Efficiency and Voltage Regulation ( Transformers ) MCQs – Transformers MCQs
Latest Transformers MCQs
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Transformers MCQs – Efficiency and Voltage Regulation ( Transformers ) MCQs
The most occurred mcqs of Efficiency and Voltage Regulation ( ) in past papers. Past papers of Efficiency and Voltage Regulation ( Transformers ) Mcqs. Past papers of Efficiency and Voltage Regulation ( Transformers ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Efficiency and Voltage Regulation ( Transformers ) Mcqs. The Important series of Efficiency and Voltage Regulation ( Transformers ) Mcqs are given below:
Efficiency
1. When will be the efficiency of a transformer maximum?
a) Copper losses = hysteresis losses
b) Hysteresis losses = eddy current losses
c) Eddy current losses = copper losses
d) Copper losses = iron losses
Answer: d
Explanation: When the variable copper losses of a transformer becomes equal to the fixed iron losses of a transformer then we will get maximum efficiency. From these losses we’ll get the value of current required.
2. Efficiency of a power transformer is near to the ___________
a) 100 per cent
b) 98 per cent
c) 50 per cent
d) 25 per cent
Answer: b
Explanation: The efficiency of the transformer obtained from various experiments conducted on various loads showed the efficiency greater than 90% always. Transformer thus, can be said highly efficient device.
3. On which factors transformer routine efficiency depends upon?
a) Supply frequency
b) Load current
c) Power factor of load
d) Load current and power factor of load
Answer: d
Explanation: Efficiency of the transformer can be calculated by the output power divided by input power. Both of these powers include power factor in their calculations while load current and load voltage is also required in calculations.
4. Normal transformers are designed to have maximum efficiency at ___________
a) Nearly full load
b) 70% full load
c) 50% full load
d) No load
Answer: a
Explanation: Every device is manufactured to get maximum efficiency at the rated loads, i.e. full load. Thus, transformer will give the maximum efficiency at nearly full load. Internal losses are so adjusted to get maximum efficiency.
5. At which load condition maximum efficiency of a distribution transformer will be achieved?
a) At no load
b) At 60% full load
c) At 80% full load
d) At full load
Answer: b
Explanation: The main difference between power transformer and distribution transformer is distribution transformer is designed for maximum efficiency at 60% to 70% load as these transformers normally doesn’t operate at full load all the time.
6. Power transformers other than distribution transformers are generally designed to have maximum efficiency around ______
a) No-load
b) Half-load
c) Near full-load
d) 10% overload
Answer: c
Explanation: Similar to normal transformers power transformers are also designed to get maximum efficiency at load which is near to the full load of a transformer specified. Only in the case distribution transformer maximum efficiency is achieved at 60% of full load.
7. For a transformer given, operating at constant load current, maximum efficiency will occur at ______
a) 0.8 leading power factor
b) 0.8 lagging power factor
c) Zero power factor
d) Unity power factor
Answer: d
Explanation: Maximum efficiency for a transformer will be achieved at full load. While in the case of power factor also every device is set to get maximum efficiency at unity power factor. Thus, one will have maximum efficiency if load is nearly equal to full load and at unity power factor.
8. Why efficiency of a transformer, under heavy loads, is comparatively low?
a) Copper loss becomes high in proportion to the output
b) Iron loss is increased considerably
c) Voltage drop both in primary and secondary becomes large
d) Secondary output is much less as compared to primary input
Answer: a
Explanation: At heavy loads current drawn by the transformer circuit increases, as we know, variable copper losses are proportional to the square of the current. Thus, we’ll get higher copper loss in proportion to the output.
9. The efficiencies of transformers compared to electric motors of the same power are ___________
a) About the same
b) Much smaller
c) Much higher
d) Can’t comment
Answer: c
Explanation: Transformer is a highly efficient device compare to all other electrical instruments. In motor we need to add windage and friction losses along with the copper losses and iron losses thus, we’ll get lee efficiency for motor compare to transformer.
10. A transformer having maximum efficiency at 75% full load will have ratio of iron loss and full load copper loss equal to ___________
a) 4/3
b) 3/4
c) 9/16
d) 16/9
Answer: c
Explanation: Condition for maximum efficiency is, Copper loss= Iron loss, i.e. Pc= I2 R = Pi. transformer can be operated at any load but maximum efficiency occurs at a particular load condition only. Let x be that load factor corresponds to maximum efficiency. Given that, maximum efficiency will occur at 3/4 load. The load factor= (3/4)2.
11. What is the correct formula of efficiency of a device?
a) Input /output
b) Output/losses
c) 1- (losses/ (output + losses))
d) Cannot be determined
Answer: c
Explanation: Efficiency of any device is equal to the ratio of output power to the input power. Here, one can write input power is equal to the addition of output power with losses. Thus, expressing all these terms mathematically will give the answer.
