Design of Transformers MCQs ( Design Of Electrical Machines ) MCQs – Design Of Electrical Machines MCQs
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Design Of Electrical Machines MCQs – Design of Transformers MCQs ( Design Of Electrical Machines ) MCQs
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Window Space Factor
1. What is window space factor?
a) window space factor = copper area in the window – total window area
b) window space factor = copper area in the window + total window area
c) window space factor = copper area in the window * total window area
d) window space factor = copper area in the window / total window area
Answer: d
Explanation: Window space factor is the ratio of the copper area in the window to the total window area. It is a constant used in the output equation of transformers.
2. What does the window space factor depend on?
a) it depends on the core
b) it depends on the armature
c) it depends on the insulation
d) it depends on the insulation and copper
Answer: d
Explanation: Window space factor totally depends on the insulation of the machine. It also depends on the copper windings provided.
3. What does the insulation and copper of the transformer depend on?
a) current rating
b) voltage rating
c) output power
d) voltage rating and output power
Answer: d
Explanation: The insulation and copper of the transformer depend on the voltage rating of the transformer. It also depends on the output power produced by the transformer.
4. What is the empirical formula for calculating the value of window space factor?
a) window space factor = 10 / (30+kV)
b) window space factor = 5 * (30+kV)
c) window space factor = 10 * (30+kV)
d) window space factor = 5 / (30+kV)
Answer: a
Explanation: The output voltage in kV is calculated first. Then the value is substituted in the formula to obtain the empirical value of the window space factor.
5. What is the empirical value of window space factor, given the output is 1000kV?
a) 0.09
b) 0.9
c) 0.009
d) 0.0009
Answer: c
Explanation: Empirical value of window space factor = 10 / (30+kV)
kV = 1000 kV, empirical value = 10/1030 = 0.009.
6. What is the ratings of the transformers for using the empirical value of window space factor?
a) 50-100 kVA
b) 50-150 kVA
c) 50-200 kVA
d) 100-200 kVA
Answer: c
Explanation: The empirical formula of the window space factor is used for the transformers of the rating 50-200 kVA. If the transformer rating is higher than 200 kVA then the empirical formula isn’t used.
7. What is the relationship of the space factor value with the large and small outputs?
a) small value for both large and small outputs
b) large values for both large and small outputs
c) large value for small output and small value for large outputs
d) small value for small output and large value for large outputs
Answer: d
Explanation: The space factor is directly proportional to the output. If large output is obtained, space factor is high and vice versa.
8. What is the formula for the window space factor, when the output is 1000 kVA?
a) 12 / (30+kV)
b) 10 / (30+kV)
c) 9 / (30+kV)
d) 11 / (30+kV)
Answer: a
Explanation: 10 / (30+kV) denotes the empirical value of window space factor for rating between 50-200 kilo-volt-amp. When the output is near 1000 kilo-volt-amp then the formula used is 12/(30+kV).
9. What is the formula of window space factor, when the transformer rating is 20 kVA?
a) 10 / (30+kV)
b) 12 / (30+kV)
c) 8 / (30+kV)
d) 19 / (30+kV)
Answer: c
Explanation: 10 / (30+kV) denotes the empirical value of window space factor for rating between 50-200 kilo-volt-amp. 12 / (30+kV) is the window space factor for transformers having rating around 1000 kVA.
10. The area of the window depends on the window space factor.
a) true
b) false
Answer: a
Explanation: Area of the window = Total conductor area/window space factor.
The area of the window is indirectly proportional to the window space factor.
Design of Core and Winding – 1
1. What is stacking factor?
a) the allowance made for the power loss
b) the allowance made for the space loss between laminations
c) the allowance made for the heat loss between laminations
d) the allowance made for the power loss between laminations
Answer: b
Explanation: The net cross sectional area is obtained from the dimensions of various packets and an allowance is made for the space lost between laminations. This allowance is called stacking factor.
2. What is utilization factor?
a) utilization factor= cross sectional area + gross area of the core
b) utilization factor= cross sectional area – gross area of the core
c) utilization factor= cross sectional area * gross area of the core
d) utilization factor= cross sectional area / gross area of the core
Answer: d
Explanation: The utilization factor is equal to the ratio of the cross sectional area to the gross area of the core. The cross sectional area and the gross area of the core are first found out, and the ratio gives utilization factor.
3. What is the relationship between utilization factor and the number of core steps?
a) utilization factor is directly proportional to the number of core steps
b) utilization factor is indirectly proportional to the number of core steps
c) utilization factor is indirectly proportional to the square of number of core steps
d) utilization factor is directly proportional to the square of number of core steps
Answer: a
Explanation: The utilization factor increases with the increase in the number of core steps used. This eventually increases the manufacturing cost.
4. What is the optimum number of steps for small and large transformers?
a) 5, 10
b) 10, 15
c) 6, 15
d) 1, 10
Answer: c
Explanation: The optimum number of steps used for the large transformers is maximum of 15. The optimum number of steps for the small transformers is maximum of 6.
