Power System Fault Analysis & Symmetrical Components – Solved Numericals + Quick Revision Notes (2026 Exam)

Power System Fault Analysis & Symmetrical Components
(Solved Numericals + Quick Revision) ⚡ 2026 Exam
QUICK REVISION SHEET (VERY IMPORTANT 🔥)

⚡ Key Formulas

 Fault MVA = Base MVA / p.u. reactance
 Fault Current = Fault MVA / (√3 × kV)
 Fault current is lagging
 Higher reactance → lower fault current

⚡ Important Points

 Use same base MVA for entire system
 Use positive sequence network for symmetrical fault
 Fault MVA is independent of base MVA

⚡ Symmetrical Components

 Positive → same sequence
 Negative → reverse sequence
 Zero → same phase

Symmetrical Faults

Definition

A symmetrical fault is a balanced 3-phase fault where all phases are equally affected.

Steps for Calculation

1. Convert all reactances into per unit
2. Draw reactance diagram
3. Reduce network using Thevenin’s theorem
4. Calculate Fault MVA and Fault Current

Fault MVA and Fault current (Steady State)
Calculate total per unit reactance upto fault point, by network reduction of positive sequence
network or reactance diagram. Take same base kVA or MVA for complete system.
𝐵𝑎𝑠𝑒 𝑀𝑉𝐴
Fault MVA=𝑝.𝑢. ( 𝑀𝑉𝐴)
𝑋𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

𝐹𝑎𝑢𝑙𝑡 𝑀𝑉𝐴×1000
Fault current = (Ampere)
√3×𝐵𝑎𝑠𝑒 𝑘𝑉

Fault current is of lagging power factor.


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,Problem 1: A three phase 5MVA, 6.6 kV generator having 10% sub-transient reactance. A
three phase short circuit occurs at its terminals. Calculate fault MVA and fault current.

Solution: Let Base MVA = 5 and
Base kV = 6.6
Reactance = 10% = 0.1 p.u.

[Generator reactance is based on its own voltage and kVA ratings.]

𝐵𝑎𝑠𝑒 𝑀𝑉𝐴 5
Fault MVA=𝑝.𝑢. = 0.10 = 50 𝑀𝑉𝐴
𝑋𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

𝐹𝑎𝑢𝑙𝑡 𝑀𝑉𝐴×1000 50×1000
Fault current = = = 4373.86 𝐴∠ − 900
√3×𝐵𝑎𝑠𝑒 𝑘𝑉 √3×6.6

⚡ Important Insight

 Lower reactance → higher fault current

Problem 2: Four 11 kV, 25MVA alternators having 15% sub-transient reactance are
operating in parallel when a three phase fault occurs on the generator bus. Calculate the three
phase fault MVA fed into the fault.

Solution: Let Base MVA = 25 and
Base kV = 11
15
The equivalent per unit reactance of four alternators operating in parallel is 4 = 3.75 % =
0.0375 𝑝. 𝑢.
𝐵𝑎𝑠𝑒 𝑀𝑉𝐴 25
Fault MVA= = = 666.67 𝑀𝑉𝐴
𝑝.𝑢. 𝑋𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 0.0375



Problem 3: Two 11 kV, three phase, 3 MVA alternators having 16% sub-transient reactance
are operating in parallel. The alternators supply power to a transmission line through a 6
MVA transformer of ratio 11/22 kV and having a leakage reactance of 5%. Calculate fault
MVA and fault current for three phase fault on (a) generator side and (b) transmission line
side.

Solution:




Considering one Base MVA


2

, Let Base MVA for the complete system = 6 MVA and
Base kV = 11 kV for alternator side
= 22 kV for transmission line side
6
Now, per unit reactance of alternator = 𝑗0.16 × 3 = 𝑗0.32 𝑝. 𝑢.

The reactance diagram is shown in Figure below.




Calculate Thevenin’s equivalent reactance for the faults.

(a) Fault on alternator side, F1.
𝑗0.32
Equivalent of two reactances of alternators (j0.32 each) in parallel = =
2
𝑗0.16 𝑝. 𝑢.
𝐵𝑎𝑠𝑒 𝑀𝑉𝐴 6
Fault MVA=𝑝.𝑢. = 0.16 = 37.5 𝑀𝑉𝐴
𝑋𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

𝐹𝑎𝑢𝑙𝑡 𝑀𝑉𝐴×1000 37.5×1000
Fault current = = = 1968.23 𝐴∠ − 900
√3×𝐵𝑎𝑠𝑒 𝑘𝑉 √3×11



(b) Fault on transmission line side, F2.
𝑗0.32
Equivalent of two reactances of alternators (j0.32 each) in parallel = =
2
𝑗0.16 𝑝. 𝑢.
Total equivalent reactance upto fault = j0.16+ j0.05 = j0.21 p.u.


𝐵𝑎𝑠𝑒 𝑀𝑉𝐴 6
Fault MVA=𝑝.𝑢. = 0.21 = 28.57 𝑀𝑉𝐴
𝑋𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

𝐹𝑎𝑢𝑙𝑡 𝑀𝑉𝐴×1000 28.57×1000
Fault current = = = 749.76 𝐴∠ − 900 (lag)
√3×𝐵𝑎𝑠𝑒 𝑘𝑉 √3×22

Considering different Base MVA

Let Base MVA for the complete system = 3 MVA and

Base kV = 11 kV for alternator side
= 22 kV for transmission line side


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