Electrical Machines MCQ Quiz - Objective Question with Answer for Electrical Machines - Download Free PDF
Last updated on Jun 19, 2025
Latest Electrical Machines MCQ Objective Questions
Electrical Machines Question 1:
Copper losses in a transformer are proportional to:
Answer (Detailed Solution Below)
Electrical Machines Question 1 Detailed Solution
Copper Losses in a Transformer
Definition: Copper losses in a transformer refer to the power losses that occur due to the resistance of the winding conductors when current flows through them. These losses are one of the primary contributors to the overall efficiency reduction in a transformer. Copper losses are also known as I2R losses, as they are directly related to the current flowing through the winding and the resistance of the winding material.
Correct Option Analysis:
The correct answer is:
Option 1: Copper losses in a transformer are proportional to the square of the load current.
Explanation:
Copper losses in a transformer are mathematically expressed as:
Power Loss = I2 × R
Where:
- I is the current flowing through the winding (load current).
- R is the resistance of the winding conductor.
From the formula above, it is evident that copper losses are proportional to the square of the current (I2). This relationship arises because the power dissipated as heat in the winding resistance is dependent on the square of the current. Therefore, as the load current increases, the copper losses increase quadratically.
In transformers, the load current varies with the load connected to the transformer. As the load current increases, the copper losses increase significantly due to the squared relationship. This is why transformers are typically designed to operate efficiently at their rated load, as excessive copper losses can lead to overheating and reduced efficiency.
Important Considerations:
- Copper losses are directly related to the winding material's resistance. High-quality materials with low resistivity (such as copper or aluminum) are used to minimize these losses.
- While copper losses depend on the load current, they do not depend on the voltage, frequency, or turns ratio of the transformer.
- Reducing the winding resistance by using thicker conductors or reducing the length of the winding can help minimize copper losses.
Additional Information:
To further understand the analysis, let’s evaluate the other options:
Option 2: Copper losses are proportional to the square of the frequency.
This option is incorrect. Copper losses are not dependent on the frequency of the electrical supply. While frequency does affect core losses (hysteresis and eddy current losses), copper losses are solely dependent on the load current and the winding resistance.
Option 3: Copper losses are proportional to the square of the turns ratio.
This option is also incorrect. The turns ratio of a transformer determines the voltage and current transformation between the primary and secondary windings but does not directly influence copper losses. Copper losses depend on the load current flowing through the windings, not on the turns ratio itself.
Option 4: Copper losses are proportional to the square of the voltage.
This option is incorrect. Copper losses are not directly related to the voltage. While the voltage determines the current flowing through the load (based on the load's impedance), copper losses are specifically proportional to the square of the load current, not the voltage.
Conclusion:
Copper losses in a transformer are a critical factor affecting its efficiency and performance. Understanding that these losses are proportional to the square of the load current is essential for designing and operating transformers effectively. By minimizing winding resistance and optimizing the transformer for its rated load, these losses can be reduced, improving the overall efficiency of the transformer.
Electrical Machines Question 2:
For small capacity motors (up to 3 Hp), what voltage can be applied at startup?
Answer (Detailed Solution Below)
Electrical Machines Question 2 Detailed Solution
Voltage Application at Startup for Small Capacity Motors (Up to 3 Hp):
Correct Option: Full normal voltage can be applied at startup for small capacity motors (up to 3 Hp).
Explanation:
Small capacity motors, such as those up to 3 Hp, are designed to handle the application of full normal voltage at startup without adverse effects. These motors typically have robust designs and lower power ratings, which allow them to start directly with full voltage. This method of starting is known as "direct-on-line" (DOL) starting.
When full normal voltage is applied directly to the motor at startup, the motor achieves its rated speed almost instantly, resulting in quicker operation. This technique is suitable for smaller motors because they require relatively low starting current and do not pose a significant risk to the electrical system or the motor itself.
Advantages of Applying Full Normal Voltage:
- Simplicity: Direct application of full voltage eliminates the need for complex starting mechanisms, making it a simple and cost-effective method.
- Quick Startup: The motor reaches its rated speed rapidly, reducing the time needed for startup and enhancing operational efficiency.
- Compact Design: Small capacity motors are designed to withstand the initial surge of current during startup, ensuring reliable operation.