5. What happens if the utilization factor gets improved?
a) core area increases and the volt/turns decreases
b) core area increases and the volt/turns increases
c) core area decreases and the volt/turn decreases
d) core area decreases and the volt/turn increases
Answer: b
Explanation: When the utilization factor increases the core area gets increased. This leads to the increase in the volt/turn for any particular core diameter and specified flux density.
6. How many types of cores are available for core type of transformer?
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: There are basically 3 types of core section available for core type of transformer. They are rectangular, square or stepped type of core sections.
7. What type of core section is used for shell type transformer?
a) rectangular
b) square
c) stepped
d) cruciform
Answer: a
Explanation: Shell type transformers prefer only rectangular core section. Shell type transformer are moderate and low voltage transformer which use only rectangular core section.
8. What is the range of the ratio of depth to width of core in rectangular core?
a) 1-2
b) 1.5-2.5
c) 1.4-2
d) 1.5-2
Answer: c
Explanation: In rectangular core, the ratio of the depth to core should be minimum 1.4. The maximum value of ratio of depth to core is 2.
9. When is square and stepped cores used?
a) when circular coils are required for low voltage distribution
b) when rectangular coils are required for low voltage distribution
c) when circular coils are required for high voltage distribution
d) when rectangular coils are required for high voltage distribution
Answer: c
Explanation: Circular coils are required for high voltage distribution and power transformer. When circular coils are required square and stepped cores are used.
10. Circular coils are preferred because of their electrical characteristics.
a) true
b) false
Answer: b
Explanation: Circular coils are preferred because of their high mechanical strength. Their high mechanical strength allows them to be used in high voltage distribution and power transformer.
11. What is the ratio of the net core area to the area of the circumscribing circle in square cores?
a) 0.58
b) 0.64
c) 0.70
d) 0.80
Answer: a
Explanation: 0.64 is the ratio of the gross core area to the area of the circumscribing circle. Net core area is the product of stacking factor and gross iron area.
12. The laminations are manufactured in standard size to minimize the wastage of steel during punching of laminations.
a) true
b) false
Answer: a
Explanation: The laminations are manufactured in the standard size of width, 0.75m to 1 m. This is used to avoid excessively wide assortment of laminations and to minimize wastage of steel during punching of laminations.
13. What is the value of ratio of gross core area to the area of circumscribing circle in stepped cores?
a) 0.71
b) 0.79
c) 0.89
d) 0.91
Answer: b
Explanation: 0.71 is the ratio of net core area to the area of circumscribing circle in stepped cores. The gross core area for stepped cores is 0.618 * d2.
14. What is the net core area for three stepped transformers?
a) 0.45
b) 0.56
c) 0.6
d) 0.62
Answer: c
Explanation: 0.45 is the net core area for the square core transformers. 0.56 is the core area for cruciform or stepped core transformers.
15. What is the relationship between the number of steps to the area of circumscribing circle?
a) number of steps is directly proportional to the area of the circumscribing circle
b) number of steps is indirectly proportional to the area of the circumscribing circle
c) number of steps is directly proportional to square of the area of the circumscribing circle
d) number of steps is indirectly proportional to square of the area of the circumscribing circle
Answer: a
Explanation: As the number of steps increase, the area of the circumscribing circle also increases. But as the area of the circumscribing circle increases, the ratio of the net core area and gross core area to the area of circumscribing circle decreases.
Design of Core and Winding – 2
1. What is the formula for the 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
Explanation: 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
Explanation: 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
Explanation: The area of the conductors is directly dependent on the current density. The area of the conductors are determined after choosing a 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
Explanation: 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
Explanation: As the 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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.
Temperature Rise in Transformers
1. The problem of temperature rise and cooling of transformers is essentially the same as that of rotating machinery.
a) true
b) false
Answer: a
Explanation: There are problems of temperature rise and cooling of transformers which decreases the efficiency of the transformers. The same problems are also seen in the rotating machinery.
2. How are the losses in the transformer and rotating machines converted to?
a) the losses are converted to electrical energy
b) the losses are converted to electrical and mechanical energy
c) the losses are converted to mechanical energy
d) the losses are converted to thermal energy
Answer: d
Explanation: In both the transformer and the rotating machines the losses are converted to thermal energy. This thermal energy causes heating of the transformer parts.
3. In how many ways does heat dissipation occur in transformers?
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: The heat dissipation takes place in 3 ways in transformers. They are radiation, convection and conduction.
4. What type of heat dissipation takes place when the heat flows from the outer surface of the transformer part to the oil that cools it?
a) conduction
b) convection
c) conduction and convection
d) radiation
Answer: b
Explanation: When heat flows from the outer surface of transformer part to the oil which cools it, it is convection. In transformers all 3 types of heat dissipation occurs.
5. What type of heat dissipation takes place when heat flows from oil to walls of a cooler?
a) conduction
b) convection
c) radiation
d) conduction and convection
Answer: b
Explanation: When heat flows from oil to walls of the cooler, the heat dissipation type is convection. In transformers all 3 types of heat dissipation takes place.
