Under no load condition if the applied voltage to an induction motor is reduced from the rated voltage to half the rated value,

Option 2 : Both speed and stator current decrease

In an induction motor,

\(T\ \propto\ \frac{{s{V^2}}}{{R_2^1}}\)

For same torque, s ∝ 1/V^{2}

Therefore, the speed will decrease with a decrease in voltage to half the rated value.

When an induction motor is on no-load and it is operating at rated frequency but at less than the rated voltage,

- Air gap flux will be less than rated flux as ϕ ∝ V/f
- Both I
_{μ}and I_{w}decreases and hence the no-load current also decreases - Iron losses decrease as these are directly proportional to the square of the voltage
- Mechanical losses remain constant.
- Constant losses will decrease
- No-load stator copper loses will decrease
- No-load power factor will increase
- Both starting torque and maximum torque are proportional to the square of the voltage and they will decrease.

**Concept:**

**Synchronous speed** in rpm of a three-phase induction motor is given by,

\({N_s} = \frac{{120f}}{P}\)

Where f is the frequency in Hz

P is the number of poles

A rotor of a three-phase induction motor rotates at a speed close to the synchronous speed but not equal to the synchronous speed.

The difference between the synchronous speed and the rotor speed is known as **slip speed**. It is given by

Slip speed = N_{s} - N_{r}

**Slip** is defined as the ratio of slip speed to the rotor speed.

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)

⇒ N_{r} = N_{s} (1 – s)

**Speed regulation** of induction motor is the ratio of the change in speed from no load to full load to the full load speed.

Speed regulation = (No load speed – Full load speed) / (Full load speed)

**Calculation:**

Given that,

Number of poles (P) = 4

Frequency (f) = 50 Hz

Synchronous speed, \({N_s} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

**No load:**

Slip at no load, s_{0} = 1 %

Rotor speed at no load, N_{0 }= N_{s} (1 – s) = 1500(1 – 0.01) = 1485 rpm

**Full load:**

Slip at no load, s = 5 %

Rotor speed at no load,

Speed at full load, N = N_{s} (1 – s) = 1500(1 – 0.05) = 1425 rpm

% speed regulation \(= \;\frac{{1485 - 1425}}{{1425}} \times 100 = 4.21\;\% \)

Option 2 : Rotor torque will then be zero

- In an induction motor when Voltage is applied to the stator of the induction motor, the current will flow through the winding, resulting in a magnetic field flux around the free space across the stator.
- The rotor is placed in such a way that this magnetic field induces a current in the rotor. This induced current will flow through the rotor winding causing another flux in the rotor.
- This flux in the rotor will lag stator flux.
- Due to this difference in flux, the rotor will experience a torque and will start to rotate at a speed less than synchronous speed due to lagging.

∴ Theoretically, the Induction motor can never run at synchronous speed. However, if by some external force, or system fault such as voltage surge, somehow **speed of induction motor becomes equal to the synchronous speed, then there will be no more lagging between both the fluxes and no more current will be induced in the rotor winding. This will result in no torque on the rotor**, and it will stop moving due to this.

The given symbols show a/an:

Option 1 : rheostat

**Rheostat:**

A rheostat is a type of variable resistor, whose resistance can be changed for varying the amount of electric current flowing through an electrical circuit. The word rheostat is made of two words (‘rheo’ meaning flow of current in Greek and ‘stat’ meaning stationary instrument). When placed in an electric circuit, the flow of electricity changed through two terminals: one terminal near the slider/adjustable contact and the other connected near the bottom.

A rheostat is internationally denoted by the following symbol:

**The rheostat is generally used in applications where high voltage or current is required such as:**

- Changing the light intensity of a light bulb. An increase in the resistance of the rheostat decreases the flow of electric current, leading to the dimming of lights and vice-versa.
- Generators
- Motor speed
- Heater and oven temperature control
- Volume control

__Additional Information__

Some important electronic components symbols given below-

An 8-pole, 50 Hz, three-phase, slip-ring induction motor has an effective rotor resistance of 0.08 Ω per phase. Its speed at maximum torque is 650 RPM. The additional resistance per phase that must be inserted in the rotor to achieve maximum torque at start is ____________ Ω. (Round off to 2 decimal places.)

Neglect magnetizing current and stator leakage impedance. Consider equivalent circuit parameters referred to stator.