- Cost-Effective: No additional equipment, such as soft starters or variable frequency drives, is required, reducing overall costs.
Disadvantages:
- Although this method is suitable for small motors, applying full normal voltage to larger motors can cause excessive current draw, leading to potential damage to the motor and the electrical system.
- Direct-on-line starting may result in a mechanical shock to the connected equipment, which could lead to wear and tear over time.
Important Information
To further understand the analysis, let’s evaluate the other options:
Option 1: Zero voltage.
This option is incorrect because applying zero voltage at startup would prevent the motor from starting entirely. Motors require a certain level of electrical energy to overcome inertia and initiate rotation. Without voltage, the motor remains stationary and cannot perform any operation.
Option 2: Double the normal voltage.
Applying double the normal voltage to a motor at startup is highly unsafe and impractical. Excessive voltage can lead to overheating, insulation breakdown, and permanent damage to the motor. Additionally, the electrical system may experience significant stress, causing potential hazards such as short circuits or equipment failure.
Option 3: Half the normal voltage.
This option involves applying reduced voltage at startup, which is not recommended for small capacity motors. While reduced voltage methods, such as soft starters or star-delta starters, are used for larger motors to limit inrush current, small motors are designed to handle full normal voltage without requiring such techniques. Applying half voltage may result in slower startup, lower torque, and inefficient operation.
Option 4: Full normal voltage.
This is the correct option. As explained above, small capacity motors (up to 3 Hp) are designed to start directly with full normal voltage, ensuring quick and efficient operation without compromising the motor's performance or safety.
Conclusion:
For small capacity motors, the application of full normal voltage at startup is the most suitable approach. It ensures simplicity, cost-effectiveness, and efficient operation. Other options, such as zero voltage, double voltage, or half voltage, are either impractical or unsuitable for these motors. Understanding the characteristics of small motors and their ability to handle direct-on-line starting is crucial for selecting the appropriate starting method.
Electrical Machines Question 3:
Why do inverted V-curves have an inverted "V" shape ?
Answer (Detailed Solution Below)
Electrical Machines Question 3 Detailed Solution
Explanation:
Inverted V-Curves
Definition: Inverted V-curves are graphical representations of the relationship between the armature current and excitation in a synchronous machine, such as a generator or motor. The term "inverted V-curve" arises from the characteristic shape of the graph, which resembles an inverted "V" or a downward-facing triangle.
Correct Option Analysis:
Option 2: Power factor peaks at optimal excitation (unity PF).
This is the correct explanation for why inverted V-curves have an inverted "V" shape. The power factor (PF) of a synchronous machine is dependent on its excitation level. When the machine operates at optimal excitation, the power factor reaches unity (1), meaning the voltage and current are perfectly in phase. At this point, the armature current is minimized because the machine operates efficiently without excessive reactive power. As the excitation deviates from this optimal level—either under-excited or over-excited—the power factor decreases, leading to an increase in armature current. This behavior creates the characteristic inverted "V" shape in the graph of armature current versus excitation.
Explanation of Power Factor:
Power factor is a measure of how effectively electrical power is converted into useful work output. It is the cosine of the angle between voltage and current. A synchronous machine achieves unity power factor (cosθ = 1) when the excitation is adjusted such that the reactive power is minimized, and the machine operates purely with active power. At unity power factor, the armature current is at its lowest, and the machine operates most efficiently.
How Inverted V-Curves are Formed:
- When the excitation is less than the optimal level (under-excitation), the machine requires reactive power from the system, causing the armature current to increase.
- When the excitation is greater than the optimal level (over-excitation), the machine supplies reactive power to the system, which also increases the armature current.
- The minimum armature current occurs at the optimal excitation level, where the power factor is unity.
Importance of Inverted V-Curves:
- Inverted V-curves help operators understand the relationship between excitation and armature current, enabling them to adjust the excitation for efficient operation.
- They are essential in identifying the conditions for unity power factor operation, which minimizes losses and ensures stable system performance.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: Mechanical load affects excitation linearly.
This option is incorrect. The inverted V-curve shape is not caused by the mechanical load affecting excitation linearly. While mechanical load does influence synchronous machine operation, the characteristic inverted "V" shape arises specifically from the relationship between armature current and excitation, which is driven by the reactive power and power factor behavior, not the mechanical load.
