Question

Derive an expression for maximum efficiency of a transformer

Answer 1

Answer:

We know that,

n= input/output + losses

Considering the secondary side,

n = V,I, cos p2/V½I2 cos 02 + Wi+I;Ro2

Diff w.rt. 12,

For maximum efficiency, dn/dI2 = 0

dn = (V2I2 cos 02/V2I2 cos 02 + W1

+ 122R02)Y2 cos 02

dIV2 In cos 02/V2 Icos

02(V2 cos 2 +W2R02

/(V2I2 cos 02/V212 cos 02 + W1+ ; Ro2)2

.. (V2 I2 cos 2/V,I2 cos o2+ W1

=

+I Ro2)Y2 cos 2 = V2I2 cos

p2(V2 cos 2 + W2R02)

Therefore W1 = 122 Ro2

Similarly on primary side,

Wiki{Ro1

Comment: When copper loss=iron loss, efficiency

of the transformers is maximum.

Load maximum efficiency

$2(V2 cos 02 +W2R02)

Therefore W1 = I22R02

0.4KB/s

0.4KB/s * l a T00

Similarly on primary side,

Wii{Rom

Comment: When copper loss=iron loss, efficiency

of the transformers is maximum.

Load maximum efficiency

for maximum efficiency, W1=1^2_2R02

12(max.eff) = (W1/R02)

Multiplying both the sides by V2

V2I2(max efficiency) = V2 W1/RO2

Load VA

(Max efficiency)

= V2I2/W1/IR2R02

= V212 W1/Wai

Load KVA max efficiency =

Full load KVA/W1/Wai

#Answerwithquality

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