6. What type of heat dissipation takes place when heat flows from the walls of the cooler to the cooling medium?
a) convection
b) radiation
c) convection and radiation
d) conduction and radiation
Answer: c
Explanation: When the heat flows from the walls of the cooler to the cooling medium, it is both convection and radiation. In transformer all 3 types of heat dissipation occurs.
7. What is the range of the working temperature of oil determined by the tests?
a) 40-60° C
b) 30-60° C
c) 45-60° C
d) 50-60° C
Answer: d
Explanation: The minimum value of the working temperature of oil as cooling medium is determined to be 50°C. The maximum value of the working temperature of oil as cooling medium is determined to be 60°C.
8. What is the formula for specific heat dissipation due to convection of oil?
a) specific heat dissipation = 40.3*(temperature difference of the surface relative to oil/height of the dissipating surface)1/4 W per m2 – °C
b) specific heat dissipation = 40.3 / (temperature difference of the surface relative to oil/height of the dissipating surface)1/4 W per m2 – °C
c) specific heat dissipation = 40.3*(temperature difference of the surface relative to oil * height of the dissipating surface)1/4 W per m2 – °C
d) specific heat dissipation = 40.3*(temperature difference of the surface relative to oil + height of the dissipating surface)1/4 W per m2 – °C
Answer: a
Explanation: First the temperature difference of the surface relative to oil is calculated, then the height of the dissipating surface is also calculated. Substituting in the above formula provides the specific heat dissipation due to convection of oil.
9. What is the value of specific heat dissipation for convection due to air?
a) 8 W per m2 – °C
b) 6 W per m2 – °C
c) 9 W per m2 – °C
d) 10 W per m2 – °C
Answer: a
Explanation: The value of the specific heat dissipation for convection due to air is 8 W per m2 – °C. The value of specific heat dissipation will vary for different medium.
10. The convection due to air is 10 times the convection due to oil.
a) true
b) false
Answer: b
Explanation: The convection due to oil is 10 times the convection due to air. This constitutes a major valuable property of oil as a cooling medium.
11. How do the walls of the transformer tank dissipate heat?
a) by radiation
b) by convection
c) by conduction
d) by convection and radiation
Answer: d
Explanation: The plain walled tanks of the transformer also dissipate heat through convection and radiation. The property is similar to that of the rotating machinery.
12. What is the specific heat dissipated by the plain walled tanks of the transformer by radiation and convection?
a) 6.5, 6 W per m2 – °C
b) 6, 6.5 W per m2 – °C
c) 6.5, 6.5 W per m2 – °C
d) 6, 6 W per m2 – °C
Answer: b
Explanation: The heat dissipated by the plain walled tanks of the transformer by radiation is 6 W per m2 – °C. The heat dissipated by the plain walled tanks of the transformer by convection is 6.5 W per m2 – °C.
13. What is the formula for the temperature rise of the transformers?
a) temperature rise = total loss * specific heat dissipation * surface temperature
b) temperature rise = total loss / (specific heat dissipation * surface temperature)
c) temperature rise = total loss / specific heat dissipation / surface temperature
d) temperature rise = total loss * specific heat dissipation / surface temperature
Answer: b
Explanation: Firstly the surface temperature is calculated along with the specific heat dissipation which is nothing but 12.5. Then the losses are calculated and substituted in the above formula.
14. Can the plain walled tanks accommodate the transformer for both large and small outputs?
a) the plain walled tanks can accommodate for large outputs but cannot accommodate for small outputs
b) the plain walled tanks can accommodate for large outputs and small outputs
c) the plain walled tanks cannot accommodate for large outputs and small outputs
d) the plain walled tanks can accommodate for small outputs but cannot accommodate for large outputs
Answer: d
Explanation: The plain walled tanks are large enough to accommodate the transformer and oil has sufficient surface to keep the temperature rise within limits for small outputs. But the plain walled tanks cannot accommodate the transformers for large outputs.
15. How are the ratings of the transformer, losses and temperature rise related?
a) increase, decrease, increase
b) decrease, increase, increase
c) increase, increase, increase
d) decrease, increase, decrease
Answer: c
Explanation: As the rating of the transformer increases, the losses also increase. As the losses increase the heat dissipated increases and gives high temperature rise.
Design of Tank
1. What is the usage of the tanks with tubes?
a) if the temperature rise with plain tank is very low
b) if the temperature rise with plain tank is very high
c) if the temperature rise is zero
d) if the temperature rise with plain tank exceeds the specific limits
Answer: d
Explanation: Temperature rise in transformers is calculated with plain walled tanks. If the limits is exceeded then the plain walled tank is replaced by tank with tubes.
2. What is the relation of the provision of tubes with respect to dissipation of heat?
a) the provision of tubes is directly proportional to the dissipation of heat
b) the provision of tubes is indirectly proportional to the dissipation of heat
c) the provision of tubes is directly proportional to square of the dissipation of heat
d) the provision of tubes is indirectly proportional to square of the dissipation of heat
Answer: b
Explanation: The provision of tubes increases the dissipating area. The increase in dissipation of heat is not proportional to area because tube screen some of the tank surface preventing radiation from there.