12. A 500 kVA transformer is having efficiency of 95% at full load and also at 60% of full load; both at unity power factor. Then Pi is ___________
a) 16.45 kW
b) 9.87 kW
c) 14.57 kW
d) Can’t be calculated
Answer: b
Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + Pi + k2*PC)]. Thus, η= 500*1/ (500 + Pi +PC) = 0.95. also from the second condition given η= 500*0.6/ (500*0.6 + Pi +0.6^2*PC) = 0.95. Thus, solving simultaneously we get 9.87 kW.
13. A 500 kVA transformer is having efficiency of 95% at full load and also at 60% of full load; both at unity power factor. Then Pc is ___________
a) 16.45 kW
b) 9.87 kW
c) 14.57 kW
d) Can’t be calculated
Answer: a
Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + Pi + k2*PC)]. Thus, η= 500*1/ (500 + Pi +PC) = 0.95. also from the second condition given η= 500*0.6/ (500*0.6 + Pi +0.62*PC) = 0.95. Thus, solving simultaneously we get an answer 16.45 kW.
Transformer Losses And Testing MCQs
14. For a power transformer operating at full load it draws voltage and current equal to 200 V and 100 A respectively at 0.8 pf. Iron and copper losses are equal to 120 kW and 300kW. What is the efficiency?
a) 86.44%
b) 96.44%
c) 97.44%
d) 99.12%
Answer: c
Explanation: Power output= VI cosθ= 200*100*0.8 = 16000 W (Independent of lag and lead). While total losses are equal to iron loss+ k2*copper losses =120+ 300= 420 W. Efficiency is equal to 1- 420/(16000+420)= 97.44%.
Voltage Regulation – 1
1. The highest voltage for transmitting electrical power in India is _______
a) 33 kV
b) 66 kV
c) 132 kV
d) 00 kV
Answer: d
Explanation: Transmission voltage in power transfer in India (highest) is 750KV AC and these lines are erected by Power Grid Corporation for interstate connections throughout India. However, work on 800KV is in the progress. DC transmission voltage (highest) in India is 600KV.
2. A transformer can have zero voltage regulation at _______
a) Leading power factor
b) Lagging power factor
c) Unity power factor
d) Zero power factor
Answer: a
Explanation: At leading power factor the voltage regulation is given by I*(Rcosφ- Xsinφ). Thus, at a particular condition of angle φ we may get zero voltage regulation. While in lagging power factor case we have + sign in the above formula.
3. What will happen to a given transformer if it made to run at its rated voltage but reduced frequency?
a) Flux density remains unaffected
b) Iron losses are reduced
c) Core flux density is reduced
d) Core flux density is increased
Answer: d
Explanation: E=4.44fNAB is the emf equation for a transformer, now as E is kept constant we can say frequency is inversely proportional to the B value. Thus, as frequency increases we will get less core flux density and vice-versa.
4. In an actual transformer the iron loss remains practically constant from no load to full load because ___________
a) Value of transformation ratio remains constant
b) Permeability of transformer core remains constant
c) Core flux remains practically constant
d) Primary voltage remains constant
Answer: c
Explanation: The reason behind core-iron loss being constant is that hysteresis loss and eddy current loss both are dependent on the magnetic properties of the material which is used in the construction and design of the core of the transformer.
5. Negative voltage regulation indicates ___________
a) Capacitive loading only
b) Inductive loading only
c) Inductive or resistive loading
d) Cannot be determined
Answer: a
Explanation: The sign -ve arises in the voltage regulation calculations when, the load connected to the transformer is leading in the nature. The only condition when we’ll get negative voltage regulation when second term is higher than first term.
6. When will a transformer have regulation closer to zero?
a) On full-load
b) On overload
c) On leading power factor
d) On zero power factor
Answer: c
Explanation: Since voltage regulation of a transformer in the leading loading condition is not additive in nature, at particular power factor in leading we can get zero voltage regulation. While, in lagging condition we’ll get ultimately non-zero VR.
7. A good voltage regulation of a transformer indicates ______________
a) output voltage fluctuation from no load to full load is least
b) output voltage fluctuation with power factor is least
c) difference between primary and secondary voltage is least
d) difference between primary and secondary voltage is maximum
Answer: a
Explanation: Voltage regulation is defined as rise in the voltage when the transformer is thrown off from full load condition to no-load condition. Thus, least voltage regulation means output fluctuations depending on the load are very less.
8. Which of the following acts as a protection against high voltage surges due to lightning and switching?
a) Horn gaps
b) Thermal overload relays
c) Breather
d) Conservator
Answer: a
Explanation: Arcing horns in a transformer form a spark gap across the insulator with a lower breakdown voltage than the air path along the insulator surface, so an overvoltage it will cause the air to break down and the arc to form between the arcing horns, diverting it away from the surface of the insulator.
