Question

The open circuit test and the short circuit test were performed on the primary side of a 10 kVA, 2000/240 V, 50 Hz transformer and the following data were obtained

$OCtest:V_{OC}=240V;I_{OC}=0.6A,P_{OC}=80W$
$SCtest:V_{SC}=112V;I_{SC}=5A,P_{SC}=160W$

Find the approximate equivalent circuit referred to the primary side,

• A. $R_{eq}=6.4\Omega ;X_{eq}=j21.46\Omega ;R_C=49.96k\Omega ;X_m=j33.38k\Omega$
• B. $R_{eq}=0.092\Omega ;X_{eq}=j0.31\Omega ;R_C=720k\Omega ;X_m=j480k\Omega$
• C. $R_{eq}=0.092\Omega :X_{eq}=j0.31\Omega ;R_C=49.96k\Omega ;X_m=j33.34k\Omega$
• D. $R_{eq}=7.2\Omega :X_{eq}=j21.46\Omega ;R_C=720\Omega ;X_m=j480\Omega$

Answer 1

The correct option is A $R_{eq}=6.4\Omega ;X_{eq}=j21.46\Omega ;R_C=49.96k\Omega ;X_m=j33.38k\Omega$

From the given data it can be seen that, OC test is performed on the LV side. Thus the equivalent core-loss resistance as referred to the low-voltage side is

$R_C=\frac{{240}^2}{80}=720\Omega$

$\text{The apparent power under no load ;}$
$S_{OC}=V_{OC}{I}_{OC}=240\times 0.6$
$=144VA$
$\text{Thus, the reactive power is}$
$Q_{OC}=\sqrt{{144}^2-{80}^2}$
$=119.73VAR$

The magnetizing reactance as referred to the low voltage side is
$X_m=\frac{{240}^2}{119.73}=481.082\Omega$

The core-loss resistance and the magnetizing reactance as referred to HV side are obtained as follows:

$\frac{N_1}{N_2}=\frac{2000}{240}=8.33$

$R_{CHV}=(8.33{)}^2\times 720=49.96k\Omega$
$X_{mHV}=(8.33{)}^2\times 481.082=33.378k\Omega$

$\text{SC best is performed on HV side, thus}$

$R_{eqHV}=\frac{P_{sc}}{I_{sc}^2}=\frac{160}{5^2}=6.4\Omega$

$Z_{eqHV}=\frac{V_{sc}}{I_{sc}}=\frac{112}{5}=22.4\Omega$

$X_{eqHV}=\sqrt{{22.4}^2-{6.4}^2}=21.46\Omega$