Transformer Core Losses MCQ Quiz - Objective Question with Answer for Transformer Core Losses - Download Free PDF

Last updated on May 13, 2025

Latest Transformer Core Losses MCQ Objective Questions

Transformer Core Losses Question 1:

The eddy current loss in a transformer is caused by:

  1. resistance of the primary winding
  2. voltage drop across the windings
  3. saturation of the core material
  4. circulating currents within the core due to changing magnetic flux

Answer (Detailed Solution Below)

Option 4 : circulating currents within the core due to changing magnetic flux

Transformer Core Losses Question 1 Detailed Solution

Eddy's 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, these EMFs circulate current within the body of the material. These circulating currents are called Eddy currents. They are produced when the conductor experiences a changing magnetic field.
     

It is given by: \(P_e=K_eB_m^2t^2f^2V\)

 

where Ke = Eddy current loss coefficient

Hysteresis loss:

  • Hysteresis loss in a transformer refers to the energy loss that occurs in the transformer's core material due to the repeated magnetization and demagnetization (cycling) when alternating current (AC) flows through it.
  • Hysteresis loss in a transformer can be minimized by using soft magnetic materials for the core like permalloy or silicon iron.
     

It is given by: \(P_h=K_hfB_m^{1.6}\)

 

 

Transformer Core Losses Question 2:

Power input to a 20 kVA, 200 V/400 V, single-phase two winding transformer during an open circuit test is measured to be 200 W. What is its core loss approximately when the transformer is fully loaded?

  1. 100 W
  2. 400 W
  3. 0
  4. 200 W

Answer (Detailed Solution Below)

Option 4 : 200 W

Transformer Core Losses Question 2 Detailed Solution

Core loss in the Transformer

The open-circuit test of a transformer is used to determine the core loss (iron loss), which consists of hysteresis and eddy current losses.

These losses are constant regardless of the load on the transformer because they depend only on the applied voltage, which remains the same in normal operation.

Explanation

  • Transformer rating: 20 kVA
  • Primary voltage (low-voltage side): 200 V
  • Secondary voltage (high-voltage side): 400 V
  • Power input during open-circuit test: 200 W


Since core loss is independent of load, it remains constant at all loading conditions, whether it is performed at no load or full load conditions.

Thus, the core loss when the transformer is fully loaded is still 200 W.

Transformer Core Losses Question 3:

When B is the flux density, then eddy current loss varies as-

  1. B
  2. B1.6
  3. B2
  4. B3.2

Answer (Detailed Solution Below)

Option 3 : B2

Transformer Core Losses Question 3 Detailed Solution

Losses in the transformer

Core Losses Or Iron Losses

  • Eddy current loss and hysteresis loss depend on the magnetic properties of the material used for the construction of the core. So, these losses are also known as core losses or iron losses.
  • Eddy's current loss: \(\rm W_e=kB_{max}^2f^2t^2V\)
  • Hysteresis loss: \(\rm W_e=kB_{max}^{1.6}fV\)

 

Copper Loss

  • Copper losses are due to the resistance of the wire in the primary and secondary windings and the current flowing through them.
  • These losses can be reduced by using wire with a large cross-sectional area in the manufacturing of the coils.
     

Stray Loss

  • The reason for the types of loss is the occurrence of the leakage field. When compared with copper and iron losses, the percentage of stray losses is less, so these losses can be neglected.
     

Dielectric Loss

  • The oil of the transformer is the reason for this loss. Oil in a transformer is an insulating material. When the oil in the transformer deteriorates then the transformer’s efficiency will be affected.

Transformer Core Losses Question 4:

In which of the following devices are eddy currents utilised beneficially? 

  1. Electric heaters
  2. Electric motors 
  3. Transformer cores
  4. Induction furnaces

Answer (Detailed Solution Below)

Option 4 : Induction furnaces

Transformer Core Losses Question 4 Detailed Solution

Explanation:

Eddy currents, which are loops of electrical current induced within conductors by a changing magnetic field in the conductor, have various applications. One of the beneficial uses of eddy currents is in induction furnaces.

Induction Furnaces

Definition: Induction furnaces are electrical furnaces where the heat is generated by inducing eddy currents in the material to be melted. These furnaces use the principle of electromagnetic induction to heat and melt metals.

Working Principle: In an induction furnace, a high-frequency alternating current (AC) is passed through a coil, creating a rapidly changing magnetic field around the coil. When a conductive material, such as metal, is placed within this magnetic field, eddy currents are induced in the material. These eddy currents flow through the resistance of the material, generating heat due to the Joule heating effect. This heat is sufficient to melt the metal, allowing it to be used in various industrial processes.

