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# Multiple choice question for engineering

## Set 1

1. What is the formula for current in each of rotor bar?
a) current in rotor bar = 2 * slot pitch * window space factor * stator torque * stator current * power factor * rotor slots
b) current in rotor bar = 2 * slot pitch * window space factor / stator torque * stator current * power factor * rotor slots
c) current in rotor bar = 2 * slot pitch * window space factor * stator torque * stator current * power factor / rotor slots
d) current in rotor bar = 2 * slot pitch / window space factor * stator torque * stator current * power factor / rotor slots

Answer: c [Reason:] For the calculation of the current through each rotor bar, firstly we find out slot pitch, window space factor and rotor slots. Then the stator torque and stator current are obtained with respect to stator side.

2. What is the relation between rotor mmf and stator mmf?
a) rotor mmf = 0.85 * stator mmf
b) rotor mmf = 0.80 * stator mmf
c) rotor mmf = 0.75 * stator mmf
d) rotor mmf = 0.70 * stator mmf

Answer: a [Reason:] First the stator mmf is calculated. Then it is multiplied by 0.85 to obtain the rotor mmf.

3. What is the relation of the rotor resistance with respect to the starting torque?
a) rotor resistance is indirectly proportional to the starting torque
b) rotor resistance is directly proportional to the starting torque
c) rotor resistance is indirectly proportional to the square of the starting torque
d) rotor resistance is directly proportional to the square of the starting torque

Answer: b [Reason:] The rotor resistance is directly proportional to the starting torques. High resistance leads to high starting torque.

4.What is the relation of the rotor resistance to efficiency and losses?
a) as rotor resistance, losses increase, efficiency increases
b) as rotor resistance, losses increase, efficiency decreases
c) as rotor resistance, losses decrease, efficiency has no change
d) as rotor resistance, losses decrease, efficiency decreases

Answer: b [Reason:] As the rotor resistance increases, the I2R losses increases and cause heating effects. This increase in losses decreases the efficiency.

5. What is the relationship between current density, conductor area and resistance?
a) higher the current density, higher the conductor area, higher the resistance
b) higher the current density, higher the conductor area, lower the resistance
c) higher the current density, lower the conductor area, higher the resistance
d) lower the current density, lower the conductor area, lower the resistance

Answer: c [Reason:] Higher the current density leads to lower conductor area, as current density is the ratio of current per area. As the conductor area decreases, resistance increases.

6. What is the formula for the calculation of rotor resistance?
a) rotor resistance = resistance of the bars + resistance of end rings
b) rotor resistance = resistance of the bars – resistance of end rings
c) rotor resistance = resistance of the bars * resistance of end rings
d) rotor resistance = resistance of the bars / resistance of end rings

Answer: a [Reason:] First the resistance of the bars are obtained. Next the resistance of the end rings are calculated and the sum gives the rotor resistance.

7. What is the range of current density in rotor bars?
a) 4-9 A per mm2
b) 4-6 A per mm2
c) 4-7 A per mm2
d) 5-6 A per mm2

Answer: b [Reason:] The minimum value of the current density in the rotor bars is 4 A per mm2. The maximum value of the current density in the rotor bars is 6 A per mm2.

8. What is the formula for the area of each bar?
a) area of each bar = current of the rotor bars + current density in rotor bars
b) area of each bar = current of the rotor bars / current density in rotor bars
c) area of each bar = current of the rotor bars * current density in rotor bars
d) area of each bar = current of the rotor bars – current density in rotor bars

Answer: b [Reason:] For calculating the area of each bar, current flowing across the rotor bars should be first calculated. Then the current density in rotor bars should be calculated next and the ratio gives the area of each bar.

9. Can closed slots are preferred for small machines?
a) true
b) false

Answer: a [Reason:] Closed slots are preferred for small machines. It is because the reluctance of the air gap is not large owing to absence of slot openings.

10. What is the relation of closed slots with leakage reactance?
a) closed slots give no leakage reactance
b) closed slots give high leakage reactance
c) closed slots give low leakage reactance
d) closed slots give negative leakage reactance

Answer: b [Reason:] It is an advantage that closed slots give large leakage reactance. If the leakage reactance is large, the current at the starting can be limited.

11. What is the relation of closed slots with leakage reactance and overload capacity?
a) closed slots give high leakage reactance, and increases the overload capacity
b) closed slots give high leakage reactance, and decreases the overload capacity
c) closed slots give low leakage reactance, and decreases the overload capacity
d) closed slots give low leakage reactance, and increases the overload capacity

Answer: b [Reason:] The closed slots have the main advantage of giving high leakage reactance. A high leakage reactance gives the advantage that the current at the starting can be limited.

