Question

A single phase transformer has a rating of 10 kVA, 50 Hz, 1100 V/ 220 V. In an open circuit test with the high voltage side open, the following are obtained:
Voltage applied to the low voltage winding = 220 V
Measured power = 363 W
Measured current = 2.75 A
Magnetizing reactance \(X_m\) referred to the high voltage winding is ________ \(\Omega\). (rounded off to the nearest integer)

Hint:

Find the no-load parameters on the LV side first, then multiply the reactance by the square of the turns ratio to refer it to the HV side.

Answer 1

Correct Answer: 2500

Step 1: Identify what the open circuit test gives.
The high voltage side is left open and the test is run from the low voltage side, so all the readings, voltage \(V=220\) V, power \(P=363\) W and current \(I=2.75\) A, are measured on the LV winding. From these three readings the no load branch (magnetizing branch) parameters referred to the LV side can be found.

Step 2: Find the no load power factor.
The power factor at no load is
\[ \cos\phi_0=\frac{P}{VI}=\frac{363}{220\times2.75}=\frac{363}{605}=0.6 \] So \(\sin\phi_0=\sqrt{1-0.6^2}=\sqrt{1-0.36}=\sqrt{0.64}=0.8\).

Step 3: Split the no load current into its two components.
The no load current has a magnetizing (reactive) component and a core loss (in phase) component. The magnetizing component is
\[ I_m=I\sin\phi_0=2.75\times0.8=2.2\text{ A} \]

Step 4: Find the magnetizing reactance referred to the LV side.
\[ X_{m,LV}=\frac{V}{I_m}=\frac{220}{2.2}=100\ \Omega \]

Step 5: Refer this reactance to the HV side.
Since the test was done from the LV side, \(X_{m,LV}\) is on the LV side. To move an impedance from one side of a transformer to the other, multiply by the square of the turns ratio (HV over LV). The turns ratio is
\[ a=\frac{V_{HV}}{V_{LV}}=\frac{1100}{220}=5 \] So
\[ X_{m,HV}=a^2\times X_{m,LV}=5^2\times100=25\times100=2500\ \Omega \]

Final Answer:
The magnetizing reactance referred to the high voltage winding is \[ \boxed{2500\ \Omega} \]