Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 8 M

Year: Dec 2013

Condition for maximum efficiency

We know that,

$Ƞ = output/ output + losses$

Considering the secondary side,

$Ƞ= V_2I_2\cosф2/ V_2I_2\cosф2 + W1 + I^2_2Ro2$

Diff w.r.t. I2,

For maximum efficiency, $dn/dI2 = 0$

$dn= (V_2I_2\cosф2/ V_2I_2\cosф2 + W1 + I22Ro2) Y2 \cos ф2 \\ dI= V_2I_2\cosф2/ V_2I_2\cosф2(V2 \cosф2 + W2Ro2/ (V_2I_2\cosф2/ V_2I_2\cosф2 + W1 + I^2_2Ro2)2 \\ ∴ (V_2I_2\cosф2/ V_2I_2\cosф2 + W1 + I^2_2Ro2) Y2\cos ф2 = V_2I_2\cosф2 (V2 \cos ф2 + W2 Ro2)$

Therefore $W1= I22Ro2$

Similarly on primary side,

$W1 = I_1^2Ro1$

Comment: When copper loss= iron loss, efficiency of the transformers is maximum.

Load maximum efficiency

∴ for maximum efficiency, W1= I^2_2Ro2

$I2 (\max. eff) = \sqrt{(W1/R02)}$

Multiplying both the sides by V2

$V_2I_2(\max efficiency) = V2 \sqrt{W1/R02}$

Load VA

(Max efficiency)

$= V_2I_2\sqrt{W1/IR 2 Ro2} \\ = V_2I_2 \sqrt{W1/Wai}$

Load KVA max efficiency =

Full load $KVA \sqrt{W1/Wai}$