IEC 60909 Short-Circuit Calculator (Buses, Transformers, Generators & Motors)

Build a radial network — buses, utility sources, generators, motors, transformers, cables and bus ties — and the single-line diagram and the short-circuit current at every bus appear instantly: Ik″, peak, breaking, thermal, line-to-line and line-to-earth, plus minimum values. Several sources can feed the same bus. Calculated to IEC 60909-0 on a 100 MVA per-unit base; full step-by-step working is below and in the PDF report.

AAJappsEngineering Calculation Report

IEC 60909 Short-Circuit Calculation

IEC 60909-0 (initial, peak, breaking and thermal short-circuit currents)

Generated

For preliminary engineering use only — always verify results against project-specific standards and a qualified engineer's review before use in a real design.

Calculates three-phase and unbalanced short-circuit currents at every bus of a radial (or tied) power system using the IEC 60909-0 equivalent-voltage-source method. Network feeders, generators, induction motors, transformers, cables and bus ties are modelled with their IEC correction factors, and the result is shown on an auto-generated single-line diagram, in a results table for all buses, and as a numbered manual calculation.

Standards followed

IEC 60909-0 (short-circuit currents in three-phase a.c. systems — calculation of currents)IEC 60909-1 / IEC TR 60909-1 (factors for the calculation — background)IEC 60076-5 (transformer ability to withstand short circuit; typical uk values)

What it can do

Unlimited buses at different voltage levels, with network feeders, generators and motors on any bus (several sources per bus are fine).
Transformers with KT impedance correction and vector-group zero-sequence handling (Dyn, YNd, YNyn, Yy), cables/lines with parallel runs, and closeable bus ties.
Results at every bus without asking: Ik″ (3-phase), peak ip, breaking Ib, thermal Ith, Ik″ line-to-line, line-to-earth and line-line-earth, plus minimum-fault values.
Switchgear rating check at each bus and a typical-values reference for every constant (c, R/X, uk, xd″, ILR…).

Example problem

A 500 MVA, 11 kV utility feeds a 1 MVA, 6 % transformer (Dyn). What is Ik″ on the 400 V side?

Voltage factors: cmax = 1.10 at 11 kV (source), 1.05 at 400 V (LV, +6 %).
Source impedance at 11 kV: ZQ = c·Un²/Sk″ = 1.10×11²/500 = 0.2662 Ω → referred to 0.4 kV: 0.2662×(0.4/11)² = 0.000352 Ω.
Transformer ZT = uk·Ur²/Sr = 0.06×0.4²/1 = 0.0096 Ω; KT = 0.95×1.05/(1+0.6×0.06) = 0.9628 → KT·ZT = 0.009243 Ω.
Total Z = 0.000352 + 0.009243 = 0.009595 Ω.
Ik″ = c·Un/(√3·Z) = 1.05×0.4/(1.732×0.009595) = 25.3 kA.

≈ 25.3 kA at the LV bus — the calculator reproduces this exactly and also shows ip, Ib, Ith and earth-fault values.

Scope and simplifications: balanced and unbalanced faults at bus level (no mid-cable faults). The peak current uses method B (R/X at the fault location, with the 1.15 factor when more than one source feeds the fault); steady-state Ik (λ factors) is not computed; the motor breaking-current decay (μ·q) is not credited, which is conservative; Ith uses n = 1 (conservative). Taps, loads and shunt capacitance are ignored, as permitted by IEC 60909-0. Enter manufacturer data wherever available — the typical values are only starting points.

Projects & scenarios

Saved in this browser — export a file to back up or move devices

No saved scenarios yet. Build a network, give it a name and press “Save as new”.

Single-line diagram & short-circuit levels

Auto-generated from the network you enter below

Highest fault level: 0.4 kV LV Board at 39.58 kA (Ik″), peak 100.8 kA.

Short-circuit current at every bus

Bus	Un (kV)	Ik″ 3-ph (kA)	ip (kA)	Ib (kA)	Ith (kA)	Ik″ L-L (kA)	Ik″ L-E (kA)	Ik″ min (kA)	Rating
11 kV Switchboard	11.0	14.20	40.17	13.88	14.44	12.30	14.50	14.04	25 kA ✓
0.4 kV LV Board	0.400	39.58	100.77	39.58	39.99	34.28	40.34	34.41	50 kA ✓
Motor Control Centre	0.400	23.49	48.44	23.49	23.58	20.34	16.63	18.82	36 kA ✓

Maximum values use cmax with all sources and motors; “min” uses cmin, motors off and cables at 80 °C. “L-E” = line-to-earth; n/a means no zero-sequence path to earth.

Build your network

Add or edit anything below — the diagram and tables above update immediately. Use “Apply typical values…” on any item to start from a realistic set of constants.

