Question

A 20 kVA, 2000 V/200 V, two winding transformer is to be used as a step up auto transformer with a constant source voltage of 2000 V. If the efficiency of the two winding transformer at full load 0.7 p.f. lagging is 97%, the efficiency of the auto transformer at same p.f at full load will be ___________%

• A. 99.72

Answer 1

The correct option is A 99.72


$\text{From the diagram shown :}$
$\text{We can write, Transformation ratio}$

$a_{auto}=\frac{2200}{2000}=\frac{11}{10}$

Also we know that, rating of auto transformer ;

$S_{auto}=\frac{S_{T.W}}{(1-\frac{1}{a})}=\frac{20kVA}{(1-\frac{10}{11})}=220KVA$

$\text{Efficiency of two winding transformer :}$

${\eta }_{T.W}=\frac{20\times {10}^3\times 0.7}{20\times {10}^3\times 0.7+P_L}$

$\text{Where}{P}_L\text{= total losses}$

$0.97=\frac{14\times 1000}{14000+P_L}$

$\Rightarrow \text{Solving we get}{P}_L\text{= 433 W}$

$\text{Now efficiency for auto transformer :}$

${\eta }_{auto}=\frac{220\times 0.7}{220\times 0.7+0.433}$

$=\frac{154}{154+0.433}\times 100\%$

${\eta }_{auto}=99.72\%$