12. What is the relation between surface of rotor and the operation?
a) smooth surface leads to the quiet operation
b) rough surface leads to the quiet operation
c) smooth surface leads to the noisy operation
d) rough surface leads to the noisy operation

Answer: a [Reason:] For the design of the rotor bars of the three phase induction machine, smooth surface is preferred. Smooth surface helps in the silent operation.

13. Rectangular shaped bars and slots are preferred to circular bars and slots?
a) true
b) false

Answer: a [Reason:] Rectangular shaped bars and slots are preferred to circular bars and slots. This is because while using the rectangular shaped bars, rotor resistance increases and this leads to the improvement of starting torque.

14. What is the relation between clearances and slots?
a) high clearances are provided for salient slots
b) low clearances are provided for skewed slots
c) low clearances are provided for salient slots
d) high clearances are provided for skewed slots

Answer: d [Reason:] When the skewed slots are being used, higher clearances are provided. High clearances can lead to the smooth and efficient operation of the machine for skewed slots.

15. What is the range of clearance that can be left between rotor bars and the core?
a) 0.1-0.4 mm
b) 0.2-0.4 mm
c) 0.15-0.4 mm
d) 0.4-0.6 mm

Answer: c [Reason:] The range of clearance is chosen based on whether the slots are skewed or not. The range is usually chosen between 0.15-0.4 mm.

## Set 2

1. What is the work of the frame of dc machines?
a) to reduce the voltage
b) to reduce the flux
c) to carry the flux
d) to carry the current

Answer: c [Reason:] The main function of the frame of dc machines is to carry the flux. Thus the frame must be large enough to carry flux.

2. Why is the length of the yoke made larger?
a) to protect the armature windings
b) to cover the armature windings
c) to cover the field windings
d) to cover and protect the field windings

Answer: d [Reason:] The length of the yoke is usually made larger than the pole cores. It is because to cover and protect the field windings.

3. What is the formula for the depth of the yoke?
a) depth of yoke = thickness/2
b) depth of yoke = thickness
c) depth of yoke = 2*thickness
d) depth of yoke = 3*thickness

Answer: b [Reason:] The depth of yoke is equal to the thickness of the yoke. It is calculated to give the required cross-section for the magnetic circuit.

4.In large machines, the thickness is relatively larger to the diameter?
a) true
b) false

Answer: b [Reason:] The thickness is used in the calculation of the depth of the yoke. In large machines, the thickness is relatively smaller to the diameter.

5. What is the formula in order to check the rigidity?
a) moment of inertia ≥ (weight of magnetic frame * radius2 * 10-6) / 225
b) moment of inertia ≤ (weight of magnetic frame * radius2 * 10-6) / 225
c) moment of inertia = (weight of magnetic frame * radius2 * 10-6) / 225
d) moment of inertia < (weight of magnetic frame * radius2 * 10-6) / 225

Answer: a [Reason:] The moment of inertia, the weight of magnetic frame and the radius is calculated first. The machine is highly rigid if the moment of inertia is greater than or equal to the product of weight of magnetic frame and square of radius divided by 225.

6. What is the formula for the thickness of the ac machines?
a) thickness = 40 * inner diameter of frame/12
b) thickness = 40 + inner diameter of frame/12
c) thickness = 40 – inner diameter of frame/12
d) thickness = 40 * inner diameter of frame*12

Answer: a [Reason:] The thickness of the ac machines depend upon the inner diameter of the frame. On obtaining the inner diameter of frame and on substitution gives the thickness of ac machines.

7.What is the formula for the breadth of the ac machine?
a) breadth = 6 + 0.01 * inner diameter of frame
b) breadth = 6 – 0.01 * inner diameter of frame
c) breadth = 6 * 0.01 * inner diameter of frame
d) breadth = 6 / 0.01 * inner diameter of frame

Answer: a [Reason:] The breadth of the ac machines also depends upon the inner diameter of the frame. On substituting the values the breadth is calculated.

8. What is the formula for the checking of rigidity of induction machines?
a) moment of inertia ≥ radius / length of stator core * 90
b) moment of inertia ≥ radius * length of stator core * 90
c) moment of inertia ≥ radius * length of stator core / 90
d) moment of inertia ≤ radius / length of stator core * 90

Answer: c [Reason:] The radius and length of the stator core along with the moment of inertia is calculated. If the moment of inertia is greater than or equal to the product of length and radius divided by 90, the machine is more rigid.

