Induction Motors

Crawling in induction motor
In this article, I will explain the crawling phenomenon and how to overcome this phenomenon.

Firstly, I want to explain some definitions that we will need to understand our topic well.

When we apply AC current to terminals of the stator winding in induction motors, the stator winding produces a rotating magnetic field; assume the rotation speed of this field is synchronous speed NS.

As the rotor cuts, these magnetic field lines will rotate but with specified speed (Nr) which is fewer than the synchronous speed (NS) In the ideal case, this speed (Nr) is equal to the synchronous speed.

The difference between the two speeds is called Slip.

S = Syn. Speed (NS) – Rotor speed (Nr)
What is crawling in the induction motor?
Crawling (or magnetic locking phenomena or teeth locking phenomena) is when induction motor (mainly squirrel cage) fails to reach its full speed (rotate with very low speed near 1/7 of synchronous speed!!!!).

Now we will start to analyze the reasons that this phenomenon occurs:

The supply source for the induction motor (AC source) has some harmonics.
Also, the flux wave that the stator winding produces is not a pure sine wave (has some odd harmonics).
These harmonics divided is into some odd harmonics. That is 3rd harmonic, 5th harmonic, and 7thharmonic … 13thharmonic… Etc.
It assumes that the main wave rotates at synchronous speed.
In 3 phase system, 3rd harmonic is absent so, 5th and 7th harmonics are more effective (other harmonics can be neglected).
The rotating field of 5th harmonic will rotate at (1/5) from synchronous frequency.
Similarly the 7th harmonic will rotate at (1/7) from synchronous frequency.
Now we have 3 kinds of torques and speeds. Firstly, the main (fundamental) torque (rotating with synchronous speed NS), secondly, the 5th harmonic torque (rotating with NS/5 speed) and Lastly, the 7th harmonic torque (rotating with NS/7 speed).
The torque produced by 5th and 7th harmonics generates reverse torque opposing main torque.
The net developed torque will be fewer than required so, the motor will rotate at (1/7) of its synchronous speed.

How to overcome the crawling phenomenon?
We can overcome this phenomenon by:

Using a soft starting method for producing a pure sine wave.
Good design for the stater winding to reduce harmonics.

Cogging phenomena in induction motor:
Here I will explain some conditions that lead to magnetic locking phenomena:

Induction motors contain slots or teeth for their rotor and stator. When these teeth of the rotor are equal to the teeth of the stator, the motor fails to start. This because of cogging phenomena.

This phenomenon also increased with decreasing applied voltage over stator winding and equality of teeth for both rotor and stator.

Cogging can also occur when the supply voltage has some harmonics; when harmonics frequently coincide with slot frequency it causes torque modulation that causes cogging in the induction motor.

In the case of permanent magnet motors, cogging results in torque ripples that occur when magnetic reluctance forces acting mainly in the teeth of the stator.
How this phenomenon occurs
When the conditions of cogging are found at any induction motor (particularly squirrel cage induction motor), the reluctance of the magnetic path is minimum.

This because the rotor and stator teeth comes in front of each other. So, the rotor of the induction motor tends to remain fixed and not start to rotate.
How to Reduce Cogging phenomena
Because of the above-mentioned drawbacks of cogging, electrical engineers have suggested ways of reducing or eliminating cogging phenomena as follows:

We can reduce cogging by changing the number of stator teeth and rotor teeth (should varies).
We can also use the skewed rotor to keep an overlap on all slots that will eliminate this phenomenon. Skew the rotor means that we should arrange the stack of rotor laminations so that the rotor slots are “skewed” or angled with respect to the axis of rotation.
We can minimize cogging by good motor design, but the problem still maintained in sensitive applications

Note that: this phenomenon is less in wound rotor motors because of the higher starting torque of wound rotor motors.
Circle diagram of induction motor
To understand the circle diagram of the induction motor we should first know what is the circle diagram.

Firstly, the circle diagram is a graphic representation of the performance of the machine. Also, it’s drawn in terms of the locus of the input voltage and current. Similarly, it’s based on the approximate equivalent circuit.
Importance of the circle diagram
A picture is worth 1000 words, that’s the importance of the circle diagram in a simple description. But we use the circle diagram because:

It is a phasor diagram drawn for than more one circuit condition in one plan; The phasor diagram provides us with the between voltage and current at only a single circuit condition.
It provides us with the power output, power factor, speed, slip, torque, efficiency, and copper loss of the induction motor in a graphical representation.
The diagram is also easier to remember and understand compared to the theoretical and mathematical descriptions.

