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Core loss of single-phase, 230/115 V, 50 Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage and variable frequency source while keeping the secondary open-circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is again measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are, respectively
1. 508 W and 542 W
2. 468 W and 582 W
3. 498 W and 552 W
4. 488 W and 562 W

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Correct Answer - Option 1 : 508 W and 542 W


Hysteresis losses: These are due to the reversal of magnetization in the transformer core whenever it is subjected to the alternating nature of magnetizing force.

\({W_h} = \eta B_{max}^{1.6}fv\)

\({B_{max}} \propto \frac{V}{f}\)


x is the Steinmetz constant

Bm = maximum flux density

f = frequency of magnetization or supply frequency

v = volume of the core

At a constant V/f ratio, hysteresis losses are directly proportional to the frequency.

Wh  f

Eddy current losses: Eddy current loss in the transformer is I2R loss present in the core due to the production of eddy current.

\({W_e} = K{f^2}B_m^2{t^2}V\)

\({B_{max}} \propto \frac{V}{f}\)


K - coefficient 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 - Volume of magnetic material in m3

At a constant V/f ratio, eddy current losses are directly proportional to the square of the frequency.

We  f2

Iron losses or core losses or constant losses are the sums of both hysteresis and eddy current losses.

Wi = W + We

At constant V/f ratio, Wi = Af + Bf2

\({230 \over50} = {138 \over30}=4.6 \;(Constant \;{V\over f})\)


The table below shows the given data.



Core losses

Case 1

50 Hz

1050 W

Case 2

30 Hz

500 W

Case 3

50 Hz

Wh , We?


Now, the equations for Case 1 and Case 2 are given below

Case 1: 1050 = A (50) + B (50)2

Case 2: 500 = A (30) + B (30)2

By solving the above two equations,

A = 10.1667, B = 0.2166

Case 3: Hysteresis losses (Wh) = Af =  10.1667 × 50 = 508 W

Eddy current losses (We) = Bf2 = 0.2166 × (50)2 = 542 W

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