The following losses are occurs in the transformer.

- The core loss depends upon the supply voltage.
- The coreloss practically remains constant for all loads because the core flux remains constant at any load conditions.
- This loss includes hysteresis loss and eddy current loss.

- This loss is due to rapid reversal of magnetism due to alternating flux.
- It depends on the loop area of the material. The hysteresis loss is given by

W

_{h}= hB_{max}^{1.6}fV Joule / second or watt
Where h = Steinmetz constant, its value depends on the core material

B

_{max}= Maximum flux density in weber / meter^{2}
f = Supply
frequency in Hz

V = Volume of the core in meter

^{3}- An emf is induced in the core due to alternating flux in the core according to the laws of electromagnetic Induction.
- Although this emf is small, it sets up large current in the core due to small core resistance.
- The loss due to this current is known as eddy current loss. The eddy current loss is given by

We = K B

_{max}^{1.6}f^{2}t^{2}
Where K = Constant

B

_{max}= Maximum flux density in weber / meter^{2}
f = Supply frequency in Hz

t = Thickness of lamination in mm

- The eddy current loss varies as ( I ) Square of maximum flux density ( II ) Square of supply frequency ( III ) Square of thickness of lamination.
- If the supply voltage keeps constant, the core loss remains constant. The core loss does not depend on the load current.

- The value of core loss can find out by open circuit test.

These losses are due to ohmic resistances of the
transformer winding.

- Copper losses in the primary winding = I
_{1}^{2}R_{1} - Copper losses in the secondary winding = I
_{2}^{2}R_{2} - Total copper losses = I
_{1}^{2}R_{1}+ I_{2}^{2}R_{2} - Total copper losses at full load ( W
_{c}) = I_{1}^{2}R_{01}

=
I

_{2}^{2}R_{02}
Where R

_{01}= Transformer winding resistance as referred to primary side
R

_{02 }= Transformer winding resistance as referred to secondary side**Copper Losses W**

_{c}α ( I )^{2}- It means that copper loss varies as square of the current.

**Total copper losses at half load**

W

_{c}α ( I / 2 )^{2 }R
=I

^{2}R / 4- Therefore total copper loss at half load is reduced four time that of full load copper losses.

**Total copper losses at twice full load**

W

_{c}α ( 2I )^{2 }R
= 4I

^{2}R- Therefore total copper loss at twice full load is four times that of full load copper losses.
- The copper loss depends on the value of square load current and its value can find out by short circuit test.

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