ElectricalPower SystemshardCourseworkPE / Advanced

Actual Impedance from a Per-Unit Value

A transformer is rated $100\ \text{kVA}$ , $480\ \text{V}$ , with a per-unit impedance of $5\%$ on its own base.

Find the actual impedance in ohms referred to the $480\ \text{V}$ side.

Given

S	100 kVA	Rated apparent power
V	480 V	Rated voltage
Zpu	0.05 -	Per-unit impedance

Hint 1

Per-unit values are normalised against a base impedance derived from the ratings.

Hint 2

Z_base = V²/S, with S in VA and V in volts.

Hint 3

Then Z_actual = Z_pu × Z_base, remembering 5% = 0.05.

Worked solution — try the problem first

Base impedance:

$Z_{base} = \frac{V_{base}^2}{S_{base}} = \frac{480^2}{100{,}000} = \frac{230{,}400}{100{,}000} = 2.304\ \Omega$

Note the voltage is squared and the apparent power is in volt-amperes, not kVA.

Actual impedance:

$Z_{actual} = Z_{pu} \times Z_{base} = 0.05 \times 2.304 = 0.1152\ \Omega$

$\boxed{Z = 0.1152\ \Omega}$

Why per-unit exists. Expressed this way, impedance is the same number on both sides of a transformer — no referring through turns ratios squared. In a network with several voltage levels that removes an entire class of error, which is why utilities work almost exclusively in per-unit.

For a three-phase base, $Z_{base} = V_{LL}^2/S_{3\phi}$ — the same expression, provided line-to-line voltage is paired with total three-phase power.

Concepts:Per-unit systemBase impedancePower system analysis

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