Question

The primary and secondary windings of a 40 kVA, 6600/250 V single phase transformer have resistance of $10\Omega and0.04\Omega$ respectively. The total leakage reactance is 30 $\Omega$ as referred to the primary winding. Find half load regulation at p.f. of 0.8 lagging.

• A. 1.41%
• B. 2.20%
• C. 4.41%
• D. 3.6%

Answer 1

The correct option is B 2.20%

$V_1=6600V,V_2=250V,$
$R_1=10\Omega ,R_2=0.04\Omega ,X_{01}=30\Omega andcos\varphi =0.8$

$K=\frac{V_2}{V_1}=\frac{250}{6600}=0.0378$,

$I_2=\frac{40\times {10}^3}{250}=160A$

Reffered to secondary side
$R_{02}=K^2{R}_1+R_2$

$=(0.0378{)}^2\times 10+0.04=0.0542\Omega$

$X_{02}=K^2{X}_{01}=0.0429\Omega$
Regulation

$=\frac{I_2{R}_{02}cos\varphi +I_2{X}_{02}sin\varphi }{E_2}\times 100$

$=\frac{80[0.0542\times 0.8+0.0428\times 0.6]}{250}\times 100$

= 2.2%