Dashboard PE Power Exam Prep Circuit Analysis Per-unit & three-phase Per-unit system + base conversion

Per-unit system + base conversion

Why expressing every quantity as a fraction of a base makes power-system math beautiful — and how to convert between bases without panic.

Sophomore ~10 min

Step 1 — Why per-unit: same impedance, different ohms on every voltage level

0.55×
S_base 100 MVA V_LL_base 13.8 kV Z_base 1.9 Ω

Reference notes

Use Next → on the narrator above to walk through six steps that build the per-unit system from the ground up: why it exists, the four base quantities, conversion in single- and three-phase, base change between systems, and why per-unit makes transformers invisible.

Why per-unit exists

A real power system has equipment at many voltage levels — a generator at 13.8 kV, a transmission line at 230 kV, a distribution feeder at 12.47 kV, a low-voltage panel at 480 V. The same physical fault current produces wildly different ampere readings on each level, and the same physical impedance produces wildly different ohm values. Comparing equipment across levels in actual units is mental gymnastics.

The per-unit system fixes this by choosing a common power base (typically a system-wide value like 100 MVA) and using each level's rated line-to-line voltage as that level's voltage base. Every quantity is then expressed as a fraction of its base, becoming dimensionless. Per-unit voltages cluster around 1.0; per-unit impedances of standard equipment fall in well-known ranges (transformer 5–10%, transmission line 10–20%, synchronous machine subtransient 15–25%); the whole system becomes an apples-to-apples comparison.

The four base quantities

There are four base quantities — V, S, I, Z — but only two are independent. By convention we pick V_base and S_base, then derive the other two from the laws of three-phase circuits.

Vbase = VLL   (line-to-line, in kV typically)
Sbase = S   (three-phase apparent power, in MVA)
Ibase = S / (√3 · VLL)
Zbase = VLL² / S

The Zbase formula is the workhorse — get a feel for plugging in: at 100 MVA and 13.8 kV, Zbase = (13.8)² / 100 = 1.9044 Ω. At 100 MVA and 230 kV, Zbase = (230)² / 100 = 529 Ω. The same per-unit impedance is therefore far larger in ohms on the high-voltage side — which is exactly why per-unit makes the comparison sensible.

Critical convention: in three-phase work, V_base is always line-to-line and S_base is always three-phase. If you mix in phase voltage or single-phase power by accident, you'll be off by factors of √3 or 3.

Converting between actual and per-unit

Per-unit conversion is the simplest part of the whole system. Divide every actual quantity by its base:

Vpu = Vactual / Vbase · Ipu = Iactual / Ibase · Zpu = Zactual / Zbase · Spu = Sactual / Sbase

Worked example: a 500 kVA, 11 kV / 415 V transformer has 5% per-unit leakage impedance on its own ratings. What's the actual impedance referred to the HV side?

  1. Zbase,HV = (11 × 10³)² / (500 × 10³) = 1.21 × 10⁸ / 5 × 10⁵ = 242 Ω
  2. Zactual = Zpu × Zbase = 0.05 × 242 = 12.1 Ω

Base change: same physical impedance, different bases

Equipment data sheets quote per-unit impedance on the equipment's own ratings. A power-system study uses a single system-wide base. You'll need to convert the equipment per-unit value to the system base:

Zpu,new = Zpu,old · (Vold / Vnew)² · (Snew / Sold)

The voltage ratio is squared because Z_base ∝ V². This is the one part of the formula students miss most often on the exam. The power ratio is linear because Z_base ∝ 1/S.

Worked example: a transformer rated 50 MVA, 13.8 kV has 8% leakage impedance on its own base. Convert to a system base of 100 MVA, 13.8 kV.

Zpu,new = 0.08 · (13.8 / 13.8)² · (100 / 50) = 0.08 · 1 · 2 = 0.16 pu

A larger MVA base produces a larger per-unit reading for the same physical impedance.

Why per-unit is invariant across an ideal transformer

This is the single biggest reason to use the system. Consider any transformer and pick its nameplate as a base. The HV side has Zbase,HV = VLL,HV² / S; the LV side has Zbase,LV = VLL,LV² / S. They differ by the turns ratio squared.

The same physical leakage impedance, referred to the HV side, is large (in ohms). Referred to the LV side, it is small (in ohms), again by the turns ratio squared. But:

Zpu,HV = Zactual,HV / Zbase,HV = Zactual,LV · N² / (Zbase,LV · N²) = Zpu,LV

The two N² factors cancel exactly. The per-unit impedance is the same regardless of which side you measured or referred to. In a per-unit one-line diagram, the transformer disappears entirely — it's just a per-unit impedance between two buses. That's the bookkeeping miracle of the per-unit system.

Quick sanity-check ranges

If your per-unit result falls outside these ranges, double-check the bases — most errors are a missed √3 or a wrong S_base.

Take-away. Per-unit replaces a system full of inconsistent ohms and amperes with dimensionless fractions of well-chosen bases. Pick V_LL and S_3φ; derive I_base and Z_base from them. Convert between bases with Zpu,new = Zpu,old · (Vold/Vnew)² · (Snew/Sold) — voltage squared, power linear. And in per-unit, every transformer becomes a simple series impedance. That's why every power-system study runs in per-unit.

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