Test performed to compute data required for drawing circle diagram:
To draw the circle diagram of induction motor we do three tests:

No-load test:

In this test, the motor runs at the rated supply voltage without any load so there would be no losses.

We assume the slip =0, the supply line voltage is V0, and the line current is I0;

We perform this test to calculate the angle between voltage and current which would be large because we have a high inductive reactance at no load.

Φ0= P0/(√(3 ) V0I0).

Blocked rotor test:

In this test, we block the rotor and apply a reduced supply voltage to obtain the rated current at the terminal of the motor, the angle between the voltage and current, and also the rotor and stator copper loss.

Φsc= Psc / √(3 ) VscIsc.

ISN ( current drawn when applying the rated voltage at blocked rotor condition)= Isc.V0/Vsc.

PSN (power input of the rated voltage in blocked rotor condition) = Psc (V0/Vsc)^2.

Resistance test:

In the resistance test, we apply the voltmeter-ammeter method to calculate the per-phase equivalent stator resistance.
Construction of circle diagram
Firstly, to draw the circle diagram we apply the Kirchhoff’s current law at the approximate equivalent circuit to have:

I1= I0+I’2.

From this equation we start to draw;

The no-load current lags the voltage V1 by the angle Φ0 which ranges between 60 to 80 degrees. They get the no-load current and the no-load angle from the no-load test as we say.

At no load we have:

I0=V1/Zn1.

While: Zn1= (R0 ||jX0).

And also, we have no load loss:

P0=V1I0 cos Φ0.

When we refer the rotor current to the stator we have:

Of course, I’2 lags the voltage V1 and the angle between them will be:

circle diagram of induction motor

While the phase angle between current and voltage If we combine the above two equations we will have: combining equation

We also need to know:

Rs: stator resistance.
Xs: stator leakage reactance.
R’r: rotor resistance referred to the stator and rotor slip.
X’r: leakage reactance referred to the stator and rotor slip.
Rc: core and mechanical losses.
Xm: magnetization reactance.
Vs: impressed stator voltage.
I0 (no load current) =OO’.
IBL ( blocked rotor current) =OA.
I1 ( operating current) =OV.
Φ0: no-load angle.
ΦBL: blocked rotor angle.
Pmax: maximum output power.
SPmax: slip related to maximum power.
Tmax: maximum power factor.
STmax: slip related to maximum power.
ƞ1: Efficiency.
S1: slip.
PF1: power factor.
Φ1: PF angle at operating current.
AB: present rotor power input.

Circle diagram of induction motor
From the circle diagram we can obtain:

Input power.
Rotational losses.
Stator copper loss.
Rotor copper loss.
Output power.
Output torque.
Starting torque.
Slip.
Speed.
Efficiency.
Also, the power factor.

Losses of induction motor
To talk about losses of induction motor we talk about constant(fixed) losses and variable losses:
Constant (fixed) losses
From the name, fixed losses remain constant over the normal working range of the motor. Likewise, we can determine these losses by the no-load test of the induction motor;

Fixed losses also subdivided into:

Iron (core) losses:

Core losses of the stator consist of eddy current losses and hysteresis losses that dependent on the frequency and voltage of the supply. The supply frequency is constant so the stator core losses are constant.

However, we can minimize the eddy current losses by using lamination on the core which decreases the area. So, the resistance increases, and the eddy losses decrease. Similarly, we can also minimize hysteresis losses by using high-grade silicon steel.

Mechanical losses:

Firstly, we can consider mechanical losses constant losses for a giving running condition. As such, these losses include friction and windage losses.

Brush friction losses:

This is in general are proportional to the rotor speed.
Variable losses
Variable (copper) losses occur due to the current flowing in stator and rotor windings. Thus, when the load changes the current flowing in the stator and the rotor changes and so the losses changes.

We can determine these losses by the blocked rotor test of the induction motor.
Power flow diagram of induction motor
The power flow illustrates the input power to the stator:

Pis= √3 .VL.IL.cos Φi =3.VSP.ISP.cos Φi.

While:

VL: The line voltage supplied to the stator of the motor.

IL: The line current.

Cos Φi: input power factor.

