All transformers have copper and core losses. Copper loss is power lost in the primary and secondary windings of a transformer due to the ohmic resistance of the windings.

Copper loss, in watts, can be found using below Equation.

**Copper Loss = I**_{P}^{2} R_{P} + I_{S}^{2} R_{S}

_{P}

^{2}R

_{P}+ I

_{S}

^{2}R

_{S}

where

I_{P} = primary current

I_{S} = secondary current

R_{P} = primary winding resistance

R_{S} = secondary winding resistance

Core losses are caused by two factors: hysteresis and eddy current losses. Hysteresis loss is that energy lost by reversing the magnetic field in the core as the magnetizing AC rises and falls and reverses direction.

Eddy current loss is a result of induced currents circulating in the core.

The efficiency of a transformer can be calculated using below Equations.

where

PF = power factor of the load

**Example 1:**

**A 5:1 step-down transformer has a full-load secondary current of 20 amps. A short circuit test for copper loss at full load gives a wattmeter reading of 100 W. If R _{P} = 0.3Ω, find R_{S} and power loss in the secondary.**

Solution :

**Copper Loss = I _{P}^{2} R_{P} + I_{S}^{2} R_{S }**= 100 W

To find I_{P}

I_{P }= (N_{S}/N_{P}) I_{S}

I_{P }= (1/5) 20 = 4 amps

To find R_{S}

I_{S}^{2} R_{S }= 100 – I_{P}^{2} R_{P}

R_{S }= (100 – I_{P}^{2} R_{P} ) / I_{S}^{2}

R_{S }= (100 – 4^{2 } x 0.3 ) / 20^{2}

R_{S }= 0.24

Power loss in secondary = I_{S}^{2} R_{S }= 20^{2} x 0.24 = 96 W

**Example 2:**

** An open circuit test for core losses in a 10 kVA transformer gives a reading of 70 W. If the PF of the load is 90%, find efficiency at full load.**

Solution :

How is transformer rating in kva?

Since Transformer doesnt produce any power rather than just only transform the Voltage and Current into high to low and low to high, for which Voltage and Ampere is related. Thats why its only rating in VA. and for larger transformer KVA and MVA is used for the rating of a transformer.

where did the copper loss of 100 W came from i the second problem?