# Why is a transformer rated in KVA?

A transformer is a static device which transfer power from one circuit to another circuit without changing the frequency. It can rise or lower the voltage with corresponding decrease or increase of current.
An important factor in the design and operation of electrical machines is the relationship between the life of the insulation and operation temperature of the machine. Therefore, temperature rise resulting from the losses is a determining factor in the rating of the machine. The rating is the load carrying capability of the machine. It shows the maximum value of the voltage at which the machine is designed, and the current consumption occurs at that voltage. The machines are always rated in watts. But the alternator and the transformer are the only machines which are rated in volt-amp (VA).
This is because there are two types losses in transformer – Copper losses and Iron (Core) Losses.

Copper Losses: -  These are the losses which occur due to the resistance in the winding. They can be determined by short circuited test. They vary as square of the current. If “I1” and “I2” is the current in the Primary winding and Secondary winding respective. R1 and R2 is the resistance of Primary and secondary winding respectively. The copper losses are given by
PCu = I21R1 + I22R2
Core Losses or Iron Losses: -  These losses consist of hysteresis losses and eddy current losses and occur in transformer due to the alternating flux. These can be determined by open circuit test.
As seen copper losses of a transformer depend on current and iron losses depend on voltage. Hence total loss depends on volt-ampere(VA) and not the phase angle between current and voltage i.e. it is independ on load power factor. For this reason, the transformer is rate Kilovolt ampere) KVA and not in Kilowatt(KW)

Lets us prove mathematically the losses of transformer does not depend on the load power factor.
Suppose we have transformer of following data
Transformer rating in kVA = 100kVA
Primary Voltages = 11KV
Primary Current = 5.25 A
Secondary Voltages = 430V
Secondary Current = 133.33 A.
Equivalent resistant on Secondary (R) = 45Ω
Iron losses = 30W

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Case I
Suppose a resistive load is connected to the secondary of the transformer at unity power factor θ = 1
Then total losses of transformer would be sum of copper losses and iron
Losses

Total Losses = I²R + Iron losses

Total Losses = (133.332 x 45) + 30W = 800kW

The output of the transformer
P = √3 x V x I x Cos θ
Putting value from secondary side
P = √3 x 430 x 133.33 x 1 = 100kW.
The KVA Rating of the transformer is
KVA = P(KW) / Power Factor
KVA = 100 / 1 = 100

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Case II
Now take the power factor, Cos θ = 0.6
Again, total losses of transformer would be sum of copper losses and iron losses, i.e.

Total Losses = I²R + Iron losses

Total Losses = (133.332 x 45) + 30W = 800kW
But The transformer output will be:
P = √3 x V x I x Cos θ
Putting value from secondary side
P = √3 x 430 x 133.33 x .6 = 60 KW
Now rating of transformer
KVA = P(KW) / Power Factor
KVA = 60 / .6 = 100
From the above example, it is clear that the rating of transformer
is same (100KVA) in both cases but different output in power (100kW and 60kW) due to different power factor values after connecting different
kind of load which is not predictable for transformer manufactures where the losses are same in both cases