Transformer MCQs Part 18

Contents

- 1 An iron-core transformer that is working at maximum flux density of 0.9 Wb/m^2 is replaced by silicon steel working at a maximum flux density of 1.2 Wb/m^2. If total flux is to remain unchanged, the reduction volume expressed as percentage of original volume (keeping the voltage and frequency per turn same) is
- 2 A transformer has zero efficiency at
- 3 The iron losses and full load Cu losses in a 50 kVA transformer are 400 and 800 W respectively. The efficiency of transformer at half full load at 0.85 PF lagging will be
- 4 The voltage per turn of primary of transformer is _________ voltage per turn of its secondary:

## An iron-core transformer that is working at maximum flux density of 0.9 Wb/m^2 is replaced by silicon steel working at a maximum flux density of 1.2 Wb/m^2. If total flux is to remain unchanged, the reduction volume expressed as percentage of original volume (keeping the voltage and frequency per turn same) is

An iron-core transformer that is working at maximum flux density of 0.9 Wb/m^2 is replaced by silicon steel working at a maximum flux density of 1.2 Wb/m^2. If total flux is to remain unchanged, the reduction volume expressed as percentage of original volume (keeping the voltage and frequency per turn same) is:

- 10%
- 25%
- 35%
- 40%

Correct answer: 2. 25%

Solution:

From formula: E = 4.44 * f * N * Φm

Voltage per turn = 4.44 * f * Φm

Voltage per turn = 4.44 * f * Bm * a

voltage per turn and frequency are same, Bm a = constant

a ∝ 1/Bm

a1∝ 1/0.9 = 1.11

a2∝ 1/1.2 = 0.83

Percentage reduction in core volume = Reduction in core area

(1.11 – 0.83)/1.11) * 100 = (0.25)

=25%

## A transformer has zero efficiency at

A transformer has zero efficiency at:

- No-load
- Half full-load
- Full-load
- None of these

Correct answer: No-load

## The iron losses and full load Cu losses in a 50 kVA transformer are 400 and 800 W respectively. The efficiency of transformer at half full load at 0.85 PF lagging will be

The iron losses and full load Cu losses in a 50 kVA transformer are 400 and 800 W respectively. The efficiency of transformer at half full load at 0.85 PF lagging will be:

- 95.66%
- 97.40%
- 98.33%
- 99.36%

Correct answer: 2. 97.40%

Solution:

Total Full load losses = 400 + 800 = 1200 W = 1.2 kW

———————————-

Full load output = 50 * 0.85 = 45 kW

Full load input = 45 + 1.2 = 46.2 kW

———————————

Full load efficiency = 45/46.2 * 100 = 97.40%

## The voltage per turn of primary of transformer is _________ voltage per turn of its secondary:

The voltage per turn of primary of transformer is _________ voltage per turn of its secondary:

- Lesser than
- Greater than
- Equal to
- None of these

Correct answer: 3. Equal to