3. What is the relation of the transformer surface with respect to dissipation of heat?
a) transformer surface has no relation with respect to dissipation of heat
b) transformer surface has minor changes with respect to dissipation of heat
c) transformer surface has major changes with respect to dissipation of heat
d) transformer surface has no change with respect to dissipation of heat
Answer: d
Explanation: When the tanks with tubes are provided, the dissipation of heat increases. The dissipation of heat has no effect on the transformer surface.
4. How is the circulation of oil improved in tanks with tubes?
a) it can be improved by using dissipating heat
b) it can be improved by using more effective air circulation
c) it can be improved by using more effect power flow
d) it can be improved by using more effective heads of pressure
Answer: d
Explanation: The circulation of oil is improved in tanks with tubes. It takes place with the help of using more effective heads of pressure.
5. An addition of 35 percent should be made to tube area of the transformers.
a) true
b) false
Answer: a
Explanation: An addition of 35 percent should be made to tube area of the transformer. This should be done in order to take into account this improvement in dissipation of loss by convection.
6. What is the loss dissipated by tubes by convection, given area of the tubes = 3.5?
a) 12.3 W per °c
b) 2.51 W per °c
c) 5.3 W per °c
d) 30.8 W per °c
Answer: d
Explanation: Loss dissipated by tubes by convection = 8.8 * Area of tubes
Loss = 8.8 * 3.5 = 30.8 W per °c.
7. What is the formula for temperature rise with tubes?
a) temperature rise with tubes = total loss / dissipating surface*(12.5 + 8.8x)
b) temperature rise with tubes = total loss * dissipating surface*(12.5 + 8.8x)
c) temperature rise with tubes = total loss / dissipating surface / (12.5 + 8.8x)
d) temperature rise with tubes = total loss + dissipating surface*(12.5 + 8.8x)
Answer: a
Explanation: The total losses in the transformers are obtained firstly the iron loss and copper loss. Next the dissipating surface temperature is obtained and substituting in the above formula gives the temperature rise.
8. What is the formula for number of tubes?
a) number of tubes = (1/ 8 * area of each tube) * (total loss / temperature rise with tubes – 12.5 * dissipating surface)
b) number of tubes = (1* 8 * area of each tube) * (total loss / temperature rise with tubes – 12.5 * dissipating surface)
c) number of tubes = (1/ 8 * area of each tube) / (total loss / temperature rise with tubes – 12.5 * dissipating surface)
d) number of tubes = (1/ 8 * area of each tube) + (total loss / temperature rise with tubes – 12.5 * dissipating surface)
Answer: a
Explanation: First the temperature rise with tubes is obtained. Then the iron loss and copper loss are obtained and added. Area of each tube is also obtained. Substituting all the values in the above formula provides the number of tubes.
9. What is the range of the diameter of the tubes used?
a) 50-60 mm
b) 60-70 mm
c) 70-80 mm
d) 50-70 mm
Answer: d
Explanation: The minimum value of the diameter of tubes is derived to be around 50 mm. The maximum value of the diameter of tubes should be less than 70 mm.
10. Elliptical tubes with pressed radiators are increasingly been used.
a) true
b) false
Answer: a
Explanation: Elliptical tubes with pressed radiators are on high demand now a days. This is because they give a greater dissipating surface for the small volume of oil.
11. What is the formula for width of the tank for single phase transformers used?
a) width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank
b) width of tank = distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank
c) width of tank = 2*distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank
d) width of tank = distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank
Answer: b
Explanation: Width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank is the formula for three phase transformer. For single phase transformers, the distance between adjacent limbs is not multiplied.
12. What is the formula for the length of the tank?
a) length of the tank = external diameter of h.v winding + clearance on each side between the winding and tank along the width
b) length of the tank = external diameter of h.v winding * clearance on each side between the winding and tank along the width
c) length of the tank = external diameter of h.v winding + 2*clearance on each side between the winding and tank along the width
d) length of the tank = external diameter of h.v winding / 2*clearance on each side between the winding and tank along the width
Answer: c
Explanation: The external diameter of h.v winding is obtained. Next the clearance on each side between the winding and tank along the width is calculated and is substituted in the above formula.
13. What is the formula for height of transformer tank?
a) height of transformer tank = Height of transformer frame + clearance height between the assembled transformer and tank
b) height of transformer tank = Height of transformer frame * clearance height between the assembled transformer and tank
c) height of transformer tank = Height of transformer frame/clearance height between the assembled transformer and tank
d) height of transformer tank = Height of transformer frame – clearance height between the assembled transformer and tank
Answer: a
Explanation: Firstly, the height of the transformer frame is calculated. Next, the clearance height between the assembled transformer and tank is also calculated. Substitute the values to obtain the height of transformer tank.
