⚡Fault Analysis Part 2 - Per Unit System
⚡ Welcome to Part 2 of our Fault Analysis series. Before we can calculate any fault current, there's one foundational tool every power systems engineer relies on: the Per-Unit (p.u.) System. Skip this, and every fault calculation in the rest of the series will feel confusing — so let's build it properly.
Why Do We Need a Per-Unit System?
Power systems have multiple voltage levels connected through transformers — generation at 11 kV, transmission at 220 kV, distribution at 11 kV or 400V, and so on. Doing calculations directly in actual (ohmic) values across all these levels is messy and error-prone.
The per-unit system solves this by expressing all quantities as a fraction of a chosen base value, so:
- Transformer turns ratio disappears from calculations — no need to refer impedances across transformers manually
- Equipment of different sizes becomes easy to compare (a per-unit impedance of 0.1 means the same relative thing whether the machine is 10 MVA or 500 MVA)
- Calculations become simpler and less error-prone across multi-voltage networks
The Basic Formula
Per-unit value of any quantity is defined as:
Choosing Base Quantities
We typically choose two independent base quantities, and the rest are derived from them:
| Quantity | Relationship |
|---|---|
| Base Power (S_base) | Chosen directly — usually same across the whole system (e.g. 100 MVA) |
| Base Voltage (V_base) | Chosen per voltage level (changes across transformers) |
| Base Current (I_base) | S_base / (√3 × V_base) — for 3-phase systems |
| Base Impedance (Z_base) | V_base² / S_base |
Changing Base — The Conversion Formula
Equipment nameplates give impedance in per-unit on their own rated base — but your system study might use a different common base. Convert using:
This single formula is one of the most frequently tested numericals across SSC JE, GATE, and ESE — memorize it, and practice a few variations.
Why It Simplifies Transformers Specifically
In actual units, a transformer's impedance looks completely different depending on whether you measure it from the primary or secondary side (turns ratio squared difference). In per-unit, if base voltages on each side are chosen in the same ratio as the transformer's turns ratio, the per-unit impedance is identical on both sides. This is exactly why per-unit is the default language of power system studies — transformers effectively "disappear" from the math.
Worked Example
A transformer is rated 20 MVA, 11 kV, with 8% impedance on its own rating. Find its per-unit impedance on a system base of 100 MVA, 11 kV.
Z_pu(new) = 0.08 × 5 × 1 = 0.4 p.u.
Quick Revision Points
- Per-unit value = Actual value ÷ Base value
- Choose S_base and V_base independently; I_base and Z_base are derived
- Z_base = V_base² / S_base
- Base conversion: multiply by (new S/old S) and (old V/new V)²
- Per-unit impedance is same on both sides of a transformer if base voltage ratio = turns ratio
Practice MCQs
Q1. Base impedance is calculated as:
a) V_base / I_base
b) V_base² / S_base
c) S_base / V_base
d) I_base × V_base
Q2. A transformer's per-unit impedance stays the same on both sides when:
a) The turns ratio is 1:1
b) The base voltage ratio equals the transformer's turns ratio
c) The transformer is star-star connected
d) The transformer is unloaded
Q3. If S_base is doubled (with V_base unchanged), the new per-unit impedance:
a) Halves
b) Doubles
c) Stays the same
d) Becomes zero
Q4. Which quantity is typically chosen directly as a base rather than derived?
a) Base impedance
b) Base current
c) Base power
d) Base admittance
(Answers: 1-b, 2-b, 3-b, 4-c)
Coming up in Part 3: Symmetrical (Three-Phase) Fault Analysis — the most severe fault type, and the simplest to calculate since it only needs the positive sequence network.
Save this post for revision before your next mock test!