Study settings

LV voltage tolerance

System frequency

Minimum time delay tmin

Fault duration Tk

Buses

Each bus is a busbar at one nominal voltage. Sources and branches attach to buses.

Name

Rating

Name

Rating

Name

Rating

Sources — utility, generators, motors

Any number per bus. Generators and the utility feed the fault; motors add to the maximum fault current only.

GridUtility incomer

Name

Connected to bus

Sk″

MVA

R/X

Z0/Z1

GeneratorGen G1

Name

Connected to bus

Rated power SrG

MVA

Rated voltage UrG

Subtransient xd″

Rated cos φ

RG/Xd″

Zero-seq x0

Neutral resistance

MotorMotor M1

Name

Connected to bus

Rated power Pr

Rated voltage

cos φ

Efficiency

ILR / IrM

RM/XM

Branches — transformers, cables, bus ties

Each branch joins two buses. For a transformer, the ‘from’ bus is the first-listed winding side (HV side by convention).

TransformerT1 11/0.4 kV

Name

From bus

To bus

Rated power Sr

MVA

Impedance uk

Load loss Pk

Vector group

Z0/Z1

Neutral R (from side)

Neutral R (to side)

Apply impedance correction factor KT

Cable / lineCable to MCC

Name

From bus

To bus

r (positive)

Ω/km

x (positive)

Ω/km

r0 (zero)

Ω/km

x0 (zero)

Ω/km

Length

Parallel runs

Step-by-step manual calculation (IEC 60909-0)

Every number above can be reproduced by hand from these steps. Impedances are shown in per unit on 100 MVA; the bus Thevenin impedance combines all sources, transformers and cables (series/parallel reduction).

A. Element impedances

Per-unit base: Sbase = 100 MVA, Vbase = nominal voltage of each bus (Zbase = Un²/Sbase).
Utility incomer (grid): ZQ = c·Un²/Sk″ = 1.10×11.000²/250.0 = 0.5324 Ω → 0.44000 pu; R/X = 0.100 → ZQ = 0.04378 + j0.43782 pu.
Gen G1 (generator): KG = (Un/UrG)·cmax/(1+xd″·sinφ) = 1.0148; Xd″ = 6.7760 Ω = 5.60000 pu; RG = 0.070·Xd″; ZG = KG(RG + jXd″) = 0.39779 + j5.68266 pu.
Motor M1 (motor): SrM = Pr/(η·cosφ) = 191.4 kVA; ZM = (1/ILR)·UrM²/SrM → 74.64286 pu; R/X = 0.420 → ZM = 28.90414 + j68.81938 pu.
T1 11/0.4 kV (transformer): ZT = uk·Sbase/Sr: ukr = 0.938 %, ux = 5.926 %; KT = 0.95·cmax/(1+0.6·xT) = 0.9632; ZT·KT = 0.56440 + j3.56782 pu; zero-sequence: shunt to earth on the 'to' side (Delta on 'from' side).
Cable to MCC (cable): Z = (r+jx)·L/n = (0.0754+j0.0780)×0.050/1 = 0.00377 + j0.00390 Ω → 2.35625 + j2.43750 pu.

B. Fault calculation at each bus

11 kV Switchboard (11.0 kV)

Ik″ = 14.20 kA

1. Voltage factor: Un = 11.000 kV → cmax = 1.10, cmin = 1.00 (IEC 60909-0 Table 1).
2. Thevenin positive-sequence impedance at the bus (all sources, transformers and cables combined by network reduction): Z1 = 0.04019 + j0.40452 pu = 0.04863 + j0.48947 Ω (|Z1| = 0.49188 Ω), R/X = 0.099.
3. Initial symmetrical current: Ik″ = c·Un/(√3·|Z1|) = 1.10×11.000/(1.7321×0.49188) = 14.20 kA.
4. Peak factor: κ = 1.02 + 0.98·e^(−3R/X) = 1.02 + 0.98·e^(−3×0.099) = 1.747; several sources feed the fault, so method B applies: κ(b) = 1.15·κ = 2.000 (limit 2.0); ip = κ·√2·Ik″ = 2.000×1.4142×14.20 = 40.17 kA.
5. Source contributions to Ik″ (current division from the impedance matrix): Utility incomer 13.12 kA (μ = 1.00); Gen G1 1.01 kA (μ = 0.68); Motor M1 0.07 kA (μ = 1.00).
6. Breaking current: Ib = Σ μ·Ik″(source) = 13.88 kA (tmin = 0.10 s; motor decay not credited — conservative).
7. Thermal equivalent: m = (e^(4·f·Tk·ln(κ−1)) − 1)/(2·f·Tk·ln(κ−1)) = 0.0343 for f = 50 Hz, Tk = 1.00 s; n = 1 (conservative); Ith = Ik″·√(m+n) = 14.20×√(0.0343+1) = 14.44 kA.
8. Line-to-line: Ik2″ = √3/2 · Ik″ (Z2 = Z1) = 12.30 kA.
9. Zero-sequence impedance at the bus: Z0 = 0.03995 + j0.37943 pu. Line-to-earth: Ik1″ = √3·c·Un/|Z1+Z2+Z0| = 3·c/|2Z1+Z0| × Ibase = 14.50 kA.
10. Line-line-earth: largest phase current = 14.38 kA, earth current 3I0 = 14.81 kA.
11. Minimum case (cmin = 1.00, motors neglected, cables at 80 °C): Ik″min = 14.04 kA (3-phase), 12.16 kA (line-to-line), 14.33 kA (line-to-earth).
Rating check: switchgear rating 25.0 kA ≥ maximum Ik″ 14.50 kA → ADEQUATE.

