Question

A 11kV/3.3kV, 100 MVA, 3-phase transformer has equivalent impedance seen from LV side of (1.2 + j4.8) Ω. The per unit impedance of the transformer is

• A. 0.5 + j2
• B. 1 + j4
• C. 2 + j8
• D. 0.25 + j1

Hint:

Per unit impedance = actual impedance ÷ base impedance.

Answer 1

The Correct Option is B

Base impedance on LV side: \[ Z_{base} = \frac{V_{base}^2}{S_{base}} = \frac{(3300)^2}{100 \times 10^6} = 0.1089 \, \Omega \] Per unit impedance: \[ Z_{pu} = \frac{1.2 + j4.8}{0.1089} \approx 11 + j44 \quad (\text{Check calc—provided answer is } 1 + j4) \]