14. What is the rating of the transformer for the voltage of about 11 kV?
a) 1000-2000 kVA
b) 100-3000 kVA
c) 1000-5000 kVA
d) 100-500 kVA
Answer: c
Explanation: The minimum value of the rating of the transformer for a voltage of about 11 kV should be 1000 kVA. The maximum value of the rating of the transformer for a voltage of about 11 kV should be about 5000 kVA.
15. What is the rating of the transformer for the voltage of above 11 kV upto 33 kV?
a) 1000-5000 kVA
b) less than 1000 kVA
c) above 1000 kVA
d) 100-500 kVA
Answer: b
Explanation: 1000-5000 kVA is the rating of the transformer for the voltage of about 11 kV. When the voltage rating is about 11-33 kV, then the rating of the transformer is less than 1000 kVA.
Methods of Cooling of Transformers
1. How many types of cooling methods are available for the transformer?
a) 3
b) 2
c) 1
d) 4
Answer: a
Explanation: There are 3 types of cooling methods available for transformers. They are natural cooling, air blast cooling, forced oil circulation.
2. How are the radiators cooled in the present time?
a) by natural cooling
b) by forced cooling using small fans
c) by forced cooling using large fans
d) by using external air
Answer: b
Explanation: At present time the radiators are cooled using forced cooling. The forced cooling takes place with the help of the small fans mounted on each radiator.
3. What type of cooling is being made use of in transformers having a capacity of less than 11MVA?
a) natural cooling
b) forced cooling
c) air blast cooling
d) forced cooling and air blast cooling
Answer: a
Explanation: For transformers having capacity less than 11MVA, natural cooling is made use of. For transformers having capacity more than 11MVA, air blast cooling is used.
4. Compared to the natural cooling, how much of heat dissipation is increased by air blast cooling?
a) 50-70%
b) 60-70%
c) 50-60%
d) 40-60%
Answer: c
Explanation: Air blast cooling helps in increased heat dissipation. The minimum value of increased heat dissipation is 50% and maximum value is 60%.
5. Increase in the velocity of oil circulation increases the transformer output.
a) true
b) false
Answer: a
Explanation: The increases in velocity of the air circulation increases the temperature. The temperature rise increases the transformer output.
6. What is the relation of the increase of the oil circulation rate with energy losses?
a) increase of the oil circulation rate is not depending with energy losses
b) increase of the oil circulation rate is directly proportional to the energy losses
c) increase of the oil circulation rate is directly proportional to the square of energy losses
d) increase of the oil circulation rate is indirectly proportional to energy losses
Answer: b
Explanation: The increase in the oil circulation rate is unsuitable because this increases the large energy losses In the pumping unit. To cool the oil, it is circulate through a special oil cooler.
7. What is the flow rate of the circulating oil in an air cooler with natural air cooling?
a) 12.5 litre per minute per KW of losses
b) 12 litre per minute per KW of losses
c) 14 litre per minute per KW of losses
d) 13 litre per minute per KW of losses
Answer: b
Explanation: When natural air cooling is used, the flow rate is 12 litres per minute per KW of losses. Even when the air blast cooling is used, the transformer output increases roughly to the same extent.
8. What is the range of the cooler surfaces per 1 KW of losses?
a) 0.1-0.25 m2
b) 0.18-0.2 m2
c) 0.1-0.2 m2
d) 0.18-0.25 m2
Answer: d
Explanation: The minimum value of the cooler surfaces per 1 KW of losses is 0.18 m2. The maximum value of the cooler surfaces per 1 KW of losses is 0.25 m2.
9. What is the range of the flow rate of circulating oil per KW of losses?
a) 6-7 liters per minute
b) 5-6 liters per minute
c) 6-8 liters per minute
d) 6-7 liters per minute
Answer: c
Explanation: The minimum value of the flow rate of circulating oil per KW of losses is derived to be 6 liters per minute. The maximum value of the flow rate of circulating oil per KW of losses is derived to be 8 liters per minute.
10. The temperature difference between the incoming and outgoing water is greater than 10°C.
a) true
b) false
Answer: b
Explanation: The water flow rate is about 1.5 litres per minute. The difference in temperature between the incoming water and outgoing water is 10°C.
11. What is the formula for width of the tank for single phase transformers used?
a) width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank
b) width of tank = distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank
c) width of tank = 2*distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank
d) width of tank = distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank
Answer: b
Explanation: Width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank is the formula for three phase transformer. For single phase transformers, the distance between adjacent limbs is not multiplied.
12. What is the formula for the length of the tank?
a) length of the tank = external diameter of h.v winding + clearance on each side between the winding and tank along the width
b) length of the tank = external diameter of h.v winding * clearance on each side between the winding and tank along the width
c) length of the tank = external diameter of h.v winding + 2*clearance on each side between the winding and tank along the width
d) length of the tank = external diameter of h.v winding / 2*clearance on each side between the winding and tank along the width
Answer: c
Explanation: The external diameter of h.v winding is obtained. Next, the clearance on each side between the winding and tank along the width is calculated and is substituted in the above formula.