Advantages:

  • Precise Temperature Control: Induction furnaces allow for precise control of the temperature, making it possible to achieve the exact melting point required for different metals and alloys.
  • High Efficiency: The generation of heat directly within the material to be melted ensures high energy efficiency, reducing energy consumption and operational costs.
  • Cleaner Operation: Induction furnaces produce less environmental pollution compared to traditional fossil fuel-based furnaces, as they do not involve combustion processes.
  • Uniform Heating: The induced eddy currents ensure uniform heating throughout the material, leading to consistent melting and better quality of the final product.
  • Quick Start-Up: Induction furnaces have a rapid start-up time, allowing for quicker initiation of the melting process compared to conventional furnaces.

Applications: Induction furnaces are widely used in the metallurgical industry for melting and refining various metals, including steel, copper, aluminum, and precious metals. They are also used in foundries for casting operations and in the manufacturing of high-quality metal products.

The correct option is: Option 4: Induction furnaces

Transformer Core Losses Question 5:

The stator of a 3-phase induction motor is laminated in order to

  1. reduce eddy current losses
  2. reduce hysteresis loss
  3. reduce copper losses in the stator winding
  4. reduce the weight of the stator

Answer (Detailed Solution Below)

Option 1 : reduce eddy current losses

Transformer Core Losses Question 5 Detailed Solution

  • In a 3-phase induction motor, the stator is laminated to reduce eddy current losses.
  • Laminating the stator core minimizes the paths for eddy currents, thus reducing the circulating currents induced by the alternating magnetic field in the stator. This helps in lowering the energy lost as heat due to eddy currents.
  • Although lamination also slightly reduces hysteresis losses, the primary purpose is to prevent eddy current losses.
     

Concept:

  • The stator is a stationary part of the induction motor. A stator winding is placed in the stator of the induction motor and the three-phase supply is given to it.
  • It carries a 3-phase winding and is fed from a 3-phase supply
  • The stator of the induction motor consists of three parts
  1. Stator Frame
  2. Stator core
  3. Stator winding
     

Stator Frame:

  • It is the outer part of the three-phase induction motor.
  • Its main function is to support the stator core and the field winding.
  • It acts as a covering, and it provides protection and mechanical strength to all the inner parts of the induction motor.
  • The frame is either made up of die-cast or fabricated steel.
  • The frame of three phase induction motor should be strong and rigid as the air gap length of three phase induction motor is very small.
  • Otherwise, the rotor will not remain concentric with the stator, which will give rise to an unbalanced magnetic pull.
     

Stator core:

  • The main function of the stator core is to carry the alternating flux. In order to reduce the eddy current loss, the stator core is laminated.
  • These laminated types of structures are made up of stamping which is about 0.4 to 0.5 mm thick.
  • All the stamping is stamped together to form a stator core, which is then housed in a stator frame.
  • The stamping is made up of silicon steel, which helps to reduce the hysteresis loss occurring in the motor.
     

Stator winding

  • The slots on the periphery of the stator core of the three-phase induction motor carry three-phase windings
  •  The three phases of the winding are connected either in star or delta depending upon which type of starting method we use. We start the squirrel cage motor mostly with star-delta starter and hence the stator of the squirrel cage motor is delta connected.
  • It is supplied with 3 phase AC supply

F2 Vinanti Engineering 17.05.23 D5

Additional Information

  • A commutator is a rotary electrical switch in certain types of electric motors and electrical generators that periodically reverses the current direction between the rotor and the external circuit. It is used to convert AC to DC
  • The air gap of a motor is the gap between the stator teeth or core and the rotor magnets. This gap is a key component in the motor design and affects the overall strength of the magnetic circuit and motor efficiency.

Top Transformer Core Losses MCQ Objective Questions

Eddy currents are induced by changing magnetic fields. Where do they flow with respect to the magnetic field?