9. What is the formula for the radius at the centre of gravity?
a) radius at the centre of gravity = inner diameter1.5/ 6.3
b) radius at the centre of gravity = inner diameter2/ 6.3
c) radius at the centre of gravity = outer diameter1.5/ 6.3
d) radius at the centre of gravity = outer diameter2/ 6.3

Answer: c [Reason:] The outer diameter of stator core is first calculated. On substituting the values the radius at the centre of gravity is obtained.

10. What is the formula of the centrifugal force?
a) centrifugal force = weight of revolving body * 39.43 * speed2* radius of circular path
b) centrifugal force = weight of revolving body / 39.43 * speed2* radius of circular path
c) centrifugal force = weight of revolving body * 39.43 / speed2* radius of circular path
d) centrifugal force = weight of revolving body * 39.43 * speed2 / radius of circular path

Answer: a [Reason:] The weight of revolving body, speed, radius of circular path is calculated. On substitution the centrifugal force is obtained.

## Set 3

1. What is the formula for number of turns in primary winding?
a) number of turns of primary winding = Voltage of primary windings * voltage per turn
b) number of turns of primary winding = Voltage of primary windings / voltage per turn
c) number of turns of primary winding = Voltage of secondary windings * voltage per turn
d) number of turns of primary winding = Voltage of secondary windings / voltage per turn

Answer: b [Reason:] For calculating the number of turns of primary windings first we calculate the voltage across the primary windings. Then the voltage per turn is calculated and the ratio gives the number of turns.

2. What is the formula for obtaining the current in the primary winding?
a) current in primary winding = kVA per turn * 103 * primary voltage
b) current in primary winding = kVA per phase * 103 * primary voltage
c) current in primary winding = kVA per turn * 103 / primary voltage
d) current in primary winding = kVA per phase * 103 / primary voltage

Answer: d [Reason:] For obtaining the current in primary winding, the kVA output per phase is obtained. Then the primary voltage is calculated, and the ratio of both gives the current in primary windings.

3. What does the area of conductors in primary and secondary windings depend on?
a) current
b) voltage
c) power
d) current density

Answer: d [Reason:] The area of the conductors is directly dependent on the current density. The area of the conductors are determined after choosing the suitable current density.

4. What does the permissible current density depend upon?
a) local heating
b) efficiency
c) output power
d) local heating and efficiency

Answer: d [Reason:] The permissible current density depends upon the local heating as the heating should not affect the output. It also depends on the efficiency of the transformer.

5. What is the relationship between temperature and the current density?
a) current density is directly proportional to the temperature
b) current density is directly proportional to the square of the temperature
c) current density is indirectly proportional to the square of the temperature
d) current density is indirectly proportional to the temperature

Answer: a [Reason:] As current density increases, the temperature also increases. As the temperature increases it can cause damage to the insulation.

6. What is the relationship between the losses and the maximum efficiency on the current density?
a) current density increases, losses decrease, efficiency increases
b) current density increases, losses increase, efficiency increases
c) current density decreases, losses decrease, efficiency increases
d) current density decreases, losses increase, efficiency increases

Answer: c [Reason:] As the current density decreases, the losses decrease. As the losses decrease the maximum efficiency increases.

7. What is the range of current density for small and medium power transformers?
a) 1-2 A per mm2
b) 1-2.5 A per mm2
c) 1.1-2.2 A per mm2
d) 1.1-2.3 A per mm2

Answer: d [Reason:] In small and medium power transformers, the lowest value of current density is 1.1. The highest permissible value is 2.3 for small and medium power transformers.

8. What is the range of current density for large power transformer with self oil cooled type?
a) 1-2 A per mm2
b) 1.5-2.5 A per mm2
c) 2.2-3.2 A per mm2
d) 2-3 A per mm2

Answer: c [Reason:] For large transformers with self oil cooled type, the highest permissible value of current density is 3.2. The minimum current density value required is 2.2.

9. What is the condition for minimum loss condition?
a) current density in primary < current density in secondary
b) current density in primary > current density in secondary
c) current density in primary = current density in secondary
d) current density in primary >= current density in secondary

Answer: c [Reason:] The condition for the minimum loss should be the current density in primary should be equal to the current density in secondary. Any different condition, could lead to high amount of loss.

10. The current density in relatively better cooled outer winding is made 10 percent greater than the inner winding?
a) true
b) false

Answer: b [Reason:] In practical case, the current density in relatively better cooled outer winding is made greater than that in the inner winding. It is usually made 5 percent greater in practical.

