Question

A 50kVA, 2500 V/250V, 2 winding transformer is to be used as a step up auto transformer with a constant source voltage of 2500 V. The efficiency of 2 winding transformer at full load, o.8 pf lagging is 96%, then the efficiency (in percentage) of auto transformer at rated load at 0.6 pf lagging is

• A. 99.49

Answer 1

The correct option is A 99.49

$a_{auto}=\frac{2750}{2500}=\frac{11}{10}$

$s_{auto}=\frac{50}{1-\frac{11}{10}}=550kVA$

$\text{For 2 winding transformer,}$

$\%\eta =0.96\times \frac{50\times {10}^3\times 0.8}{50\times {10}^3\times 0.8\times P_L}$

$P_L=1666.67W$

At rated load, total losses remains same for both two winding transformer and auto transformer.

$\text{For auto transformer at rated load}$
$\%\eta =\frac{550\times {10}^3\times 0.6}{550\times {10}^3\times 0.6+1666.67}\times 100\%$

$\%\eta =99.49\%$