**EFFICIENCY OF TRANSFORMER**

- 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%
Where

- 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 are zero so the transformer efficiency is much higher than any other rotating device.

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

- The efficiency of transformer increases as the load current increases from no load to full load.
- It becomes maximum at full load.
- The efficiency is practically constant from about 115% load current to 120% load current.
- The transformer efficiency becomes poor at light loaded and over loaded conditions.

**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 %

= ( 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 %
= ( V

_{1}I_{1}Cos F / V_{1}I_{1}Cos F ) – ( 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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