Transformer MCQs Part 14

Contents

- 1 Calculating transformer secondary turns numerical
- 2 Which of following is major reason of occurrence of leakage flux in transformer
- 3 Short circuit test helps us to determine
- 4 In a transformer, the amount of copper in primary as compare to secondary is:
- 5 Certain transformer has an efficiency of 80% and works at 100 V, 4 kW. If secondary voltage is 240 V, considering the power factor to be unit, what is value of primary current

## Calculating transformer secondary turns numerical

The no-load ratio of a 50 Hz single phase transformer is 6000/220 V. The value of maximum flux in core is 0.06 Wb. What is the number of secondary turns?

- 10
- 12
- 17
- 26

Correct answer: 17

From formula: Es = 4.44 * f * Ns * Φm

220 = 4.44 * 50 * Ns *0.06

Ns = 17

## Which of following is major reason of occurrence of leakage flux in transformer

Which of following is major reason of occurrence of leakage flux in transformer:

- Air is not a good magnetic insulator
- The input voltage is sinusoidal in nature
- Iron core has high permeability
- None of these

Correct answer: 1. Air is not a good magnetic insulator

## Short circuit test helps us to determine

Short circuit test helps us to determine:

- Iron losses at no load
- Cu losses at no load
- Both of these
- None of these

Correct answer: 2. Cu losses at no load

## In a transformer, the amount of copper in primary as compare to secondary is:

In a transformer, the amount of copper in primary as compare to secondary is:

- Greater
- Smaller
- Almost same
- None of these

Correct answer: 3. Almost same

Explanation: The number of turns on primary and secondary windings of a transformer depend on their respective voltages, the high voltage winding has far greater number of turns as compared to low voltage winding, whereas the current in high voltage winding is very smaller, thus enabling to use a smaller size conductor. As a result of this, the amount of copper in primary and secondary is almost same.

## Certain transformer has an efficiency of 80% and works at 100 V, 4 kW. If secondary voltage is 240 V, considering the power factor to be unit, what is value of primary current

Certain transformer has an efficiency of 80% and works at 100 V, 4 kW. If secondary voltage is 240 V, considering the power factor to be unit, what is value of primary current.

- 10 A
- 20 A
- 30 A
- 40 A

Correct answer: 4. 40 A

Explanation:

P = Vp * Ip * PF or Ip = P/(Vp * PF) = 4*1000/(100*1) = 40 A