And the losses in the stator:

Stator copper losses (losses in the stator winding resistance): I^2.R.
PSCL = 3.Isp^2*Rsp.

Stator core losses (hysteresis and eddy current losses in the stator core).
Ps (h+e)

We have an output power of the stator:

Pos=Pis-Psc-Ps (h+e)

The air gap between the stator and the rotor transfers the output power of the stator to the rotor. Hence we have air gap power (Pg).

So:

The power output of the stator= air gap power= input power to the rotor.
Pos= Pg= Pir.

When we enter the rotor there are:

Rotor copper losses (losses in the rotor resistance): I^2.R.
Prc= 3.I2^2*R2.

Rotor core losses (hysteresis and eddy current losses in the rotor core):
Pr (h+e)

Friction and windage losses (Pfw).

Stary load losses (Pmisc): which contain any other losses as losses due to the harmonic field.

Developed mechanical power (Pmd).

Developed mechanical power= rotor input- rotor copper loss.

Pmd=Pir– Prc = Pg– 3I2^2*R2.

And the output power of the motor will be:

Po= Pmd– Pfw– Pmisc.

This output power is also known as useful power or shaft power.

Rotational losses:

They are additional losses produced at starting and during acceleration. They are the sum of the friction, core, and windage losses and we can say they are constant losses with the change of speed.
Prot= Pfw+ P(h+e)+ Pmis.

So the final output power of the motor will be:

Po= Pmd– Prot = Pmd– Pfw– P(h+e)– Pmis.
The efficiency of induction motor
We care about losses of induction motor because they are our entrance to the efficiency of the induction motor. IN other words, the net efficiency of any motor is the ratio between the input power and the output power.

Ƞ = output power/ input power.

= output power/ (output power+losses).

Which means that the losses play a vital role in estimating efficiency; The efficiency of induction motor varies between 85% to 97% as the losses are:

Fiction and windage losses 5-15%.
Iron (core) losses 15-25%.
Stator losses 25-40%.
Rotor losses 15-25%.
And stray losses 10-20%.

Remember, the efficiency also varies with motor size. The efficiency increases when the motor gets bigger. However, small motors don’t have high efficiency.

Also, motors in many kilowatts have efficiency above 90%. To have a high efficiency we need a high cost.

To achieve the maximum efficiency of the induction motor, the rotor should almost run as fast as the magnetic field in the stator to have a slip close to zero.
Speed control of induction motor by 7 ways
The speed of a three-phase induction motor of a given load is determined by matching the torque-speed characteristics of the motor and load.
Speed control of induction motor characteristics

The operating point is the intersection of the two characteristics and we can fix for a given load and motor parameters.
Until the emergence of modem solid-state drives, induction motors were not favored for considerable speed control applications.
For design classes, A, B, and C, the normal operating slip range of a typical induction motor is restricted to less than 5%.
Even if they slip could be increased, the motor efficiency would drop significantly because of the corresponding increase in rotor-copper losses.
If we desire to change motor speed while carrying the same load, then we need a change in the torque-speed characteristic of the motor.

There are several methods of speed control of 3 phase induction motor.

We can classify into two main categories.

Induction motor speed control through the stator.
Induction motor speed control through the rotor.

Speed control of induction motor (Stator)
Variable frequency method (changing applied frequency):
Changing the electrical frequency applied to the stator of an induction motor will result in the following.

Firstly, it will cause the rate of rotation of its magnetic field (ns) to change in direct proportion to the change in electrical frequency. Also, the no-load point on the torque-speed characteristic curve will change with it.

The base speed is the synchronous speed of the motor at rated conditions. By using variable frequency control, it is possible to adjust the speed of the motor either above or below the base speed.

Similarly, a properly designed variable frequency induction motor drive can be very flexible. It can control the speed of an induction motor over a range from as little as 5% of base speed up to about twice base speed.

However, it is important to maintain certain voltage and torque limits on the motor to ensure safe operation as the frequency is varied.
Running at speeds below the base speed
When running at speeds below the base speed of the motor, it is necessary to reduce the terminal voltage applied to the stator for proper operation.

Nonetheless, the terminal voltage applied to the stator should be decreased linearly with decreasing stator frequency.

This is called derating and is needed so the iron core of the motor does not saturate and cause excessive magnetization currents to flow in the machine.

