1

5. A 4 pole 50 Hz 3 phase induction motor has a full load slip of 5 % the full load speed is

N = Ns( 1 - s) = (120 × f / p ) × ( 1 - s) = 1425 rpm

6. A 3 - Φ, 4 pole 50 Hz is running at 1450 rpm under full load. The full load current is 22 Amp, and the rotor resistance at standstill is 2 Ω. Calculate the mechanical power developed?

Ns = 1500 rpm
Nr = 1450 rpm
s = ( Ns - Nr ) / Ns = ( 1500 - 1450 ) / 1500 = 0.03
Pm = I22r2 × [ ( 1 - s ) / s] = (22)2 × 2 × ( 1 - 0.03 ) / 0.03 = 31299 Watt.

7. A 3 - Φ, 4 pole, 50 Hz IM is running at 1440 rpm when energized at rated voltage and frequency, the rotor induced voltage at standstill is equal to 120 Volt. Calculate the starting torque if per phase rotor resistance is 0.2 Ω and per phase leakage reactance at standstill is 1 Ω?

Ns = (120 × 50 ) / 4 = 1500 rpm
Rotor starting current,
I2.st = E2 / √( r2 + x2 ) = 120 / 1.0198 = 117.67 Amp
ωs = ( 2πNs) / 50 = 50π rad/sec.
∴ Starting torque = 3 / ( ωs × I2.st2 ) × r / 1 [ s = 1 at standstill] = 52.9 Nm.

8. The direction of rotation of a 3 phase induction motor is clockwise. When it is supplied with 3 phase sinusoidal voltage having phase sequence A-B-C. For counter clock wise rotation of the motor the phase sequence of the motor the phase sequence of the power supply should be

To reverse direction of rotation, phase sequence of the supply has to be reversed. Direction for clockwise direction, the phase sequence was A - B - C for counter clockwise direction, the phase sequence has to be A - C - B.

9. A 3-phase 412 V, 6 pole IM is fed from 50 Hz supply. It runs at a speed of 850 rpm then the rotor frequency is

Ns = 120f / P = 120 × 50 / 6 = 1000 rpm
Then slip = ( Ns − N ) / Ns = ( 1000 − 850 ) / 1000 = 0.15 = 15 %
Rotor frequency fr = sfs = ( 0.15 × 50 ) = 7.5 Hz.

10. If rotor power output 3 phase induction motor is 15 KW. Then rotor copper losses at a slip of 4 % will be

Here rotor copper loss = sPg ( s = slip, Pg = output power of motor)
⇒ ( 0.04 × 15000 ) = 600 Watt.

11. The stator loss of a 3 - Φ induction motor is 2 kW. What will be the rotor copper loss if the motor is running with a slip of 4 % and power input 90 kW?

s = 0.04, Stator input = 90 kW, Stator loss = 2 kW
Now, Rotor input = stator output = stator input - stator loss = 90 - 2 = 88 kW
∴ Rotor copper loss = s × rotor input = 0.04 × 88 = 3.52 kW.

12. The stator output of a 3 - Φ induction motor is 59 kW. The stator losses total 1 kW, total rotor copper loss 1.77 kW. What is the mechanical power developed?

Total rotor copper loss = 1.77 kW
Rotor input = stator output = 59 kW
∴ Mechanical power developed = rotor input - rotor copper loss = 59 - 1.77 = 57.23 kW.

13. If the electromotive force in the stator of an 8-pole induction motor has a frequency of 50 Hz and in the rotor 1.5 Hz. What is the slip?

Stator frequency = 50 Hz
Rotor frequency = 1.5 Hz
Slip = rotor frequency / stator frequency = 1.5 / 50 = 0.03.

14. A 3 - Φ, 4 pole 50 Hz induction motor has rated output of 10 kW at 1425 r.p.m. Calculate the full load torque?

Slip s = ( 1500 - 1425 ) / 1500 = 0.05
Power output = 10000 W
∴ Full load torque = P / { ωs( 1 - s ) } = ( 10000 × 60 ) / { 2π × 1500 ×( 1 - 0.05 ) } = 67 Nm

15. If a 3 - Φ slip ring induction motor is fed from the rotor side with stator winding short circuited, then frequency of the electric current flowing in the short circuited stator is

Induction motor is similar to a transformer with secondary short-circuited if supply is supplied across primary or primary short-circuited if supply is applied across rotor, then flux which is revolving in nature is developed. It sweeps pass the stator surface and cuts conductor of stator and so emf is induced in the stator which is revolving at natural but lesser speed than the rotor. So due to this relative speed, the rotor starts to rotate. So frequency of stator at the above specific condition is slip frequency.

16. A 3 - Φ 400 V star connected induction motor has a star connected rotor with a stator to rotor turn ratio 6.5. The rotor resistance and standstill reactance per phase are 0.05 Ω and 0.25 Ω respectively. The value of external resistance, which are inserted in rotor circuit to obtain the maximum torque will be

Here K = 1 / 6.5 because transformation ratio ( K ) is defined by,
Stand still rotor emf / phase E2 = 400 / √3 × 1 / 6.5 = 35.5 Volt
We know that starting torque is maximum when R2 = X2 i.e. when R2 = 0.25 Ω in the present case.
∴External resistance / phase required = 0.25 - 0.05 = 0.2 Ω.

17. A three phase, 400 V, 50 Hz supplies power to a 3 - phase balanced motor load If the line electric current be 25 A, power factor being 0.85, what is the value of real and reactive power consumed?

Real power = √3 × VLILcosφ = √3 × 400 × 25 × 0.85 = 14.72 kW and Reactive power = √3 × VLILsinφ = √3 × 400 × 25 × 0.53 [ since sinφ = sin(cos − 10.85) = 0.53 ] = 9.123 kVAR.

18. An induction motor running at a slip of 0.01 is operating from 50 Hz supply, if rotor inductor is one henry, then reactance of the rotor at given slip would be _____________ ohms .

For an IM at running condition we know, Xr = sXs = 0.01 × 2 × π × 50 × 1 = π.

19. A 400 V, 50 Hz, 30 Hp, 3 phase induction motor is drawing 50 an electric current at 0.8 pf lagging. The stator and rotor copper loss is 1050 watt and the core losses are 1200 W respectively. The air gap power of the motor will be

Motor input power in stator = √3 × VIcos(φ) = √3 × 400 × 50 × 0.8 = 27.71 KW. Air gap power = motor input to stator - stator copper loss - core loss = ( 27.71 - 1.5 - 1.2 ) = 25.01 KW.

20. A 6 pole 50 Hz wound rotor induction motor when supplied at the rated voltage and frequency with slip ring open circuited, developed a voltage of 100 V between any two rings. If the rotor is driven by an external. Means at 1000 rpm opposite to the direction of stator field, the frequency of voltage across slip ring will be

Ns = 120 × 50 / 6 = 1000 rpm
( Where Ns synchronous speed )
Rotor is driven in opposite direction, hence slip is = ( Ns + Nr ) / Nr = ( 1000 + 1000 ) / 1000 = 2.
Frequency of voltage across the slip rings, f' = sf = 2 × 50 = 100 Hz.