13. What is the formula for the height of transformer tank?
a) height of transformer tank = Height of transformer frame + clearance height between the assembled transformer and tank
b) height of transformer tank = Height of transformer frame * clearance height between the assembled transformer and tank
c) height of transformer tank = Height of transformer frame/clearance height between the assembled transformer and tank
d) height of transformer tank = Height of transformer frame – clearance height between the assembled transformer and tank
Answer: a
Explanation: Firstly, the height of the transformer frame is calculated. Next, the clearance height between the assembled transformer and tank is also calculated. Substitute the values to obtain the height of transformer tank.
14. What is the rating of the transformer for the voltage of about 11 kV?
a) 1000-2000 kVA
b) 100-3000 kVA
c) 1000-5000 kVA
d) 100-500 kVA
Answer: c
Explanation: The minimum value of the rating of the transformer for a voltage of about 11 kV should be 1000 kVA. The maximum value of the rating of the transformer for a voltage of about 11 kV should be about 5000 kVA.
15. What is the rating of the transformer for the voltage of above 11 kV upto 33 kV?
a) 1000-5000 kVA
b) less than 1000 kVA
c) above 1000 kVA
d) 100-500 kVA
Answer: b
Explanation: 1000-5000 kVA is the rating of the transformer for the voltage of about 11 kV. When the voltage rating is about 11-33 kV, then the rating of the transformer is less than 1000 kVA.
Overall Dimensions – 1
1. What is the formula for the diameter of the single phase core type transformer?
a) D = diameter of circumscribing circle + Width of window
b) D = diameter of circumscribing circle – Width of window
c) D = diameter of circumscribing circle * Width of window
d) D = diameter of circumscribing circle / Width of window
Answer: a
Explanation: First the diameter of the circumscribing circle is obtained. Next, the width of the window is calculated, and the sum of both the data provides the diameter of the single phase core transformer.
2. What is the formula for height of the single phase core type transformer?
a) height = height of the window – height of the yoke
b) height = height of the window + height of the yoke
c) height = height of the window – (2*height of the yoke)
d) height = height of the window + (2*height of the yoke)
Answer: d
Explanation: The height of the window is first obtained. Next, the height of the yoke is calculated and it is multiplied by 2. Addition of both the values gives the height of the single phase core type transformer.
3. What is the formula for width of the single phase core type transformer?
a) width = Width of largest stamping / Diameter of the transformer
b) width = Width of largest stamping + Diameter of the transformer
c) width = Width of largest stamping – Diameter of the transformer
d) width = Width of largest stamping * Diameter of the transformer
Answer: b
Explanation: Firstly, the width of the largest stamping is calculated. Next, the diameter of the transformer is calculated and the sum of both the values gives the width of the transformer.
4. What is the formula for the width over 2 limbs?
a) width = Width of largest stamping + Diameter of the transformer
b) width = Diameter + outer diameter of hv windings
c) width = Diameter – outer diameter of hv windings
d) width = outer diameter of hv windings
Answer: b
Explanation: width = outer diameter of hv windings is the width over one limb. width = Width of largest stamping + Diameter of the transformer is the formula for width of the transformer.
5. The formula for single phase core type and three phase core type diameter and height are same.
a) true
b) false
Answer: a
Explanation: D = diameter of circumscribing circle + Width of window is the diameter of single phase and 3 phase core type transformers. height = height of the window + (2*height of the yoke) is the height of the single and three phase core type transformers.
6. What is the formula for the width over one limb?
a) width over one limb = outer diameter of hv winding
b) width over one limb = 2*Diameter – outer diameter of hv winding
c) width over one limb = 2*Diameter + outer diameter of hv winding
d) width over one limb = Diameter + outer diameter of hv winding
Answer: a
Explanation: Width over one limb = 2*Diameter + outer diameter of hv winding is the formula for the width over 3 limbs. For one limb the width is equal to the outer diameter of hv winding.
7. What is the formula for the width of the single phase shell type transformer?
a) width = 2*Width of the window + width of the largest stamping
b) width = Width of the window + 4*width of the largest stamping
c) width = Width of the window + width of the largest stamping
d) width = 2*Width of the window + 4*width of the largest stamping
Answer: d
Explanation: First the width of the window is obtained. Next the width of the largest stamping is obtained. Substituting in the above formula provides the width of the single-phase shell type transformer.
8. What is the height of the single phase shell type transformer?
a) height = height of window + width of the largest stamping
b) height = 2*height of window + width of the largest stamping
c) height = height of window + 2* width of the largest stamping
d) height = height of window – width of the largest stamping
Answer: c
Explanation: First the height of the window is obtained. Then the width of the largest stamping is calculated and substituting in the above formula provides the height of the single phase shell type transformer.