Option 3: Stator resistance varies non-linearly with excitation.
This option is incorrect. The stator resistance in a synchronous machine is generally constant and does not vary with excitation. The inverted V-curve is a result of the interplay between excitation, reactive power, and armature current, not changes in stator resistance.
Option 4: Armature current peaks at unity power factor.
This option is incorrect and is the opposite of what occurs in an inverted V-curve. At unity power factor, the armature current is minimized, not maximized. This is because the machine operates most efficiently at this point, with minimal reactive power and optimal excitation.
Option 5: (No explanation provided in the question).
Since no explanation is provided for option 5, it is impossible to evaluate its correctness. However, based on the given options, the correct reasoning for the inverted V-curve shape is provided in option 2.
Conclusion:
The inverted V-curve shape in synchronous machines arises from the relationship between armature current and excitation. At optimal excitation, the machine achieves unity power factor, resulting in minimized armature current. Deviations from this optimal level (either under-excitation or over-excitation) increase the armature current due to changes in reactive power. This behavior creates the characteristic inverted "V" shape, which is critical for understanding and optimizing the operation of synchronous machines.
Electrical Machines Question 4:
How can eddy current loss be reduced in the armature core?
Answer (Detailed Solution Below)
Electrical Machines Question 4 Detailed Solution
Explanation:
Reducing Eddy Current Loss in Armature Core
Definition: Eddy current loss is a type of power loss that occurs in the core of electrical machines such as motors, transformers, and generators. It is caused by circulating currents induced within the conductive material of the core due to the alternating magnetic flux. These currents flow in loops within the material, producing heat and resulting in energy loss.
Working Principle of Eddy Currents:
Eddy currents are induced in a conductor when it is exposed to a changing magnetic field, as per Faraday's law of electromagnetic induction. The magnitude of these currents depends on the rate of change of the magnetic flux, the material's electrical conductivity, and the geometry of the conductor. The circulating currents create their own magnetic field, which opposes the original magnetic field (as stated by Lenz's law), leading to energy dissipation in the form of heat.
Correct Option Analysis:
The correct option is:
Option 4: By laminating the core to reduce the flow of eddy currents.
This is the most effective method for minimizing eddy current loss. The lamination process involves dividing the core into thin layers or sheets of insulated material. These laminations are stacked together, and each layer is electrically insulated from the others, typically using a thin coating of varnish or oxide. The purpose of laminating the core is to restrict the flow of eddy currents by reducing the area available for their circulation. As a result, the eddy current paths are interrupted, and their magnitude is significantly decreased.
Why Laminating the Core Works:
- The induced voltage in the core due to the alternating magnetic flux is proportional to the rate of change of flux and the area of the loop (as per Faraday's law). By reducing the cross-sectional area of the loops using laminations, the induced voltage and hence the eddy current magnitude are minimized.
- The heat generated by eddy currents is directly proportional to the square of the current. Therefore, reducing the magnitude of eddy currents through lamination significantly reduces energy losses.
- Laminations are typically made of high-resistance materials like silicon steel, which further limits the flow of eddy currents.
Advantages of Lamination:
- Significant reduction in eddy current losses, improving the efficiency of electrical machines.
- Cost-effective and straightforward technique for core design.
- Improved thermal performance due to reduced heat generation.
Applications:
Laminated cores are widely used in transformers, electric motors, generators, and other electrical machines where alternating magnetic fields are present. This technique is crucial for ensuring the efficient operation of these devices.
Additional Information:
Eddy current loss is one of the two primary core losses in electrical machines, the other being hysteresis loss. While lamination is effective for reducing eddy current loss, hysteresis loss is minimized by using magnetic materials with low hysteresis, such as silicon steel.
Important Information
To further understand the analysis, let’s evaluate the other options:
Option 1: By increasing the motor's speed.
This option is incorrect as increasing the motor's speed does not reduce eddy current loss. In fact, higher speeds result in a higher rate of change of magnetic flux, which increases the induced voltage and, consequently, the eddy currents. This leads to greater energy loss and heat generation.
Option 2: By increasing the core resistance.
