### Condition for Maximum Efficiency

1. Condition for Maximum Efficiency
When a transformer works on a constant input voltage and frequency then efficiency varies with the load. As load increases, the efficiency increases. At a certain load current, it achieves a maximum value. If the transformer is loaded further the efficiency starts decreasing. The graph of efficiency against load current  I2 is shown in the Fig.1
 Fig. 1

The load current at which the efficiency attains maximum value is denoted as I2m and maximum efficiency is denoted as ηmax.
Let us determine,
1. Condition for maximum efficiency.
2. Load current at which ηmax occurs.
3. KVA supplied at maximum efficiency.
The efficiency is a function of load i.e. load current I2 assuming cos Φ constant. The secondary terminal voltage V2 is also assumed constant. So for maximum efficiency,
dη /d I2 = 0
Now      η = (V2 I2 cos Φ2 )/(V2 I2 cos Φ2 + Pi  + I22 R2e)

...       (V2 I2 cos Φ2 + Pi  + I22 R2e)(V2 cos Φ2) - (V2 I2 cos Φ2)(V2 cos Φ2 + 2I2  R2e) = 0
Cancelling (V2 cos Φ2) from both the terms we get,
V2 I2 cos Φ2 + Pi  +I22 R2e - V2 I2  Φ2 - 2I22 R2e = 0
...       Pi  - I22 R2e= 0
...       Pi  = I22 R2e = Pcu
So condition to achieve maximum efficiency is that,
Copper losses = Iron losses
1.1 Load Current I2m at Maximum Efficiency
For ηmax,           I22 R2e = Pi  but   I2 = I2m
I2m2 R2e = Pi
I2m = √(Pi / R2e)
This is the load current at ηmax,
Let        (I2)F.L. = Full load current
...          I2m /(I2) F.L.= (1/(I2) F.L.)√(Pi / R2e)
...           I2m /(I2) F.L.=   √(Pi )/({(I2) F.L.}2 R2e)
=  √(Pi )/((Pcu) F.L.)
...            I2m = (I2 )F.L.√(Pi )/((Pcu) F.L.)
This is the load current at ηmax interms of full load current.
1.2 KVA supplied at maximum Efficiency
For constant V2 the KVA supplied is the function of.
KVA at ηmax = I2m V2= V2 (I2) F.L. x √(Pi) /((Pcu)F.L.)
KVA at ηmax = (KVA rating) x √(Pi) /((Pcu)F.L.)
Substituting condition for in the expression of efficiency, we can write expression for ηmax as,

Example : A 250 KVA single phase transformer has iron loss of 1.8 KW. The full load copper loss is 2000 watts. Calculate
i) Efficiency at full load, 0.8 lagging p.f.
ii) KVA supplied at maximum eficiency
iii) Maximum efficiency at 0.8 lagging p.f.
Solution : The given values are,
Pi = 1800 W , (Pcu)F.L. = 2000 W
i)

= 98.135%
ii)                 KVA at = KVA rating x √(Pi) /((Pcu)F.L.)
= 250 x √(1800/2000)
= 237.1708 KVA
iii)

where        Pcu = Pi =1800W

= 98.137%