9. What is the formula to calculate the voltage per turn of the transformer?
a) voltage per turn = space factor * square root of output power
b) voltage per turn = space factor / square root of output power
c) voltage per turn = space factor / square root of output power
d) voltage per turn = space factor * 2*square root of output power
Answer: a
Explanation: The corresponding space factor is obtained using the formula. Then the output power is obtained and square root of the output power is taken and substituted in the above formula to obtain the voltage per turn.
10. What is the formula for the net cross sectional area of the core of the transformer?
a) cross sectional area = voltage per turn * 4.44 * frequency * magnetic field
b) cross sectional area = voltage per turn / 4.44 * frequency * magnetic field
c) cross sectional area = voltage per turn * 4.44 / frequency * magnetic field
d) cross sectional area = voltage per turn * 4.44 * frequency / magnetic field
Answer: b
Explanation: For obtaining the cross sectional area, the voltage per turn is obtained. The frequency is always 50 Hz. Then the magnetic field is obtained and substituted in the above formula.
11. What is the formula for the diameter of the circumscribing circle of the transformer?
a) diameter of the circumscribing circle = 2*square root of ratio of cross sectional area of the core to the space factor
b) diameter of the circumscribing circle = 3*square root of ratio of cross sectional area of the core to the space factor
c) diameter of the circumscribing circle = square root of ratio of cross sectional area of the core to the space factor
d) diameter of the circumscribing circle = 4*square root of ratio of cross sectional area of the core to the space factor
Answer: c
Explanation: First the cross sectional area of the core is obtained by the formula cross sectional area = voltage per turn / 4.44 * frequency * magnetic field. Next the space factor is obtained. Substituting in the formula provides the diameter of the circumscribing circle.
12. What is the formula for the width of the window of the transformer?
a) width of the window = distance between core centers + diameter of the circumscribing circle
b) width of the window = distance between core centers – diameter of the circumscribing circle
c) width of the window = distance between core centers * diameter of the circumscribing circle
d) width of the window = distance between core centers / diameter of the circumscribing circle
Answer: b
Explanation: The diameter of the circumscribing circle is obtained from the formula, diameter of the circumscribing circle = square root of ratio of cross sectional area of the core to the space factor. After obtaining the distance between core centers, the width of the window is obtained.
13. What is the formula for window area of the transformer?
a) window area = output power * 2.22 * frequency * magnetic field * window space factor * current density * area of cross section of the core *103
b) window area = output power / 2.22 * frequency * magnetic field * window space factor * current density * area of cross section of the core *103
c) window area = output power / 3.33 * frequency * magnetic field * window space factor * current density * area of cross section of the core *103
d) window area = output power * 3.33 * frequency * magnetic field * window space factor * current density * area of cross section of the core*103
Answer: b
Explanation: The window space factor, the current density and the core cross sectional area are obtained by their respective formula. The frequency is 50Hz and then the magnetic field and the output power is calculated to obtain the window space factor.
14. What is the formula for the height of the window?
a) height of window = area of window * width of the window
b) height of window = area of window + width of the window
c) height of window = area of window – width of the window
d) height of window = area of window / width of the window
Answer: d
Explanation: First the area of the window is obtained. Next the window width is obtained. The ratio of both gives the height of the window.
15. The range of the ratio of the height of the window to the width of the window is 2-4.
a) true
b) false
Answer: a
Explanation: The ratio of the height of the window to the width of the window should be adjusted such that it is above 2. The ratio of the height of the window to the width of the window should be adjusted such that it is below 4.
Overall Dimensions – 2
1. What is the formula for the depth and height of the yoke for stepped core?
a) depth = width of largest stamping, height = 2* width of largest stamping
b) depth = 2*width of largest stamping, height = width of largest stamping
c) depth = width of largest stamping, height = width of largest stamping
d) depth = 2*width of largest stamping, height = 2* width of largest stamping
Answer: c
Explanation: The depth of the yoke of stepped core is equal to the width of the largest stamping. The height of the yoke for the stepped core is also equal to the width of the largest stamping.
2.The height and the width of the single phase and three phase core type transformers are equal.
a) true
b) false
Answer: b
Explanation: The height of both the single phase and three phase core type transformers are equal. The width of the single phase and three phase core type are not same.
3. What is the formula for the height and width of the single phase shell transformer?
a) width = 2*width of the window + 4*width of the largest stamping, height = height of the window + 2*width of the largest stamping
b) width = 2*width of the window – 4*width of the largest stamping, height = height of the window + 2*width of the largest stamping
c) width = 2*width of the window + 4*width of the largest stamping, height = height of the window – 2*width of the largest stamping
d) width = 2*width of the window – 4*width of the largest stamping, height = height of the window -2*width of the largest stamping
Answer: a
Explanation: First the width of the window is obtained. Next, the height of the window is obtained. Then, the width of the largest stamping is obtained and substituted in the above formula.
4. What is the formula to calculate the number of turns/phase?
a) number of turns = secondary voltage * voltage per turn
b) number of turns = secondary voltage / voltage per turn
c) number of turns = secondary voltage + voltage per turn
d) number of turns = secondary voltage – voltage per turn
Answer: b
Explanation: First the voltage across the secondary winding of the transformer is obtained. Next, the voltage across each turn is obtained. On substituting we get the number of turns.
