This set of Transformers Assessment Questions and Answers focuses on “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

View Answer

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

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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

View Answer

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

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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

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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

View Answer

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

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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

View Answer

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

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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

View Answer

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

View Answer

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 P_{i} is ___________

a) 16.45 kW

b) 9.87 kW

c) 14.57 kW

d) Can’t be calculated

View Answer

Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + P

_{i}+ k

^{2}*PC)]. Thus, η= 500*1/ (500 + P

_{i}+P

_{C}) = 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

View Answer

Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + P

_{i}+ k

^{2}*PC)]. Thus, η= 500*1/ (500 + P

_{i}+P

_{C}) = 0.95. also from the second condition given η= 500*0.6/ (500*0.6 + P

_{i}+0.6

^{2}*PC) = 0.95. Thus, solving simultaneously we get an answer 16.45 kW.

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%

View Answer

Explanation: Power output= VI cosθ= 200*100*0.8 = 16000 W (Independent of lag and lead). While total losses are equal to iron loss+ k

^{2}*copper losses =120+ 300= 420 W. Efficiency is equal to 1- 420/(16000+420)= 97.44%.

**Sanfoundry Global Education & Learning Series – Transformers.**

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