While increasing the core resistance can theoretically reduce eddy current losses, it is not a practical solution. Core resistance is primarily determined by the material properties and geometry of the core. Lamination is a more effective and feasible method for increasing the core's effective resistance to eddy currents without compromising the machine's performance.
Option 3: By using a high resistance core material.
Using high-resistance materials like silicon steel can help in reducing eddy current losses. However, this alone is not sufficient to completely mitigate the problem. Lamination remains the most effective method for minimizing eddy currents, even when high-resistance materials are used.
Option 5: (No option provided in this case).
This option is not applicable in the given context.
Conclusion:
Among the given options, laminating the core to reduce the flow of eddy currents is the most effective and widely used method for minimizing eddy current losses. This technique significantly improves the efficiency and performance of electrical machines by reducing energy dissipation and heat generation. Understanding and implementing effective strategies to mitigate core losses is essential for the optimal design and operation of electrical devices.
Electrical Machines Question 5:
In a Synchronous reluctance motor, flux barriers are typically filled with _____________
Answer (Detailed Solution Below)
Electrical Machines Question 5 Detailed Solution
Synchronous Reluctance Motor
Definition: A Synchronous Reluctance Motor (SynRM) is a type of electric motor that operates using the principle of reluctance torque. It does not require windings or permanent magnets on the rotor, making it a simple and cost-effective motor design. The rotor in a synchronous reluctance motor is specifically designed to have a high degree of anisotropy, meaning it has different magnetic reluctance in different directions.
Flux Barriers and Their Purpose:
In the rotor of a synchronous reluctance motor, "flux barriers" are strategically placed to guide the magnetic flux. These barriers are essentially non-magnetic regions that prevent the magnetic flux from passing through certain parts of the rotor, thereby creating a preferred path for the flux. This anisotropy is key to the operation of the motor, as it enables the generation of reluctance torque.
The flux barriers in a synchronous reluctance motor are typically filled with air or non-magnetic material (Option 4). These materials have very low magnetic permeability, which ensures that magnetic flux is restricted from passing through them. By doing so, the rotor creates distinct magnetic paths with varying reluctances, thereby improving the motor's efficiency and torque production.
Advantages of Filling Flux Barriers with Air or Non-Magnetic Material:
- Cost-Effectiveness: Air or non-magnetic materials are inexpensive compared to other materials like ferrite or copper, reducing the overall cost of the motor.
- Lightweight: Using air or lightweight non-magnetic materials helps keep the rotor's weight low, improving dynamic performance and reducing inertia.
- Improved Torque Characteristics: The anisotropic design created by these barriers enhances the motor's ability to generate reluctance torque efficiently.
- Thermal Stability: Non-magnetic materials generally exhibit good thermal properties, ensuring stable operation over a wide range of temperatures.
Applications: Synchronous reluctance motors are widely used in applications requiring high efficiency and cost-effectiveness, such as in industrial drives, pumps, and fans.
Analysis of Other Options
To further understand the correct option, let’s evaluate the other options:
Option 1: Ferrite
Ferrite is a magnetic material commonly used in transformers, inductors, and magnetic cores. However, it is not suitable for filling flux barriers in a synchronous reluctance motor. The purpose of the flux barriers is to create regions of high reluctance (low permeability), which is not achievable with ferrite since it is a magnetic material. Using ferrite would defeat the purpose of creating anisotropy in the rotor, thereby reducing the motor's efficiency and torque production.
Option 2: Copper
Copper is an excellent conductor of electricity and is commonly used in windings and electrical connections. However, it is not appropriate for filling flux barriers in a synchronous reluctance motor. Copper does not have the non-magnetic properties required for creating high reluctance regions. Additionally, using copper would increase the motor's cost and weight unnecessarily, without providing any benefits to the flux barrier design.
Option 3: Laminated Steel
Laminated steel is used in the construction of motor cores to reduce eddy current losses. However, it is a magnetic material with high permeability, making it unsuitable for flux barriers in a synchronous reluctance motor. The purpose of the flux barriers is to restrict the magnetic flux, which cannot be achieved with laminated steel. Instead, laminated steel is typically used in the stator core of the motor, where magnetic flux needs to be guided effectively.