As with any transformer, the flux in the core of an induction motor can be found from Faraday’s law of sinusoidal operation as;

If we reduce the electrical frequency ω by a factor α.
Then the flux in the core will increase by a factor 1/ α.
The magnetization current of the motor will increase.

In the unsaturated region of the motor’s magnetization curve,
the increase in magnetization current will also be about 1/ α.

In the saturated region of the motor’s magnetization curve, however, an increase in flux 1/ α requires a much larger increase in the magnetization current.

We design induction motors to operate near the saturation point on their magnetization curves. So the increase in flux due to a decrease in frequency will cause excessive magnetization currents to flow in the motor.

To avoid excessive magnetization currents, it is customary to decrease the applied stator voltage in direct proportion to the decrease in frequency whenever the frequency falls below the rated frequency of the motor.

When the voltage applied to an induction motor is varied linearly with the frequency below the base speed. The flux in the motor will remain approximately constant.

Therefore, the maximum motor torque remains fairly high. However, the maximum power rating of the motor must be decreased linearly with the decrease in frequency to protect the stator circuit from overheating.

If the voltage VL is decreased, then the maximum power P must also be decreased.

Or else the current flowing in the motor will increase, and the motor will overheat.
The resulting family of torque-speed characteristics
The family of induction-motor torque-speed characteristic curves for speeds below the base speed

When the electrical frequency applied to the motor exceeds the rated frequency of the motor, the stator voltage is held constant at the rated value.

Although saturation considerations would permit the voltage to be raised above the rated value under these conditions, it is limited to the rated voltage to protect the winding insulation of the motor.

With high electrical frequency above base speed, the resulting flux in the machine decreases, and the maximum torque decreases with it.

The principal disadvantage of electrical-frequency control as a method of speed changing was that it requires a dedicated generator or mechanical frequency changer.

The development of modem solid-state variable-frequency motor drives has overcome this limitation, and it is now the method of choice for induction-motor speed control.

Note that we can use this method with an induction motor, unlike the pole changing technique, which requires a motor with special stator windings.
Speed control of induction motor by stator voltage method
The line-voltage control of the torque developed varies with the square of the applied voltage.

The speed can be varied simply by changing the voltage applied to the stator. This method is common for squirrel-cage motors.
Speed control of induction motor using v/f

The improved performance of adjustable-speed drives is associated with a variable frequency stator supply.
The air-gap flux is directly proportional to the stator applied voltage and inversely proportional to the frequency.
A reduction in the supply frequency to achieve speed control below rated synchronous speed will be associated with an increase in the air-gap flux if the applied voltage is maintained at rated value.
To avoid saturation due to the increased flux, variable-frequency drives employ a variable voltage as well, with the object of maintaining an acceptable air-gap flux level.

This concept is generally referred to as constant (v/f) control and is used in employing squirrel-cage induction motors of all classifications.
Speed Control of induction motor by pole changing
There are three major approaches to changing the number of poles in an induction motor

The method of consequent poles
Multiple stator windings
Pole-amplitude modulation (PAM)

Method of consequent poles
The method of consequent poles is one of the earliest speed-control methods. It allows us to change n, in discrete steps.

A two-pole stator set of windings is changed to a four-pole configuration by a simple switching operation.

Of course, we can do this for any number of poles for a ratio of 2: 1. This mechanism is simple to implement in squirrel-cage rotor motors, but for the wound-rotor type, we must also rearrange the rotor windings.
Multiple stator windings
A traditional approach to overcome the limitations of the consequent pole method employs multiple stator windings with different numbers of poles and to energize only one set at a time.

This method increases the cost and weight of the motor and is, therefore, we use only when absolutely necessary.

Combining the method of consequent poles with multiple stator windings, allows us to build a four-speed induction motor.

For example, with separate four- and six-pole windings, we can produce a 60Hz motor that has synchronous speeds of 600, 900, 1200, and 1800 rpm.
The PAM method
The PAM method achieves multiple sets of poles in a single stator winding, where the resulting number of poles can be in ratios other than 2: 1.

Switching the number of poles in a winding is achieved by changing the connections at six terminals, in the same manner as in the method of consequent poles.

To create an induction motor with two close speeds, PAM windings are preferred over multiple-stator windings because they cost about 75% of two completely separate windings.

In PAM, the spatial distribution of the magneto motive-force waves in the machine stator is multiplied together.

