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)
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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.
Step 1: Set up the no load admittance on the LV side.
With the HV side open, the LV winding sees the full excitation branch. The magnitude of the no load admittance is
\[ Y_0=\frac{I}{V}=\frac{2.75}{220}=0.0125\ \text{S} \]
Step 2: Split the admittance into conductance and susceptance.
The core loss conductance comes from the power reading,
\[ G_0=\frac{P}{V^2}=\frac{363}{220^2}=\frac{363}{48400}=0.0075\ \text{S} \]
The magnetizing susceptance follows from $Y_0^2=G_0^2+B_0^2$, so
\[ B_0=\sqrt{Y_0^2-G_0^2}=\sqrt{0.0125^2-0.0075^2}=\sqrt{0.00015625-0.00005625}=\sqrt{0.0001}=0.01\ \text{S} \]
Step 3: Invert to get the LV side magnetizing reactance.
\[ X_{m,LV}=\frac{1}{B_0}=\frac{1}{0.01}=100\ \Omega \]
This matches the same 100 ohm figure found directly from the power factor triangle, so the split checks out.
Step 4: Scale up to the HV side using the turns ratio squared.
The turns ratio is $a=1100/220=5$. Referring an LV side impedance to the HV side means multiplying by $a^2$:
\[ X_{m,HV}=a^2 X_{m,LV}=25\times100=2500\ \Omega \]
\[ \boxed{2500\ \Omega} \]
