- The efficiency of the transformer at any load power factor is defined as the ratio of the output power and input power.

Efficiency

= ( Output power / Input power ) × 100%

=
( Output power / Output power + Losses ) × 100%

=
( Output power / Output power + Iron losses +
Copper losses ) × 100%

=
kVA Cos F / ( kVA Cos F + W

_{i }+ W_{cu}) × 100%
Cos F = Load power factor

W

_{i}= Iron losses
W

_{cu}= Full load copper losses
kVA = Rating of the transformer

**Why the efficiency of the transformer is much higher than any other rotating machines?**

- There are no moving or rotating parts in the transformer i.e. the mechanical losses is zero so the transformer efficiency is much more higher than any other rotating devices.

**How the efficiency of transformer varies with load current?**

- The efficiency is practically constant from about 100% load current to 120% load current. At very light load and overloaded conditions efficiency becomes poor.

**Condition For Maximum Efficiency**

- The transformer iron losses are practically constant at constant voltage, but copper losses varies as square of load current therefore the efficiency is affected by the factor load current.
- Let us find out the condition for maximum efficiency.

Efficiency

= ( Output
power / Input power ) × 100 %

h = ( Input power – Total losses / Input power ) ×
100 %

h = [ ( V

_{1}I_{1}Cos F – W_{cu}– W_{i}) / V_{1}I_{1}Cos F ] × 100 %
h = [ ( V

_{1}I_{1}Cos F – I_{1}^{2}R_{01}– W_{i}) / V_{1}I_{1}Cos F ] × 100 %
= 1 – (
I

_{1}^{2}R_{01}/ V_{1}I_{1}Cos F ) – ( W_{i}/ V_{1}I_{1}Cos F )
= 1 – ( I

_{1}R_{01}/ V_{1}Cos F ) – ( W_{i}/ V_{1}I_{1}Cos F )
As the load current is variable parameter, the
transformer efficiency becomes maximum when

dh / dI

_{1}= 0
= 0 – ( R

_{01}/ V_{1}Cos F ) + ( W_{i}/ V_{1}I_{1}^{2}Cos F )
( R

_{01}/ V_{1}Cos F ) = ( W_{i}/ V_{1}I_{1}^{2}Cos F )
I

_{1}^{2}R_{01 }= W_{i}
W

_{cu}_{ }= W_{i}**The transformer efficiency becomes maximum when total copper losses are equal to iron losses.**

**At what load current efficiency becomes maximum?**

For maximum efficiency W

_{cu}_{ }= W_{i}
I

_{1}^{2}R_{01 }= W_{i}
I

_{2}^{2}R_{02 }= W_{i}
I

_{2}= √ ( W_{i }/ R_{02})
I

_{2}= [ I_{2full}/ I_{2full}] √ ( W_{i }/ R_{02})
= I

_{2full}√ ( W_{i }/ I_{2full}^{2}R_{02})
= I

_{2full}√ ( W_{i }/ W_{cu })
= I

_{2full}√ ( Iron losses_{ }/ Full load copper losses )**At what kVA rating the efficiency of transformer becomes maximum?**

I

_{2}= I_{2full}√ ( W_{i }/ W_{cu })
V

_{2}I_{2}= V_{2}I_{2full}√ ( W_{i }/ W_{cu })
V

_{2}I_{2}/ 1000 = [ V_{2}I_{2full}/ 1000 ] √ ( W_{i }/ W_{cu })
kVA = kVA

_{full}√ ( W_{i }/ W_{cu })
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