11. How many total high voltage windings are present?
a) 1
b) 2
c) 3
d) 4

Answer: c [Reason:] There are 3 high voltage windings present. They are i) Cylindrical winding , ii) Cross-over winding iii) Continuous disc type winding.

12. The low voltage windings are generally divided into 2 types?
a) true
b) false

Answer: a [Reason:] The low voltage windings are basically divided into 2 types. They are i) cylindrical winding ii) helical winding.

13. What is the rating for cylindrical type of winding with circular conductors?
a) 5000-10000 kVA
b) 5000-8000 kVA
c) 160-10000 kVA
d) 200-10000 kVA

Answer: a [Reason:] 5000-8000 kVA is used for rectangular conductors with cylindrical winding. 160-10000 kVA is used for helical winding. 200-10000 kVA is used for continuous disc type of winding.

14. What is the voltage for cross over type of winding?
a) upto 15 kV
b) upto 33 kV
c) upto 66 kV
d) upto 6 kV

Answer: b [Reason:] Helical windings have a voltage of upto 15 kV. Whereas, the cylindrical winding with rectangular conductors have a voltage of upto 6 kV.

15. What is the maximum current per conductor for helical winding?
a) from 12 A and above 12 A
b) from 300 A and above 300 A
c) upto 40 A
d) upto 80 A

Answer: b [Reason:] The maximum current per conductor for continuous disc winding is from 12 A and above 12 A. The maximum current per conductor for cross over winding is upto 40 A and the maximum current per conductor for cylindrical winding with circular conductors is upto 80 A.

## Set 4

1. Which material has the highest conductivity of all materials?
a) Silver
b) Copper
c) Gold
d)Tungsten

Answer: a [Reason:] On a scale of 100, silver has 100 percent on high conductivity, copper has 97. When compared to silver and copper gold has only 76 percent. Tungsten is not a material of this group.

2. High conductivity materials are used in electrical machines.
a) True
b) False

Answer: a [Reason:] These materials have low resistivity. Hence they allow for the good flow of current, which in turn allows the proper operation of the machine.

3. What are the characteristics of high conductivity materials based on cost and flexibility?
a) Low cost, low flexibility
b) Low cost, high flexibility
c) High cost, low flexibility
d) High cost, high flexibility

Answer: b [Reason:] Cost should be always less, in order to help in purchase of many quantities of the material for more applications. It should also be highly flexible, in order to mould according to people’s choice.

4. What is the temperature coefficient of silver?
a) 0.00386 per0C
b) 0.0034 per0C
c) 0.00429 per0C
d) 0.0038 per0C

Answer: d [Reason:] 0.0034 per0C relates to the temperature coefficient of Gold, whereas 0.00429 per0C is the temperature coefficient of Aluminum. 0.00386 per0C corresponds to temperature coefficient of Copper.

5. Silver is not used in practical electrical machines.
a) True
b) False

Answer: a [Reason:] Silver has lots of properties which can make it suitable to be used in practical use. But the high cost factor which occurs to Silver makes it used only for important instruments.

6. What is the conductivity of Copper?
a) 0.6329*106 mho/cm
b) 0.5952*106 mho/cm
c) 0.4529*106 mho/cm
d) 0.3773*106 mho/cm

Answer: b [Reason:] 0.6329*106 mho/cm relates to the conductivity value of Silver. 0.4529*106 mho/cm relates to the conductivity value of Gold and 0.4529*106 mho/cm relates to the conductivity value of Aluminum.

7. What is the melting point of aluminum?
a) 6600C
b) 10850C
c) 9620C
d) 10640C

Answer: b [Reason:] 6600C is the melting point of Aluminum. 9620C relates to the melting point of Silver and 10850C is the melting point of Copper.

8. What is the specific gravity of aluminum?
a) 8.96 gm/cm3
b) 19.30 gm/cm3
c) 2.70 gm/cm3
d) 10.49 gm/cm3

Answer: c [Reason:] 8.96 gm/cm3 is the specific gravity of Copper. 19.30 gm/cm3 relates to the specific gravity of Gold and 10.49 gm/cm3 is the specific gravity of Silver.

9. Which two elements are used in precious instruments?
a) Copper, Silver
b) Gold, Silver
c) Copper, Aluminum
d) Gold, Aluminum

Answer: b [Reason:] Silver is used only in precious instruments because of its high cost. Gold, on the other hand, is not only costly, but also not suitable for many practical applications and can lose its properties easily.