Conclusion:
The correct option for filling flux barriers in a synchronous reluctance motor is air or non-magnetic material (Option 4). This choice ensures the creation of high reluctance regions in the rotor, which is essential for efficient motor operation. The use of air or non-magnetic materials contributes to the motor's cost-effectiveness, lightweight design, and improved torque characteristics. Other options such as ferrite, copper, and laminated steel are not suitable for this purpose, as they do not meet the requirements for creating the necessary anisotropic properties in the rotor.
Top Electrical Machines MCQ Objective Questions
In the method of speed control of induction motor by inducing emf in the rotor circuit, if the injected voltage is in phase opposition to the induced rotor emf, then:
Answer (Detailed Solution Below)
Electrical Machines Question 6 Detailed Solution
Download Solution PDFRotor emf injection method:
For below-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is 180° out of phase with rotor emf.
E2R is resultant emf in the rotor
E2R = E2 – E1
\(T \propto \frac{{sE_{2R}^2}}{{{R_2}}}\)
R2 is rotor resistance
T is the torque
s is the slip
Here, the value of rotor emf becomes less. To maintain constant torque, the value of slip will increase. Therefore, the speed will be decreased.
At this condition, effective rotor resistance increases.
For above-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is in phase with rotor emf.
E2R is resultant emf in the rotor
E2R = E2 + E1
\(T \propto \frac{{sE_{2R}^2}}{{{R_2}}}\)
R2 is rotor resistance
T is the torque
s is the slip
Here, the value of rotor emf becomes more. To maintain constant torque, the value of slip will decrease. Therefore, the speed will be increased.
At this condition, effective rotor resistance decreases.
The capacitor commonly used for ceiling fan motors has a value of 2.3 μF. The type of capacitor is used __________.
Answer (Detailed Solution Below)
Electrical Machines Question 7 Detailed Solution
Download Solution PDF- The single-phase induction motor is not self-starting. Hence, it requires an auxiliary means or equipment to start the motor.
- Mechanical methods are impractical and, therefore the motor is started temporarily converting it into the two-phase motor.
- Commonly used starting methods for a ceiling fan is a permanent capacitor or single value capacitor motor.
- The permanent capacitor motor also has a cage rotor and the two windings named as main and auxiliary windings.
- The capacitor C is permanently connected in the circuit both at the starting and the running conditions.
- It is also called as a single value capacitor motor as the capacitor is always connected in the circuit.
- A paper capacitor is used in the permanent capacitor motor as an electrolytic capacitor cannot be used for the continuous running operation of the ceiling fan.
- The cost of the paper capacitor is higher, and the size is also large as compared to the electrolytic capacitor of the same ratings.
How can a load be shifted from one DC shunt generator to another running in parallel?
Answer (Detailed Solution Below)
Electrical Machines Question 8 Detailed Solution
Download Solution PDF- When two generators are operating in parallel, the load may be shifted from one shunt generator to another merely by adjusting the field excitation
- If generator 1 is to be shut down, the whole load can be shifted onto generator 2 provided it has the capacity to supply that load
- In that case, the field current of generator 1 gradually reduces to zero
Important Points:
For stable parallel operation, the most suitable type of DC generator is a shunt generator as it has slightly drooping characteristics. If there is any tendency for a generator to supply more or less than its proper share of load it changes system voltage which certainly opposes this tendency. This restores the original division of load. Thus the shunt generators automatically remain in parallel, once they are paralleled.
The characteristics of dc shunt generator are shown below
Flemings right hand rule is used to find the
Answer (Detailed Solution Below)
Electrical Machines Question 9 Detailed Solution
Download Solution PDFFleming right-hand thumb rule:
When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction.
Fleming's right-hand rule (for generators) shows the direction of induced current when a conductor attached to a circuit moves in a magnetic field.
- The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.
- The first finger is pointed in the direction of the magnetic field. (north to south)
- Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)
Finger |
Right-hand rule |
Left-hand rule |
Thumb |
The direction of motion of conductor (input) |
The direction of the conductor (output) |
Forefinger |
Magnetic field |
Magnetic field |
Middle finger |
The direction of induced emf (output) |
The direction of current (input) |
Conclusion:
The direction of induced emf is known by Flemings right-hand thumb rule
Induction generators deliver power at ______ power factor
Answer (Detailed Solution Below)
Electrical Machines Question 10 Detailed Solution
Download Solution PDFInduction generator always works with leading power factor since it will take large amount of reactive power to produce sufficient amount of working flux so that armature reaction is always magnetizing hence it will work always with leading pf.
