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Transformer Exam Related Question

December 28, 2024

Q1)

A 40 kVA transformer has core loss of 400 W & full load copper loss of 800 W. The proportion of full load at maximum efficiency is -

A) 50%

B) 62.3%

C) 70.7%

D) 100%

Answer: 70.7%

Concept

CONCEPT

The condition for maximum efficiency in a 1ϕ transformer is:

PC = (x)2Pcufl

where, PC = Core loss

Pcufl = Copper loss at full load

x = Fraction of loading

Calculation:

Given, PC = 400 W

Pcufl = 800 W

400 = (x)2 × 800

x = 70.7%

Q2)

The no load current in transformer lags applied voltage by -

A) 90°

B) 75°

C) 0°

D) 110°

Answer: 75°

Concept

  • Ideally, a transformer draws the magnetizing current and lags the primary applied voltage by 90°.
  • But the transformer also has a core loss current component which will be in phase with applied voltage.
  • No-load current is nothing but the vector summation of these two currents.
  • Hence, the no-load current will not lag behind applied voltage by exactly 90° but it lags somewhat less than 90°.
  • It is in practice generally about 75°.
important keyAdditional Information

Circuit diagram for a transformer on no-load:

NO LOAD TRANSFORMER

Where,
V1 is the applied primary voltage
IW is the working component of current through RO (Magnetizing resistance)
Iμ is the magnetizing component of current through XO (magnetizing reactance)
N1 and N2 are primary and secondary turns ratio

  • In the case of no-load, the second terminal of the transformer is open.
  • There is no path available for the current to flow on the secondary side.
  • Hence, the transformer does not draw current from the source.
  • A small ampere of current flows through the primary transformer (no-load current I ) called excitation current (used for excitation of the core).
  • No-load current (IO) is further divided into IWand Iμ

Phasor diagram when transformer on no-load:

Phasor diagram when transformer on no-load saarkari exam

Q4)

If the transformer frequency is changed from 50 Hz to 60 Hz, the ratio of eddy current loss at 60 Hz to 50 Hz at constant voltage will be:

A) 5/6

B) 1/1

C) 25/36

D) 36/25

Answer: 1/1

Eddy Current Loss:

  • When an alternating magnetic field is applied to a magnetic material, an emf is induced in the material itself according to Faraday’s Law of Electromagnetic induction.
  • Since the magnetic material is a conducting material, this EMF circulates current within the body of the material.
  • These circulating currents are called Eddy Currents. They will occur when the conductor experiences a changing magnetic field.
  • As these currents are not responsible for doing any useful work, and it produces a loss (I2R loss) in the magnetic material known as an Eddy Current Loss.

Mathematical Expression for Eddy Current Loss:

Pe=KeB2mt2f2V

where,
Ke – co-efficient of eddy current. Its value depends upon the nature of magnetic material
Bm – maximum value of flux density in Wb/m2
t – thickness of lamination in meters
f – frequency of reversal of the magnetic field in Hz
V – The volume of magnetic material in m3

From eddy current loss equation we can write

cal formula

The eddy current loss at constant voltage is independent of transformer frequency

cal formula

The ratio of eddy current loss at 60 Hz to 50 Hz at constant voltage is 1:1

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