Question

A 1-phase 12 kVA, 2200/220 V, 50 Hz transformer has following test results:
Open circuit (LV side) : V = 220 V, I = 1.32 A, P = 80 W
Short circuit (HV side) : V = 110 V, I = 4.09 A, P = 150 W
Efficiency at full load with 0.8 lagging power factor is ____%(Answer upto 2 decimal places)

• A. 96.51

Answer 1

The correct option is A 96.51
$\text{Full load current,}{I}_{fl}=\frac{12\times {10}^3}{2200}=5.45A$
$\text{Short circuit test is conducted at I = 4.09 A}$

$\text{Full load short circuit losses,}$ $P_{cu}=150\times {\left(\frac{5.45}{4.09}\right)}^2=266.34W$

$\text{Efficiency,}$
$\eta =\frac{12000\times 0.8}{12000\times 0.8+80+266.34}=96.51\%$