Given:

Pole (P) = 8

Frequency (f) = 50 Hz

Rotor resistance (r_{2}) = 0.08 Ω

**Rotor speed at maximum torque = 650 RPM**

Let's consider the slip at maximum torque as s_{m,}

Slip at maximum torque is given by

**\(s_m = \dfrac {r_2}{x_2}\)**

Consider additional rotor resistance to be added = r_{2}'

Rotor reactance = x_{2}

Synchronous speed N_{s} = 120f/P = 120 × 50 / 8 = 750 RPM

As the speed of the motor at maximum toque condition is 650 RPM

Slip at maximum torque is given by

**\(s_m = \dfrac {N_s - N_r}{N_s} =\dfrac {750-650}{750}= 0.1333\)**

At maximum torque

\(s_m =\dfrac {r_2}{x_2} ⇒ 0.1333 = \dfrac {0.08}{x_2}\)

**x _{2} = 0.6 Ω **

To get the maximum torque at starting the slip at maximum torque should be equal to slip at starting (s=1).

⇒ \(1 =\dfrac {{r_2+r_2'}}{ x_2}\)

x_{2} = r_{2} + r_{2}'

0.6 = 0.08 + r_{2}'

**r _{2}' = 0.52 Ω **

**∴ The value of additional rotor resistance to be added (r2') at the time of starting is 0.52 Ω.**

Option 3 : below 5 Hertz

**Variable Frequency Drives:**

- Induction motors are designed for a specific voltage per frequency ratio (V/f).
- Voltage is the supply voltage to the motor, and frequency is the supply frequency.
**The V/f ratio is directly proportional to the amount of magnetic flux in the motor magnetic material (stator and rotor core laminations).****The torque developed on the motor shaft is proportional to the strength of the rotating flux.**- The type and the amount of magnetic material used in motor construction are factors to define motor power rating.

**Torque speed characteristics:**

From the characteristics,

- The pullout torque is constant at all points below the rated speed, except at low frequencies.
**At low frequencies, the pullout torque is reduced because of the effect of stator resistance.**- As the frequency approaches zero, the voltage drop due to stator resistance becomes important.
- Flux reduction causes the torque reduction to become prominent.
**This effect is known and easily mitigated by low-speed voltage boosting: increasing the V/f ratio at low frequencies to restore the flux.**

**The typical set of torque-speed curves for a drive with low-speed voltage boosting:**

**Therefore, In V/f control of induction motors, the ratio of V/f is boosted during below 5 Hertz**

Option 2 : Phase sequence of the stator voltage is reversed

**Concept of Plugging:**

**Due to the reversal of the phase sequence of the stator voltage,**the direction of the rotating magnetic field gets reversed.- This produces a torque in the reverse direction and the motor tries to rotate in the opposite direction.
- This opposite flux acts as a brake and it slows down the motor. During plugging the slip is (2 - s), if the original slip of the running motor is s.
- This method is the quickest way of braking, but during the plugging operation very high I²R losses occur in the form of heat. This heat is more than produced when the rotor is normally locked.
- So, we cannot apply plugging frequently due to the high heat produced rotor which can damage or melt the rotor bars and even may overheat the stator as well.

Therefore, Plugging of a three-phase induction motor is done by interchanging connections of any two phases of the stator supply terminals for obtaining quick braking.

Option 1 : Load is lowered by a hoisting machine

**Regenerative Braking:**

- Regenerative braking of an induction motor can only take place if the speed of the motor is greater than its synchronous speed, both rotating in the same direction
- When the load is reduced then there is a possibility for
**higher speeds and regenerative breaking.** - In regenerative braking, induction motor works as an induction generator, and power is fed back to the source
- This enhances the energy efficiency of the motor and improves the operating power factor the machine
- The main advantage of regenerative braking is that the generated power is fully used

During regenerative braking slip is negative

\(s = \frac{{{N_s} - {N_r}}}{{{N_S}}}\)

\({N_s} < {N_r}\)

Therefore, \(s < 0\)

Regenerative braking is used to control the speed of motors driving loads such as in electric locomotives, elevators, cranes, and hoists

Regenerative braking cannot be used for stopping the motor; It is only used for controlling the speed above the no-load speed of the motor

A 4-pole induction motor (main) and a 6-pole motor (auxiliary) are connected in cumulative cascade. Frequency in the secondary winding of the auxiliary motor is observed to be 1 Hz. For a supply frequency of 50 Hz the speed of the cascade set is:

Option 4 : 588 rpm

Given P1 = 4, P2 = 6

f2 = 1 Hz; f = 50 Hz

f2 = s2f1, f1 = s1f

⇒ f2 = s1s2f ⇒ s1s2 = 1/50 = 0.02

For cumulative cascade

\(\begin{array}{l} \frac{{120f}}{{{P_1}}}\left( {1 - {s_1}} \right) = \frac{{120{s_1}f}}{{{P_2}}}\left( {1 - {s_2}} \right)\\ \frac{1}{{{P_1}}} - \frac{{{s_1}}}{{{P_1}}} = \frac{{{s_1}}}{{{P_2}}} - \frac{{{s_1}{s_2}}}{{{P_2}}}\\ {s_1}\left( {\frac{1}{{{P_1}}} + \frac{1}{{{P_2}}}} \right) = \frac{1}{{{P_1}}} + \frac{{{s_1}{s_2}}}{{{P_2}}}\\ {s_1}\left( {\frac{{{P_1} + {P_2}}}{{{P_1}{P_2}}}} \right) = \frac{1}{{{P_1}}} + \frac{{{s_1}{s_2}}}{{{P_2}}}\\ {s_1}\left( {\frac{{10}}{{24}}} \right) = \left( {\frac{1}{4} + \frac{{0.02}}{6}} \right)\\ {s_1} = 0.608 \end{array}\)

