Question

A single-phase, 50 Hz, 1200/120 V transformer gave the following results of open-circuit test with voltage 120 V, current 16 A, power input 400 W, then the magnetizing component of no-load current is ______A.

• A. 15.65

Answer 1

The correct option is A 15.65
$R_0=\frac{V^2}{P}=\frac{120\times 120}{400}=36\Omega$

$I=\sqrt{{\left(\frac{V}{R}\right)}^2+{\left(\frac{V}{X}\right)}^2}$

${\left(\frac{I}{V}\right)}^2=\frac{1}{R^2}+\frac{1}{X^2}$

$\frac{1}{X^2}={\left(\frac{16}{120}\right)}^2-{\left(\frac{1}{36}\right)}^2$

$X=7.668\Omega$

$\text{The magnetizing component of no load current,}$

$I_m=\frac{V}{X}=\frac{120}{7.668}=15.65A$

$\text{Alternative Solution :}$

$\text{Given,}{V}_{NL}=120V$

$I_{NL}=16A$

$P_{NL}=400W\text{(Open circuit test data)}$

$\therefore P_{NL}=V_{NL}{I}_{NL}cos\varphi$

$400=120\times 16\times cos\varphi$

$cos\varphi =0.2083$

$sin\varphi =0.978$

$I_m=I_{NL}sin\varphi$

$=16\times 0.978=15.65A$