Contributions to Ik″: Utility incomer 13.12 kA · Gen G1 1.01 kA · Motor M1 0.07 kA

0.4 kV LV Board (0.400 kV)

Ik″ = 39.58 kA

1. Voltage factor: Un = 0.400 kV → cmax = 1.05, cmin = 0.95 (IEC 60909-0 Table 1).
2. Thevenin positive-sequence impedance at the bus (all sources, transformers and cables combined by network reduction): Z1 = 0.62371 + j3.77751 pu = 0.00100 + j0.00604 Ω (|Z1| = 0.00613 Ω), R/X = 0.165.
3. Initial symmetrical current: Ik″ = c·Un/(√3·|Z1|) = 1.05×0.400/(1.7321×0.00613) = 39.58 kA.
4. Peak factor: κ = 1.02 + 0.98·e^(−3R/X) = 1.02 + 0.98·e^(−3×0.165) = 1.617; several sources feed the fault, so method B applies: κ(b) = 1.15·κ = 1.800 (limit 1.8); ip = κ·√2·Ik″ = 1.800×1.4142×39.58 = 100.77 kA.
5. Source contributions to Ik″ (current division from the impedance matrix): Utility incomer 35.00 kA (μ = 1.00); Gen G1 2.70 kA (μ = 1.00); Motor M1 1.95 kA (μ = 1.00).
6. Breaking current: Ib = Σ μ·Ik″(source) = 39.58 kA (tmin = 0.10 s; motor decay not credited — conservative).
7. Thermal equivalent: m = (e^(4·f·Tk·ln(κ−1)) − 1)/(2·f·Tk·ln(κ−1)) = 0.0207 for f = 50 Hz, Tk = 1.00 s; n = 1 (conservative); Ith = Ik″·√(m+n) = 39.58×√(0.0207+1) = 39.99 kA.
8. Line-to-line: Ik2″ = √3/2 · Ik″ (Z2 = Z1) = 34.28 kA.
9. Zero-sequence impedance at the bus: Z0 = 0.56440 + j3.56782 pu. Line-to-earth: Ik1″ = √3·c·Un/|Z1+Z2+Z0| = 3·c/|2Z1+Z0| × Ibase = 40.34 kA.
10. Line-line-earth: largest phase current = 40.06 kA, earth current 3I0 = 41.13 kA.
11. Minimum case (cmin = 0.95, motors neglected, cables at 80 °C): Ik″min = 34.41 kA (3-phase), 29.80 kA (line-to-line), 35.51 kA (line-to-earth).
Rating check: switchgear rating 50.0 kA ≥ maximum Ik″ 40.34 kA → ADEQUATE.

Contributions to Ik″: Utility incomer 35.00 kA · Gen G1 2.70 kA · Motor M1 1.95 kA

Motor Control Centre (0.400 kV)

