Last Updated on February 11, 2022 by Electricalvolt

**Three Phase Induction Motor Interview Questions & Answers**-Three phase induction motor is very popular because it is rugged in construction. It needs very less maintenance. Read three phase induction motor interview questions & answers for the better understanding of induction motor.

**What is a magnetizing current in an induction motor?**

The magnetizing current is essential for setting up a magnetic field in the air gap of the motor.

Without magnetizing current, the magnetic field can’t be generated, and the motor operation is not possible. The stator winding is inductive in nature and when applied AC to the stator, it draws current that lags the applied voltage.

**What happens if an induction motor runs below its rated voltage?**

If the induction motor runs below its rated voltage, the flux will reduce and as a result, the torque delivering capacity of the motor will get reduced.

If the motor drives the same load at reduced voltage, it will draw more current and will get overload tripped.

**What are the applications of the squirrel cage and the slip ring induction motor?**

The slip ring induction motor is used for driving the loads which demand higher starting torque. The starting torque of the motor gets increased because of the increased rotor resistance

The external resistance is added to the rotor conductor and the resistance is reduced as the motor is accelerated. When the motor accelerates up to its base speed the external resistance is short-circuited and the motor is equivalent to a squirrel cage induction motor.

The slip ring induction motor is used for driving the rotary kiln, bucket elevator, etc.

**What is the standstill condition of the induction motor?**

When the motor is at standstill, the relative speed of the rotating magnetic field and the rotor is equal to 1. The difference in the synchronous speed of the motor and the actual speed of the rotor is unity. In other words, the slip of the motor is unity when the rotor is at standstill.

slip(s)= (Ns-Nr)/Ns

At standstill, the Speed of the motor Nr=0

s= (Ns-0)/Ns

=Ns/Ns

s=1

At the start, the slip of the motor is 1. The motor draws maximum current when it starts accelerating from its standstill position. As the motor speed pick-up, the current starts reducing.

**In the induction motor, why is the starting current as large as 6 to 8 times the full load current?**

At the start, the induction motor takes the large rotor current because the slip of the motor is unity and maximum rotor voltage induces when the motor is started.

The rotor induced voltage= slip* Stator voltage

At start, Slip= 1

The rotor voltage= The stator Voltage

The rotor of the induction motor is short-circuited.

Rotor impedance=√(Rr^{2}+sXr^{2})

The rotor impedance is highly inductive when the motor starts as Xr= 2πfL

Rotor current= sE_{s}/√(Rr^{2}+sXr^{2})

**The motor draws a large current at the start because;**

- Rotor voltage is equal to the specified rotor open circuit(OCV)
- The rotor is highly inductive

The induction motor draws about 6 times the current of its FLC at the start as the rotor is highly inductive and the voltage induced in the rotor is maximum. The rotor voltage decreases as the motor accelerates towards its base speed and the motor current decrease accordingly.

**Why is the no-load power factor of an induction motor low when compared to the transformer?**

The transformer and the induction motor function on the induction principle. The magnetic field produced in the primary of the transformer gets linked through the magnetic core. The reluctance of the core is very low because of the high permeability of the iron core. Therefore transformer takes less current to develop the rated magnetic flux.

On the other hand, the flux produced in the stator travels through the air gap length, The permeability of the air is low compared to the iron. A more magnetizing current is required to produce the required flux as the reluctance of the magnetic path is more.

That is why the magnetizing current of the motor is much more than the magnetizing current of the transformer. The higher magnetizing current of the motor at no load causes a low power factor of the induction motor. The power factor of the motor improves when the motor gets loaded.

**Why is the power factor of an induction motor always lagging, and how is it improved when loaded?**

The power factor of the induction motor is always lagging because the rotor and the stator winding have inductive impedance.

When the loading on the motor has increased the slip of the motor gets decreased. With decreased slip, the frequency of the rotor current gets decreased as fr=s*fs. As a result, the rotor inductance Xr=2πf_{r}L gets decreased. The power factor of the motor improves with the increased loading on the motor because the rotor impedance is less inductive when the motor is fully loaded.

Therefore, the economic operation of the induction motor can be achieved if the motor is fully loaded.

**Why does an induction motor work on a lagging power factor?**

In an induction motor and a transformer, the electrical energy is transferred to the secondary side through the magnetic circuit. The induction motor draws magnetizing current for setting up necessary magnetic flux in the air gap.

The magnetizing current always lags the voltage because of the inductive property of the stator coils.

An induction motor draws more magnetizing current as compared to the magnetizing current drawn by the transformer.

This is because the reluctance of the magnetic core of the motor is more as the air gap between the stator and the rotor offers more reluctance.

**How to calculate the inductance of the coils in a 3-phase motor? **

Connect the inductance meter between the two coils of the star-connected coils. Divide the measured value by 2 to get the inductance of the coil.

Or- If you don’t have an inductance meter,you can measure the inductance of the coil by the following method.

- Measure the resistance(R) of coils(phase to phase)
- Feed AC voltage (V), 50 Hz to coils and measure the current(I)
- Calculate impedance of the coils Z=V/I
- Reactance(XL) of the coils by XL=√(Z
^{2}-R^{2}) - Calculate Inductance of the coils by L= XL/2*π*f
- Divide the value of L by 2 is the inductance of the coil.

**Can induction motor be run above synchronous speed?**

We can not run the motor above its synchronous speed. Even with VFD, the actual speed of the motor is always less than the synchronous speed of the motor.

In the case of the slip ring induction motor, the motor can be run above its synchronous speed by feeding rotor power to the power source.

**A 3-phase induction motor is rated 50 HP, 220 volts, and 1745 rpm. What is the percent slip?**

**For 50 Hz Induction motor**

The synchronous speed of the induction motor of 50 Hz operation is 3000, 1500, 1000, 750, 600 RPM

If the rated speed of the motor is 1745, the synchronous speed of the motor is 3000 RPM. However, the slip 3000–1745=1225 RPM is not possible at all. The slip of the induction motor is in the range of 2–3 %.

Probably, the motor is 4 pole machine(Ns=1500 RPM), and it is being operated through VFD to achieve the 1745 RPM.

**For 60 Hz Induction motor**

Ns=120f/P

=120*60/4=1800 RPM

slip=(Ns-N)/Ns*100

Slip=(1800–1745)/1800

=(55/1800)*100

** s=3.06 %**

**What is the difference between a squirrel cage motor and a wound-rotor (slipring) induction motor?**

Both types of induction motor function on the principle of Faraday’s law of electromagnetic induction.

The squirrel cage induction motor has a rotor with short circuit end rings. The rotor has a fixed resistance. The squirrel cage induction motor has low starting torque.

In the slip ring induction motor, the rotor is an open circuit, and we add the external resistance to the rotor resistance. The total rotor resistance increases due to addition of external resistance. Therefore, starting torque of the motor improves. The resistance gradually cut off with motor acceleration towards its base speed. Once the motor attains base speed , the slip ring is short circuit and, now the motor works like a squirrel cage induction motor.

**Can a slip of an induction motor be negative? If so, then in which condition?**

The slip of the induction motor is negative when the speed of the rotor is more than the synchronous speed and the motor operates in generating mode.

s= (Ns-Nr)/Ns

If Nr> Ns

Then the slip is negative.

When the speed of the motor is reduced with v/f drive the actual speed of the rotor does not reduce because of the high inertia load. In this situation speed of the rotor is more than the synchronous speed of the motor and it works as a generator.

The generating energy dissipates in a resistor to bring the motor speed down as per the set point of speed in bed drive. This is known as dynamic braking.

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