  1. Closed loops perpendicular to magnetic field
  2. Closed loops parallel to magnetic field
  3. Open loops parallel to magnetic field
  4. Open loops perpendicular to magnetic field

Answer (Detailed Solution Below)

Option 1 : Closed loops perpendicular to magnetic field

Transformer Core Losses Question 6 Detailed Solution

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Eddy Current:

  • Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction.
  • Eddy currents flow in closed loops within conductors, in-plane perpendicular to the magnetic field.
  • By Lenz law, the current swirls in such a way as to create a magnetic field opposing the change; for this to occur in a conductor, electrons swirl in a plane perpendicular to the magnetic field.
  • Because of the tendency of eddy currents to oppose, eddy currents cause a loss of energy.
  • Eddy currents transform more useful forms of energy, such as kinetic energy, into heat, which isn’t generally useful.
  • Thus eddy currents are a cause of energy loss in alternating current (AC) inductors, transformers, electric motors and generators, and other AC machinery, requiring special construction such as laminated magnetic cores or ferrite cores to minimize them.
  • Eddy currents are also used to heat objects in induction heating furnaces and equipment, and to detect cracks and flaws in metal parts using eddy-current testing instruments.

F5 Madhuri Engineering 16.08.2022 D1 V2

  • The magnitude of the current in a given loop is proportional to the strength of the magnetic field, the area of the loop, and the rate of change of flux, and inversely proportional to the resistivity of the material.

Hysteresis and eddy current losses of single-phase transformer working on 200 V, 50 Hz is Ph and Pc respectively. The percentage decrease in Ph and Pc, when the transformer operates on 160 V, 40 Hz supply, will respectively be

  1. 32, 36
  2. 25, 50
  3. 20, 36
  4. 40, 80

Answer (Detailed Solution Below)

Option 3 : 20, 36

Transformer Core Losses Question 7 Detailed Solution

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Concept:

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

\({W_h} = \eta B_{max}^xfv\)

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

Where

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.

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

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

Where,

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 sum of both hysteresis and eddy current losses.

Wi = W + We

At constant V/f ratio, Wi = Af + Bf2

Calculation:

The table below shows the given data.

 

Voltage (V)

Frequency (f)

V/f ratio

No load losses (W)

Case 1

200 V

50 Hz

4

Ph and Pc

Case 2

160 V

40 Hz

4

 

 

The V/f ratio is constant in all the cases as shown in the above table.

Ph ∝ f

\( \Rightarrow {P_{h2}} = \frac{{{f_2}}}{{{f_1}}} \times {p_h} = \frac{{40}}{{50}}{p_h} = 0.8{P_h}\)

Percentage decrease in Ph = 20%

Pe ∝ f2

\( \Rightarrow {P_{e2}} = {\left( {\frac{{{f_2}}}{{{f_1}}}} \right)^2} \times {p_h} = {\left( {\frac{{40}}{{50}}} \right)^2}{p_e} = 0.64{P_e}\)

Percentage decrease in Pe = 36%

A single phase transformer when supplied with 220 V, 50 Hz has eddy current loss of 50 W. If the transformer is connected to a voltage of 330 V, 50 Hz the eddy current loss will be 

  1. 168.75 W
  2. 112.5 W
  3. 75 W
  4. 50 W

Answer (Detailed Solution Below)

Option 2 : 112.5 W

Transformer Core Losses Question 8 Detailed Solution

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Concept:

In a transformer iron loss or core loss can be calculated as

Iron losses = Eddy current loss + hysteresis loss

Eddy current loss, \({P_e} = K{f^2}B_m^2{t^2}V\) 

Hysteresis loss = KBη f

Where,

K = 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 magnetic field in Hz

V = Volume of magnetic material in m3

Kh = Hysteresis constant

Hence

\({P_e} \propto B_m^2{f^2}{t^2}\)

\(P_e \propto {\left( {\frac{v}{f}} \right)^2} \times {f^2}{t^2}\left(\because {{B_m}\alpha \frac{V}{f}} \right)\)

When flux density (Bm) is constant, then eddy current losses are

P∝ V2 t2  

and hysteresis losses are

\({P_h} \propto B_m^{1.6}f \propto \frac{{{V^{1.6}}}}{{{f^{0.6}}}}\)

Calculation:

Given-

f = 50 Hz

Pe1 = 50 W

V1 = 220 V

V= 330 V

In the given question

V / f is not same in both cases

Hence flux density is not constant

∴ \({\frac{P_e1}{P_e2}} \propto {\left( {\frac{(V_1)f}{fV_2}} \right)^2} \times {f^2}{t^2}\left(\because {{B_m}\alpha \frac{V}{f}} \right)\)

∴ Pe2 = \( {\left( {\frac{330}{220}} \right)^2} \times 50\)

Pe2 = 112.5 W

A transformer is connected to 2200 V, 40 Hz supply. The core loss is 800 Watts out of which 600 Watts are due to hysteresis and remaining eddy current losses. What is the core loss if the supply voltage and frequency are 3300 V and 60 Hz respectively?