The resulting output consists of components with frequencies equal to the sum and the difference of two original frequencies.

The winding of a machine normally having a P pole is modulated by making N switches in the connections on the phase groups in a given phase.

Then two magneto motive-force waves will be produced in the stator winding, one of them having P + N poles and the other having P – N poles.

If one of these waveforms can be selected over the other, then the motor will have that number of poles on its stator.

However, the same number of poles will, of course, be induced in the squirrel cage rotor.

Example of pole ratios available with the PAM technique and the resulting synchronous speed ratios for 60-Hz operation are as follows:

Pole ratio1 = 4/6 -+ 1800/1200

PR2= 6/8 -+ 1200/900

PR 3= 8/10 -+ 900/720
From a mathematical point of view:
We express the magnetomotive force produced by a conventional P-pole winding in terms of time and position as

If Q is the desired final number of poles on the machine, then P – Q is the difference between the original number of poles and the desired number of poles.
Now, modulate the original spatial waveform by switching connections at P-Q uniformly spaced points in each phase.

Note that there are two different spatial distributions of poles present in the resulting magnetomotive force.

We Can Express this magnetomotive force as

If Q is the desired number of poles in the motor, then we need to reject the other distribution.
We can accomplish it by a proper choice of the distribution and chording of the stator windings.

This analysis is approximate because we assume the modulating spatial wave is sinusoidal when it is a square wave.

Square-wave modulation introduces more spatial harmonics into the magnetomotive force distribution. However, this can be reduced by the proper choice of winding chording.

In practice, the choice of spatial modulating frequency, winding chording, winding distribution, and other factors we required to achieve a given speed ratio is an art based on the experience of the designer.
Speed control of induction motor through the Rotor
We can achieve speed control of induction motors of the wound-rotor type by inserting additional rotor resistance (Ra ). In addition to this, we can achieve torque control at a given speed using this method.

Where the subscript 1 refers to operating conditions without additional rotor resistance and the subscript 2 refers to conditions with the additional rotor resistance.

Introducing a resistance in the rotor winding will change α. So let the original torque be TI. Let the new torque with additional resistance be T2

The equation provides us with the means to derive the value of additional resistance, Ra, required to obtain the same torque (T1 = T2) at two different values of slip (or rotor speed).

To do this we set α= 1, to obtain additional resistance, Ra, required to obtain the same torque.
Rotor-Slip Energy Recovery
The theoretical rotor efficiency is

ɳr = 1 – s

Thus, it is clear concerning speed control that by increasing S results in an increase in rotor losses, which results in decreased efficiency.

There exist many methods for recovering this energy from the rotor at rotor frequency (fr = sfs), converting it to supply frequency, and subsequently returning it to the source.

The devices used to achieve this are solid-state power devices.
Speed control of induction motor through a cascading operation
In this method, we use two motors and both are installed on a common shaft.

The motor number 1 feeds from a three-phase power source.
The other motor feeds through induced e.m.f via slip rings.

Let’s assume that motor 1 is main and motor 2 is auxiliary.

then:

Ns1: synchronous speed of the motor no. 1.

P1: No. of stator poles of the motor no.1.

Ns2: synchronous speed of the motor no. 2.

P2: No. of stator poles of the motor no. 2.

N: Set speed of both motor 1 & 2.

F: supply frequency.

For motor no. 1 (main motor):

S1= (Ns1-N)/Ns1

So,

F1= S1* f

For motor no. 2 (aux. motor):

Ns2= (120*f1) / P2 = (120 * S1 * f) / P2

So,

Ns2= (120 * f * (Ns2-N)) / (P2 * Ns1)

We can conclude that:

So,

N= (120 * f) / (P1+P2)

In this equation we conclude 4 cases where we can obtain different speeds:

If the motor 1 is working only

So, Ns1= (120 * f) / P1

If the motor 2 is working only

So, Ns2= (120 * f) / P2

If both motors work (cumulative cascading)

So, N= (120 * f) / (P1+P2)

If both motors work (differential cascading)

So, N= (120 * f) / (P1-P2)
Equivalent circuit of induction motor
We have repeated over and over that we use the induction motor everywhere in the industrial and domestic applications.

This makes it very important to predict the behavior of the induction machine under various operating conditions. With the help of the equivalent circuit of induction motor that became easy.

