Circuit Analysis

C. Per-unit system

The Per-Unit System

Normalizing voltage, current, power, and impedance to common bases — the change-of-base formula and why transformers vanish from the math.

Power systems span voltage levels from 120 V to 765 kV through dozens of transformers. Working in volts and ohms means endlessly re-referring impedances across turns ratios — tedious and error-prone. The per-unit (pu) system normalizes every quantity to a fraction of a chosen base, and with bases chosen to track transformer ratios, the transformers literally disappear from the impedance diagram. Equipment also behaves comparably in pu (a transformer is 5–8% impedance whether it's 500 kVA or 500 MVA), so wrong answers look wrong.

Choosing bases

Pick ONE system-wide apparent-power base

S_{base}

(commonly 100 MVA in utility studies, or the rating of a major transformer) and a voltage base

V_{base}

for each voltage zone, stepping the voltage base across every transformer in proportion to its turns ratio. Everything else follows:

I_{base} = \frac{S_{base}}{\sqrt{3}\,V_{LL,base}}, \qquad Z_{base} = \frac{V_{LL,base}^2}{S_{base}}

Per-unit values

A quantity in per-unit is its actual value divided by the base of the same dimension. Note that with three-phase bases (

S

= total three-phase MVA,

V

= line-line kV) the familiar formulas hold without any

\sqrt{3}

or 3 appearing in pu equations — those constants are absorbed into the bases.

Why transformers vanish

Refer an impedance across an ideal transformer and it scales by the turns ratio squared; but the impedance BASE also scales by the voltage-base ratio squared. If voltage bases follow the transformer ratio, the two effects cancel exactly: a device's per-unit impedance is the same from either side. The whole network becomes one flat impedance diagram — the reason fault studies and power flows are always done in per-unit.

Change of base

Manufacturers state impedances on the equipment's own ratings (e.g., 'Z = 6% on 25 MVA, 69 kV'). To use them on the study's system base, convert — impedance scales directly with the MVA base and inversely with the voltage base squared:

Z_{pu}^{new} = Z_{pu}^{old}\left(\frac{V_{base}^{old}}{V_{base}^{new}}\right)^{2}\left(\frac{S_{base}^{new}}{S_{base}^{old}}\right)

Working a per-unit problem

The standard recipe: (1) choose

S_{base}

and one zone's

V_{base}

; (2) propagate

V_{base}

Exam strategy

Per-unit questions are usually one of three: compute a base quantity (

Z_{base}

I_{base}

), execute one change of base, or run a small fault/voltage calculation in pu. The change-of-base voltage ratio trips people: it is

(V_{old}/V_{new})^2

Key equations

Base current (3-φ)

I_{base} = \dfrac{S_{base}}{\sqrt{3}\,V_{LL,base}}

Worked examples

Base quantities

Problem. Choose $S_{base} = 100$ MVA at 13.8 kV. Find $Z_{base}$ and $I_{base}$

Common pitfalls

•Inverting the change-of-base voltage ratio — it is $(V_{old}/V_{new})^2$ with the OLD base on top. When nameplate and system voltages match, skip it entirely.

References

NCEES PE Power Reference Handbook §3.1.3 — Per Unit System
Grainger & Stevenson, Power System Analysis — ch. 1; per-unit conventions
IEEE Std 399 (Brown Book) — study practices using per-unit

A. Three-phase circuits

Three-Phase Circuits and Power

Phase sequence and notation

Wye connection

Delta connection

Balanced three-phase power

Delta–wye conversion and per-phase analysis

Unbalanced systems — a preview

Exam strategy

Key equations

Worked examples

Common pitfalls

References

B. Symmetrical components

Symmetrical Components

C. Per-unit system

The Per-Unit System

Choosing bases

Per-unit values

Why transformers vanish

Change of base

Working a per-unit problem

Exam strategy

Key equations

Worked examples

Common pitfalls

References

D. Phasor diagrams

Phasors and Complex Power

E. Single-phase circuits

Single-Phase AC Circuits

F. Direct current circuits

DC Circuit Analysis

— Single-line diagrams

Single-Line Diagrams and ANSI Device Numbers