Speed of cascade set \( = \frac{{120f}}{{{P_1}}}\left( {1 - {s_1}} \right) = \frac{{120 \times 50}}{4}\left( {1 - 0.608} \right) = 588\;rpm\)

Option 2 : At low frequency, the stator flux decreases from its rated value.

__Concept:__

(V / F) control or frequency control method:

It is basically frequency control but to maintain Bmax constant

The frequency variations should be done by keeping the (V / f) ratio constant.

The synchronous speed of the induction motor in rpm is

\({N_S} = \frac{{120f}}{P}\) rpm

NS ∝ f

NS = Synchronous speed

f = Frequency

P = No. of poles

Torque produced in the motor is

\(T = \frac{{3 \times 60}}{{2\pi {N_S}}} \times \frac{{sE_2^2{R_2}}}{{R_2^2 + {{\left( {s{X_2}} \right)}^2}}}\)

E2 = Rotor induced emf

R2 = Rotor resistance

X2 = Rotor reactance

s = slip

\({E_{ph}} = 4.44\phi nf\;\;\)

\(\begin{array}{l} E ∝ V\\ \phi ∝ {B_{max}} ∝ \frac{V}{f} \end{array}\)

__The Torque - Speed characteristics:__

__Explanation:__

\(\frac{V}{f}\) is maintained constant from f = 0 to Base value in V/f control of IM.

Now, at low frequency, the effect of resistance can not be neglected as compared to reactance and must be compensated by increasing the stator voltage.

V/f = constant which implies that Torque will be constant. But at low frequency, magnitude of air gap flux decreases due to resistive effect. As a result, the stator flux decreases from its rated value, as at low frequency, flux density increases, which results in saturation of stator. Hence to avoid saturated, stator flux should decrease.

Option 1 : Low performance applications

**Open Loop V/F Control:**

**The open-loop V/F control of an induction motor is the most common method of speed control**- Because of its simplicity and these types of motors are widely used in the industry.
- Traditionally, induction motors have been used with open-loop 50Hz power supplies for constant speed applications.
**For adjustable speed drive applications, frequency control is natural.**- However, voltage is required to be proportional to frequency so that the stator flux

**Circuit diagram of inverter-fed induction motor in the open-loop:**

- The power circuit consists of a diode rectifier with a single or three-phase ac supply, filter, and PWM voltage-fed inverter.
**Ideally, no feedback signals are required for this control scheme.**- The PWM converter is merged with the inverter block.

**Some problems encountered in the operation of this open-loop drive are the following:**

**The speed of the motor cannot be controlled precisely,**because the rotor speed will be slightly less than the synchronous speed and that in this scheme the stator frequency and hence the synchronous speed is the only control variable.**The slip speed, being the difference between the synchronous speed and the electrical rotor speed, cannot be maintained**, as the rotor speed is not measured in this scheme. This can lead to operation in the unstable region of the torque-speed characteristics.

The effect of the above can make the stator currents exceed the rated current by a large amount thus endangering the inverter- converter combination.

- These problems are to be suppressed by having an outer loop in the induction motor drive,
**In which the actual rotor speed is compared with its commanded value and the error is processed through a controller usually a PI controller**- Also, a limiter is used to obtain the slip-speed command

**Therefore, Implementation of Volts/Hertz strategy for inverter-fed induction motor in an open loop is used in Low performance applications**

**Directions: **It consists of two statements, one labelled as the ‘Statement (I)’ and the other as ‘Statement (II). Examine these two statements carefully and select the answer using the codes given below:

**Statement (I):** The speed control of induction motor by pole changing is suitable for cage motors only.

Option 1 : Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)

**Pole Changing Method:**

The synchronous speed is given by,

\({N_s} = \frac{{120f}}{P}\)

where f is frequency and

P is number of poles

We can change the speed by changing the number of poles.

- The speed control of induction motor by pole changing is suitable for cage motors only as the cage motors automatically develops number of poles equal to the poles of stator winding.
- In slip ring induction motor, the motor poles are fixed in number, and can only be changed by re-winding the rotor windings.

Therefore, both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I).

**Note:**

The number of stator poles can be changed by the following three methods.

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