The induction motor works on the same principle as the transformer. When we supply the stator with an EMF, a voltage is induced in the rotor as a result of electromagnetic induction.

So, we can say that the induction motor is a transformer with a rotating secondary.

The stator winding of the induction motor resembles the primary of the transformer. The rotor of the induction motor resembles the secondary of the transformer.

And as the induction motor runs below the synchronous or full load speed, there will be a relative difference between the speed of rotation of the motor and the synchronous speed.

This relative difference is known as the slip (s).

While:

Ns: the rotation synchronous speed. so,

Rotation synchronous speed

while f: the supply voltage frequency.

And, P: number of poles of the machine.

In electric machines, all information is connected with all types even it’s a transformer, AC, DC, motor, or even generator. And to understand the equivalent circuit we should tie everything with the other.
Equivalent circuit of three-phase induction motor
The equivalent circuit illustrates the various parameters of the machine and the losses may occur with the machine.

The induction motor has an induction and a resistor which causes losses.

The winding resistance causes copper losses in the windings, the windings inductive resistance cause a voltage drop (power factor).

And the induction motor, especially the three-phase induction motor, has two types of equivalent circuits which are:
Exact equivalent circuit
R1: the stator winding resistance.

X1: the stator winding inductance.

Rc: the copper loss component.

Xm: the winding magnetizing reactance.

R2/s: the rotor power and it includes the output mechanical power and the rotor copper losses.

So, When we draw this circuit as referred to the stator it will be:

Exact equivalent circuit referred to the stator

while :

R’2: the resistance of the rotor referred to the stator winding.

X’2: the inductance of the rotor referred to the stator winding.

R2(1-s)/s: the resistance used to show the power that is converted to useful power or mechanical output power.
Approximate equivalent circuit
In the approximate circuit, we delete one node to simplify the calculation. We shift the shunt branch towards the primary side.

And what helps us to do that is the less voltage drop between the stator resistance and inductance. Also, the difference between the supply voltage and the induced voltage is not much.

But there are some reasons that make it not appropriate to do that. the reasons are:

In the induction motor, we have a larger inductance of stator and rotor.
Also, we use distributed windings in the induction motor.
There is an air gap in the magnetic circuit of the induction motor which made the current larger compared to the transformer.

These reasons make it better to use the exact equivalent circuit, not the approximated. So, We can only use the approximate analysis for large motors.
Power relation of the equivalent circuit:
The input power: which is directed by the stator and it equals (3V1I1Cos Ө).
while: V1: the voltage applied to the stator.

and, I1: the current drawn by the stator, Cos Ө: the stator power factor.

The rotor copper losses: and they equal (slip* rotor input power).

Developed power: which equals {(1-s) * rotor input power}.
Equivalent circuit of a single-phase induction motor
The equivalent circuit of the single-phase differs from the three-phase because the single-phase motor circuit is given by double-revolving field theory.

We should resolve the stationary pulsating magnetic field into two rotating fields. The two fields are equal in magnitude but opposite in direction which makes the induced net torque at standstill equals zero.

And we have a forward rotation known as the rotation with slip(s), and a backward rotation (2-s).

So, the equivalent circuit presented as:

Zf: shows the forward impedance.

Zb: shows the backward impedance.

Note the sum of forward and backward slip = 2 … So, we say that the backward slip is (2-s).

R1: the stator winding resistance.

X1: The stator winding inductive resistance.

Xm: the magnetizing reactance.

R’2: the rotor reactance referred to the stator.

X’2: the rotor inductive reactance referred to the stator.
Calculation of power of equivalent circuit:
We first find the forward slip (Zf) and backward slip (Zb).

We find the stator current which is calculated from (stator voltage/total circuit impedance).

And the input power: which is (stator voltage*stator current*Cos (Ө)); where:
Ө: the angle between the stator current and voltage.

The power developed(Pg): which illustrates the difference between the forward field power and the backward power.

And the rotor copper losses: Pg*slip.

Finally, the output power: (Pg-s*Pg-Rotational losses). The rotational losses include the friction loss, windage loss, and core loss.

Also, we can calculate the efficiency by dividing the output power by the input power.

Also, the equivalent circuit of an induction motor may also help us to:

Choose the power supply for the circuits we designed.
Also, know the temperature coefficient.
Illustrate the noise.
Find the necessary parameters needed.