  1. 135 Watts
  2. 450 Watts
  3. 900 Watts
  4. 1350 Watts

Answer (Detailed Solution Below)

Option 4 : 1350 Watts

Transformer Core Losses Question 9 Detailed Solution

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Concept:

Iron losses (Pi) = Hysteresis loss(Ph) + Eddy current loss(Pe)

Hysteresis loss ∝ Bn f ∝  f

Bn is flux density

with flux density being kept constant → Hysteresis loss ∝ k1 f

Eddy current loss ∝ B2 f2 ∝ f2

 with flux density being kept constant → Eddy current loss ∝ K2 f2 

Where, K1 and K2 are constant for a particular transformer.

Hence total iron losses

P= k1f + k2f2

Calculation:

Given:

For 40 Hz frequency 

Core (iron) loss (Pi) = 800 watts

Ph = 600 = k1

k1 = 600/40 = 15

Pe = (800 - 600) = 200 = k2f2

k2 = (200/ 402) =0.125

For 60 Hz frequency

Pi = (15 × 60) + (0.125)(602)= 1350 watts

Addition of a small per cent of silicon 3% to iron will increase the ________ significantly; by that ________ will be reduced.

  1. eddy current loss, resistivity
  2. resistivity, eddy current loss
  3. conductivity, eddy current loss
  4. conductivity, hysteresis loss

Answer (Detailed Solution Below)

Option 2 : resistivity, eddy current loss

Transformer Core Losses Question 10 Detailed Solution

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Concept:

Losses in a transformer

Iron loss: Losses that are generated in the conducting core laminations by the alternating magnetic field due to hysteresis and eddy currents.         

These losses are constant at all load variations.

Hysteresis loss: When the core of the transformer is applied to an alternating magnetizing force and for each cycle of emf a hysteresis loop is traced out. Power is dissipated in the form of heat known as hysteresis loss.

Ph = Kη Bmnf V

Where, Ph = Hysteresis loss (W)

η  = Steinmetz hysteresis coefficient, depending on material (J/m3)

Bm = Maximum flux density (Wb/m2)

n = Steinmetz exponent, ranges from 1.5 to 2.5, depending on material

f = Frequency (Hz)

V = Volume of magnetic material (m3)

Eddy current losses: When an alternating magnetic flux links with a conducting surface, it induces electric currents in this surface as currents can’t go anywhere, but as the potential across the surface is continuously changing, the currents tend to circulate through the surface these are known as eddy currents and heat generated by it is known as eddy current losses.

Pe = KeBm2 f2 t2 V

Where, Pe = Eddy current loss (W)

Ke = Coefficient of eddy current.

Bm = Maximum value of flux density (wb/m2)

t = Thickness of lamination  (m)

f = Frequency (Hz)

V = Volume of magnetic material (m3)

Copper losses: Losses that are caused by currents flowing in the primary and secondary windings.

 These losses changes with change in loads.

Hysteresis losses can be minimized by

  • Choosing a material with less hysteresis loop area.
  • Core stamping

 

Eddy current losses can be minimized by

  • Cores are cut into thin laminated sheets.
  • The resistance of the core should be increased, Therefore by adding 3% silicon to the iron core its resistance is increased.
  • The laminations are thin with relatively high resistance.

As the load on the transformer is increased, the core losses:

  1. may decrease or increase slightly depending upon the nature of the load
  2. remain constant
  3. increase slightly
  4. decrease slightly

Answer (Detailed Solution Below)

Option 2 : remain constant

Transformer Core Losses Question 11 Detailed Solution

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Losses in the transformer:

1.) Core loss:

  • It is a loss that happens in the core of a transformer when it is subjected to the change in magnetic flux \((\phi = {V \over f})\)
  • Hence, core loss depends upon supply voltage and frequency.
  • Core losses remain constant with an increase or decrease in the load.
  • Core loss is the sum of eddy current loss and the hysteresis loss.
  • The induced emf produces a current known as eddy current. The losses due to this current are known as eddy current losses.
  • The losses in a transformer that are due to magnetization saturation in the core are known as hysteresis losses.

2.) Copper loss:

  •  The losses produced due to heat produced by electrical currents in the transformer windings are known as the copper losses.
  • These losses increase with an increase in loading conditions.

3.) Stray magnetic losses:

  • The losses occurring in the transformer due to the presence of external magnetic field nearby transformer are known as stray magnetic losses.