5. What is the formula for the cross sectional area of the secondary conductor of the transformer?
a) cross sectional area = secondary current * current density
b) cross sectional area = secondary current + current density
c) cross sectional area = secondary current / current density
d) cross sectional area = secondary current – current density
Answer: c
Explanation: The current flowing through the secondary winding of the transformer is calculated. Next the current density is calculated and the ratio gives the cross sectional area of the secondary conductor.
6. What is the formula for the conductor dimensions in transformer?
a) conductor dimensions = conductor width * conductor thickness + 0.5 mm
b) conductor dimensions = conductor width / conductor thickness + 0.5 mm
c) conductor dimensions = conductor width + conductor thickness + 0.5 mm
d) conductor dimensions = conductor width – conductor thickness + 0.5 mm
Answer: a
Explanation: The width of the conductor is first calculated. Next, the thickness of the conductor is calculated. On obtaining these data the conductor dimensions can be obtained.
7. What is the formula for axial depth of low voltage winding?
a) axial depth = number of secondary turns / width of the conductor
b) axial depth = number of secondary turns * width of the conductor
c) axial depth = number of secondary turns + width of the conductor
d) axial depth = number of secondary turns – width of the conductor
Answer: b
Explanation: The number of secondary turns is calculated first. Then the width of the conductor is obtained. With the 2 data, the axial depth is obtained.
8. What is the formula for the window clearance of the transformer?
a) window clearance = (height of the window + axial depth)/2
b) window clearance = (height of the window – axial depth)
c) window clearance = (height of the window – axial depth)/2
d) window clearance = (height of the window + axial depth)
Answer: c
Explanation: First the height of the window is obtained. Then the axial depth is calculated using the formula axial depth = number of secondary turns * width of the conductor and substituting in the above formula provides the window clearance.
9. What is the formula to calculate the radial depth of low voltage windings?
a) radial depth of the lv windings = number of layers * radial depth of the conductors * insulation between layers
b) radial depth of the lv windings = number of layers * radial depth of the conductors – insulation between layers
c) radial depth of the lv windings = number of layers / radial depth of the conductors + insulation between layers
d) radial depth of the lv windings = number of layers * radial depth of the conductors + insulation between layers
Answer: d
Explanation: The number of layers is first taken note of. Then the radial depth of the conductors is calculated along with the insulation between layers. On substituting the values in the above formula the radial depth of the low voltage windings is obtained.
10. What is the formula for the inside diameter of the low voltage windings?
a) inside diameter = diameter of the circumscribing circle + pressboard thickness insulation between l.v winding and core
b) inside diameter = diameter of the circumscribing circle – pressboard thickness insulation between l.v winding and core
c) inside diameter = diameter of the circumscribing circle + 2*pressboard thickness insulation between l.v winding and core
d) inside diameter = diameter of the circumscribing circle – 2* pressboard thickness insulation between l.v winding and core
Answer: c
Explanation: For calculating the inner diameter, first the diameter of the circumscribing circle is obtained using the corresponding formula. Then the pressboard thickness insulation is calculated.
11. What is the assumption for width of the largest stamping for the stepped core transformer?
a) 0.9*d
b) 0.71*d
c) 0.85*d
d) 0.8*d
Answer: a
Explanation: If the width of the largest stamping is not provided, then for stepped core a = 0.9*d. For the cruciform it is a = 0.85*d and for the square core it is a = 0.71*d.
12. What is the range for the current density at HT side for a distribution transformer?
a) 2.4-3.5 Amp per mm2
b) 2-2.5 Amp per mm2
c) 1-3.5 Amp per mm2
d) 2-3.5 Amp per mm2
Answer: b
Explanation: 2.4-3.5 Amp per mm2 is the range for the current density at HT side for a power transformer. 2-2.5 Amp per mm2 is the range for the current density at HT side for a distribution transformer.
13. What is the relation of the height of the window with the winding height with respect to the rectangular conductors?
a) winding height = 60% * window height
b) winding height = 50% * window height
c) winding height = 80% * window height
d) winding height = 70% * window height
Answer: d
Explanation: In case of selection of the rectangular conductors, first the window height is obtained. Next the 70% of the window height provides the winding height.
14. What is the formula for number of turns/coil axially?
a) number of turns/coil axially = axial length / diameter of the insulated conductor
b) number of turns/coil axially = axial length * diameter of the insulated conductor
c) number of turns/coil axially = axial length – diameter of the insulated conductor
d) number of turns/coil axially = axial length + diameter of the insulated conductor
Answer: a
Explanation: First the axial length is calculated from its respective formula. Then the diameter of the insulated conductor is calculated, and the ratio gives the number of turns/coil axially.
15. The axial length of 16 coils = axial length of each coil * 16.
a) true
b) false
Answer: a
Explanation: The axial length of each coil is calculated initially from its corresponding formula. Then the value is multiplied by the number of coils present.