Important Point
- induction generator is basically an induction motor, which runs above the synchronous speed
- when it acts as a generator it will supply the active power back to source, but for this supply of active power it needs reactive power as input to keep its winding excited
- in case if the induction motor is connected to the grid, it will draw the required reactive power for the excitation of windings, but if it is standalone system (ie. not connected to grid) then a capacitor bank will be always connected, and this will provide leading reactive power to keep the winding excited for the process of mechanical to electrical energy conversion
- since the reactive power is supplied by the capacitor, the induction generator is operating in leading power factor
The three characteristics shown in the given graphs, represent which of the following motors?
Answer (Detailed Solution Below)
Electrical Machines Question 11 Detailed Solution
Download Solution PDFIn a DC motor, T ∝ ϕIa
In series motor, ϕ ∝ Ia
⇒ T ∝ (Ia)2
In shunt motor, ϕ is constant
⇒ T ∝ Ia
The characteristic of DC series and shunt motor are shown below.
Characteristics of DC series motor:
Characteristics of DC shunt motor:
Answer (Detailed Solution Below)
Electrical Machines Question 12 Detailed Solution
Download Solution PDFBased on the connection of armature and field windings DC generators can be classified as:
Type of DC Machine |
Circuit diagram |
Separately excited DC generator |
|
DC shunt generator |
|
DC series generator |
|
DC short shunt compound generator |
|
DC long shunt compound generator |
|
Therefore, the machine shown in the question represents a DC long shunt compound generator.
Which of the following generators at load condition offers positive poorest voltage regulation?
Answer (Detailed Solution Below)
Electrical Machines Question 13 Detailed Solution
Download Solution PDFThe correct answer is option "3'.
Concept :- For differential compound generator it is positive (Poorest positive regulation among all)
- The voltage regulation of a generator is defined as the change in the voltage drop from no load to full load to full load voltage.
- Voltage regulation = (no-load voltage - full load voltage)/full load voltage
- In the case of a series generator, the field is connected in series with the armature. Any increase in load current causes an increase in the field and hence the terminal voltage rises.
- Hence it has negative voltage regulation and it has the poorest voltage regulation.
During on-load conditions, the differentially compounded DC generator has the poorest voltage regulation as shown.
During the no-load condition, the DC series generator has the poorest voltage regulation.
Key Points
- Voltage regulation of shunt generator is positive
- For series generator, it is negative (Poorest negative voltage regulation among all)
- For over compound generator it is negative
- For under compound it is positive
- For, flat compound generator it is zero (lowest Voltage regulation among all)
- For differential compound generator it is positive (Poorest positive regulation among all)
In case of dc shunt motors, the regenerative braking is employed when the load _________.
Answer (Detailed Solution Below)
Electrical Machines Question 14 Detailed Solution
Download Solution PDFRegenerative braking: In this type braking back emf Eb is greater than the supply voltage V, which reverses the direction of the motor armature current. The motor begins to operate as an electric generator.
Overhauling motor: A motoring motor is converting electrical energy into mechanical energy. An overhauling motor is being driven by the load and is converting mechanical energy into electricity, acting as a generator.
In case of regenerative braking, dc shunt motor acts a generator and hence regenerative braking is employed when the load has an overhauling characteristic.Determine the pitch factor for winding: 36 stator slots, 4-poles, coil spans 1 to 8.
Answer (Detailed Solution Below)
Electrical Machines Question 15 Detailed Solution
Download Solution PDFConcept:
Pitch factor for nth harmonic is given by,
\({k_c} = \cos \frac{{n\alpha }}{2}\)
Where α is short pitch angle in degrees
Calculation:
Given-
Total slots = 36,
Number of poles = 4
Coil span = 1 to 8 = 8 - 1 = 7 slots
Now, Slots per pole = 36 / 4 = 9
Number of empty slots = 9 – 7 = 2 slots
\(\therefore \alpha = 180^\circ \times \frac{2}{{9}} = 40^\circ\)
Hence pitch factor can be calculated as
Kc = cos 20°