9. Minimum voltage regulation occurs when the power factor of the load is ______________
a) Unity
b) Lagging
c) Leading
d) Zero
Answer: c
Explanation: When the leading load is connected to the transformer difference of Rcosφ and Xsinφ is multiplied with the current, thus we may get -ve, zero voltage regulations at this condition. That is minimum voltage regulation.
Voltage Regulation – 2
1. Voltage regulation of transformer is given by _____________
a) E2-V2/V2
b) E2-V2/E2
c) V2-E2/E2
d) V2-E2/V2
Answer: b
Explanation: Voltage regulation is defined as change in the voltage or rise in voltage when transformer is load is thrown off. Thus, it is the difference of the no load voltage with the full load voltage divide by full load voltage to get % increase.
2. On which load power factor zero voltage regulation will be achieved?
a) 0
b) 1
c) Leading
d) Lagging
Answer: c
Explanation: At leading power factor the voltage regulation can be negative or zero. This can be found from this equation % regulation = εxcosθ – εrsinθ. Bu substituting the appropriate value of angle one can check this mathematically.
3. A transformer has resistance and reactance in per unit as 0.01 and 0.04 pu respectively. What will be its voltage regulation for 0.8 power factor lagging and leading?
a) 3.2% and 1.6%
b) 3.2% and -1.6%
c) 1.6% and -3.2%
d) Can’t be defined
Answer: b
Explanation: Voltage regulation for lagging power factor = (R cosθ + X sinθ) × 100, Voltage regulation for 0.8 lagging power factor = (0.01×0.8+0.04×0.6) × 100 = 3.2%. Voltage regulation for leading power factor = (R cosθ – X sinθ) × 100, Voltage regulation for 0.8 leading power factor = (0.01×0.8-0.0 4×0.6) × 100 = -1.6%.
4. At which power factor one will get maximum voltage regulation?
a) 0
b) 1
c) Leading
d) Lagging
Answer: d
Explanation: At lagging power factor the voltage regulation is given by I*(Rcosφ+ Xsinφ). Thus, at a particular condition of angle φ we will get maximum voltage regulation. While in leading power factor case we have – sign in the above formula.
5. Which is the correct phasor equation indicating the transformer voltages lagging?
a) V1 = V2 + I*(R cosφ+ X sinφ)
b) V2 = V1 + I*(R cosφ+ X sinφ)
c) V1 = V2 + I*(X cosφ+ X sinφ)
d) V1 = V2 + I*(R cosφ+ R sinφ)
Answer: a
Explanation: According to the phasor diagram drawn for lagging current, we will have positive sign in the voltage regulation formula thus, V1 = V2 + I*(R cosφ+ X sinφ) gives the correct relation, while V1 indicates the primary voltage.
6. Which is the correct phasor equation indicating the transformer voltages leading?
a) V1 = V2 + I*(R cosφ- X sinφ)
b) V2 = V1 + I*(R cosφ- X sinφ)
c) V1 = V2 + I*(X cosφ- X sinφ)
d) V1 = V2 + I*(R cosφ- R sinφ)
Answer: a
Explanation: According to the phasor diagram drawn for leading current, we will have negaitive sign in the voltage regulation formula thus, V1 = V2 + I*(R cosφ- X sinφ) gives the correct relation, while V1 indicates the primary voltage.
7. What is the correct formula to get power factor angle in leading condition?
a) tan φ= X/R
b) tan φ= R/X
c) cos φ = R/√(R2+X2)
d) cos φ= R/X
Answer: b
Explanation: For leading condition derivative of voltage regulation with respect to φ is obtained and solved for the power factor angle calculations we’ll get tan φ = R/X for leading condition, for lagging condition we’ll get tan φ= X/R.
8. What is the correct formula to get power factor angle in lagging condition?
a) sin φ= X/R
b) tan φ= R/X
c) cos φ = R/√(R2+X2)
d) cos φ= R/X
Answer: c
Explanation: For lagging condition derivative of voltage regulation with respect to φ is obtained and solved for the power factor angle calculations we’ll get tan φ = R/X for leading condition, for lagging condition we’ll get tan φ= X/R. In terms of cosine function, we’ll get cos φ = R/√(R2+X2).
9. Zero voltage regulation of a transformer is achieved at 1 pf leading.
a) True
b) False
Answer: b
Explanation: Though zero voltage regulation occurs at leading power factor condition, it is not occurring at unity power factor leading. As at unity power factor leading, cos term will be equal to 1. Hence, we’ll get some non-zero VR at unity power factor.