In a transformer copper losses at full load and unity pf. are 800 W. The copper losses at full load and at 0.8 p.f. lagging will be

  1. 400 W
  2. 640 W
  3. 800 W
  4. 200 W

Answer (Detailed Solution Below)

Option 3 : 800 W

Transformer Core Losses Question 12 Detailed Solution

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Concept:

A transformer has mainly two types of losses

  • core losses
  • copper losses.

 

Core loss, which is also referred to as iron loss, consists of hysteresis loss and eddy current loss.

These two losses are constant when the transformer is charged. That means the amount of these losses does not depend upon the condition of a secondary load of the transformer. In all loading conditions, these are fixed.

Copper losses are directly proportional to the square of the load on the transformer.

\({W_{cu}} = {x^2}{W_{cufl}}\)

Here, x is the percentage of the full load of the transformer.

Wcufl­ is the copper losses at the full load.

As there is no change in the load, load current current will ramain same hence copper loss will be same as previous

The iron loss of a transformer at 400 Hz is 10 W. Assuming that eddy current and hysteresis losses vary as the square of flux density, the iron loss of the transformer at rated voltage but at 50 Hz would be ___ W.

  1. 80
  2. 640
  3. 1.25
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 640

Transformer Core Losses Question 13 Detailed Solution

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We have,

Eddy current and hysteresis losses vary as the square of flux density.

⇒ Iron Loss (Eddy Current loss + Hysteresis loss) ∝ (flux density)2

 Pi ∝ B2 .... (1)

Using the concept of the EMF equation,

V ∝ f B

B ∝ V/f .... (2)

Here B is the flux density

Using equation (1) & (2),

Pi ∝ \(\frac{V^2}{f^2}\)

Since the rated voltage is constant,

∴ Pi ∝ \(\frac{1}{f^2}\)

Case 1:

Pi = 10 W and,

f = 400 Hz

⇒ 10 ∝ \(\frac{1}{400^2}\) .... (3)

Case 2:

Pi = ?

f = 50 Hz

⇒ Pi ∝ \(\frac{1}{50^2}\) .... (3)

From equation (2) & (3),

\(\frac{P_i}{10}=\frac{400^2}{50^2}=64\)

Hence, Pi = 64 × 10 = 640 W

Match the items in List - I and List - II and choose correct answer from the given code :

List - I List - II
A. Copper loss 1. η(Bmax)1.6 fv
B. Eddy current loss 2. αN2
C. Hysteresis loss 3. I2R
D. Windage loss 4. k(Bmax)2 f2vt2

  1. A - 3, B - 4, C - 2, D - 1
  2. A - 4, B - 3, C - 1, D - 2
  3. A - 3, B - 1, C - 4, D - 2
  4. A - 3, B - 4, C - 1, D - 2

Answer (Detailed Solution Below)

Option 4 : A - 3, B - 4, C - 1, D - 2

Transformer Core Losses Question 14 Detailed Solution

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Losses in the transformer:

1.) Core loss: 

  • It is a loss that happens in the core of a transformer when it is subjected to the change in magnetic flux \((\phi = {V \over f})\)
  • Hence, core loss depends upon supply voltage and frequency.
  • Core losses remain constant with an increase or decrease in the load.
  • Core loss is the sum of eddy current loss and the hysteresis loss.
  • The induced emf produces a current known as eddy current. The losses due to this current are known as eddy current losses.
  • The eddy current loss is given by: Pe = k(Bmax)2 f2vt2
  • The losses in a transformer due to magnetization saturation in the core are known as hysteresis losses.
  • The hysteresis loss is given by: Ph = η(Bmax)1.6 fv

2.) Copper loss:

  • The losses due to the heat produced (I2R) by electrical currents in the transformer windings are known as the copper losses.
  • These losses increase with an increase in loading conditions.

3.) Windage loss:

  • Windage losses are caused by friction in the bearings or in the rotating parts of the motor or generator.
  • These losses are proportional to the winding factor (αN2).

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:

  1. 5 / 6
  2. 1 / 1
  3. 25 / 36
  4. 36 / 25

Answer (Detailed Solution Below)

Option 2 : 1 / 1

Transformer Core Losses Question 15 Detailed Solution

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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:

The eddy current power loss in a magnetic material is given by the equation shown below:
 \({P_e} = {K_e}B_m^2{t^2}{f^2}V\)watts
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

Pe\(B_m^2{f^2}\)

Bm ∝ (V / f)

Pe ∝ V2

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

\(\frac{{{P_{e60}}}}{{{P_{e50}}}} = \frac{{V_{60}^2}}{{V_{50}^2}} = \frac{1}{1}\)

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

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