10. Which property of aluminum it the most preferred element?
a) good conductivity
b) highly malleable, highly ductile
c) most abundant element
d) good corrosion resistant

Answer: c [Reason:] All the other elements among Silver, Copper, Gold have the other 3 properties along with Aluminum. But all the above mentioned materials are not highly abundant, which is also an important factor.

## Set 5

1. What is the property of insulating materials?
a) Prevents the unwanted flow of current
b) Allows the unwanted flow of current
c) Increases the unwanted flow of current
d) Decreases the unwanted flow of current

Answer: a [Reason:] Conductors, allow the flow of current through the material. Insulators are the opposite of conductors. The material doesn’t allow the flow of current through them.

2. In Transmission and Distribution sector, where should the insulators be placed?
a) Between towers and poles
b) Between towers and ground
c) Between towers and conductors
d) Between conductors and ground

Answer: c [Reason:] The insulators are used to block the flow of unwanted current. In power system, already the tower and the conductors are grounded. Thus the insulators are connected between towers and conductors.

3. What is the main cause for the failure of overhead line insulators?
a) Surges
b) Flashover
c) Arching
d) Grounding

Answer: b [Reason:] In overhead lines, there occurs a flow of abnormal over voltages. This abnormal over voltages, causes flashover. This flashover causes damage to overhead line insulators.

4. What happens when some serious phenomenon occurs in the insulators?
a) Puncher is produced in the insulator body
b) Insulator body bulges
c) Insulator body bursts
d) Insulator body tears apart

Answer: a [Reason:] The serious phenomenon is the abnormal over voltage, produced in the insulators. Due to that, flashover occurs in the insulators. This causes puncher of the insulator body.

5. Insulation Resistance should be high in insulators.
a) True
b) False

Answer: a [Reason:] Insulation Resistance is very important in the performance of insulating materials. If the insulation resistance becomes low, high flow of current occurs and can damage the material.

6. How should the properties of strength and dielectric strength in insulating materials?
a) High strength, low Dielectric strength
b) Low strength, low Dielectric strength
c) High strength, high Dielectric strength
d) Low strength, high Dielectric strength

Answer: c [Reason:] The insulator should have high strength in order to prevent the insulating materials. The insulator should have high dielectric strength, in order to hold the electric field without breaking down.

7. What is property of porosity and temperature change in insulating materials?
a) Less, less affected
b) Less, highly affected
c) High, highly affected
d) High, less affected

Answer: a [Reason:] The insulating materials should have less porosity as it should not lose the internal properties due to holes. The material should also be less affected by temperature change in order to preserve its properties.

8. What is the dielectric strength of porcelain insulators?
a) 60 kV/cm
b) 140 kV/cm
c) 50 kV/cm
d) 40 kV/cm

Answer: a [Reason:] Porcelain has a dielectric strength of 60kV/cm. 140 kV/cm denotes the dielectric strength of glass insulator.

9. What is the dielectric strength, coefficient of thermal expansion of glass with respect to porcelain insulators?
a) High, high
b) High, low
c) Low, low
d) Low, high

Answer: b [Reason:] Glass has a higher dielectric strength (140 kV/cm) when compared to porcelain (60 kV/cm) and glass has a lower coefficient of thermal expansion, when compared to porcelain.

10. Glass has lower tensile strength compared to porcelain insulators.
a) True
b) False

Answer: b [Reason:] Glass insulators have all properties better than that of porcelain. Glass has high dielectric strength, low coefficient of thermal expansion and then High tensile strength than that of porcelain.

11. What is the other name of Polymer Insulator?
a) Moisture insulator
b) Core insulator
c) Composite insulator
d) Mixed insulator

Answer: c [Reason:] It is also known as composite insulator. It is known as composite insulator because it consists of both core and the weather sheds in them.

12. How many classifications of overhead line insulators are there?
a) 3
b) 4
c) 5
d) 6

Answer: a [Reason:] There are basically3 types of overhead line insulators. They are Pin type, Suspension type and Stray Insulator type.

13. How many types of electrical insulators are present on the basis of voltage application?
a) 2
b) 3
c) 4
d) 5

Answer: a [Reason:] There are two types of insulators based on voltage application. They are Stay Insulators and Shackle Insulators.

14. How many discs are used in suspension insulators for 220kV?
a) 3
b) 4
c) 8
d) 14

Answer: d [Reason:] 3 discs are used when voltage is 33kV. 4 discs are used when voltage is 66kV. 8 discs are used when voltage application is 132kV.

15. What is the other name of shackle insulator?
a) String
b) Hanging
c) Spool
d) Post