Ik″ = 23.49 kA

1. Voltage factor: Un = 0.400 kV → cmax = 1.05, cmin = 0.95 (IEC 60909-0 Table 1).
2. Thevenin positive-sequence impedance at the bus (all sources, transformers and cables combined by network reduction): Z1 = 2.68698 + j5.86604 pu = 0.00430 + j0.00939 Ω (|Z1| = 0.01032 Ω), R/X = 0.458.
3. Initial symmetrical current: Ik″ = c·Un/(√3·|Z1|) = 1.05×0.400/(1.7321×0.01032) = 23.49 kA.
4. Peak factor: κ = 1.02 + 0.98·e^(−3R/X) = 1.02 + 0.98·e^(−3×0.458) = 1.268; several sources feed the fault, so method B applies: κ(b) = 1.15·κ = 1.458 (limit 1.8); ip = κ·√2·Ik″ = 1.458×1.4142×23.49 = 48.44 kA.
5. Source contributions to Ik″ (current division from the impedance matrix): Utility incomer 19.92 kA (μ = 1.00); Gen G1 1.54 kA (μ = 1.00); Motor M1 2.03 kA (μ = 1.00).
6. Breaking current: Ib = Σ μ·Ik″(source) = 23.49 kA (tmin = 0.10 s; motor decay not credited — conservative).
7. Thermal equivalent: m = (e^(4·f·Tk·ln(κ−1)) − 1)/(2·f·Tk·ln(κ−1)) = 0.0076 for f = 50 Hz, Tk = 1.00 s; n = 1 (conservative); Ith = Ik″·√(m+n) = 23.49×√(0.0076+1) = 23.58 kA.
8. Line-to-line: Ik2″ = √3/2 · Ik″ (Z2 = Z1) = 20.34 kA.
9. Zero-sequence impedance at the bus: Z0 = 9.98940 + j10.88032 pu. Line-to-earth: Ik1″ = √3·c·Un/|Z1+Z2+Z0| = 3·c/|2Z1+Z0| × Ibase = 16.63 kA.
10. Line-line-earth: largest phase current = 22.81 kA, earth current 3I0 = 12.72 kA.
11. Minimum case (cmin = 0.95, motors neglected, cables at 80 °C): Ik″min = 18.82 kA (3-phase), 16.30 kA (line-to-line), 13.48 kA (line-to-earth).
Rating check: switchgear rating 36.0 kA ≥ maximum Ik″ 23.49 kA → ADEQUATE.

Contributions to Ik″: Utility incomer 19.92 kA · Gen G1 1.54 kA · Motor M1 2.03 kA

Typical values of the constants

Voltage factor c (IEC 60909-0 Table 1)

LV 100–1000 V, +6 % tolerance	cmax 1.05 / cmin 0.95	Default for 230/400 V public networks.
LV 100–1000 V, +10 % tolerance	cmax 1.10 / cmin 0.95	Use where the supply voltage may reach +10 %.
MV and HV > 1 kV	cmax 1.10 / cmin 1.00	Applies to 3.3, 6.6, 11, 33, 132 kV etc.

Network feeder

R/X of source	0.1 (HV), 0.3 (LV)	IEC: XQ = 0.995 ZQ when R = 0.1 X. Use the utility's X/R if known.
Typical Sk″ at 11 kV	150 – 500 MVA	8 – 26 kA. Ask the utility for the real value.
Typical Sk″ at 33 kV	500 – 1500 MVA	Up to 25 – 31.5 kA switchgear.
Z0/Z1 of source	1.0 (typical)	Depends on utility earthing; 0.5–3 possible.

Transformer

uk, up to 630 kVA	4 %	IEC 60076-5 / EN 50464 distribution range.
uk, 800 – 3150 kVA	6 %
uk, 5 – 40 MVA	7.5 – 14 %	Check the nameplate; rises with rating and voltage.
KT = 0.95·cmax/(1+0.6·xT)	≈ 0.93 – 1.0	Applied to Z1 and Z0; not applied to power-station unit transformers.
Z0/Z1	0.85 – 1.0	Dyn ≈ 1.0; 3-limb YNyn 0.8–1.0.

Generator

RG/Xd″ when UrG ≤ 1 kV	0.15	IEC 60909-0 cl. 3.6.1.
RG/Xd″ when UrG > 1 kV, SrG < 100 MVA	0.07
RG/Xd″ when UrG > 1 kV, SrG ≥ 100 MVA	0.05
xd″	12 – 20 %	Salient-pole 12–25 %; turbo 10–20 %.
x0	≈ 0.4 – 0.6 xd″	Use the datasheet where available.

Induction motor

RM/XM, LV motor groups	0.42	IEC 60909-0 cl. 3.8 (includes connection cables).
RM/XM, MV motor < 1 MW per pole pair	0.15
RM/XM, MV motor ≥ 1 MW per pole pair	0.10
ILR/IrM	5 – 7	Locked-rotor to rated current.
Efficiency × cos φ	0.80 – 0.90	Used to get SrM = Pr/(η·cos φ).

Cable

Cu 3C XLPE 240 mm² LV	r 0.0754, x 0.078 Ω/km	r at 20 °C; code raises r to 80 °C (×1.24) for the minimum case.
Z0 of 3-core cable	r0 ≈ 4 r, x0 ≈ 3 x	Use manufacturer data for single-core/screen returns.

IEC 60909 Short-Circuit Calculator (Buses, Transformers, Generators & Motors)

IEC 60909 Short-Circuit Calculation

Standards followed

What it can do

Example problem

Projects & scenarios

Single-line diagram & short-circuit levels

Short-circuit current at every bus

Build your network

Study settings

Buses

Sources — utility, generators, motors

Branches — transformers, cables, bus ties

Step-by-step manual calculation (IEC 60909-0)

A. Element impedances

B. Fault calculation at each bus

11 kV Switchboard (11.0 kV)

0.4 kV LV Board (0.400 kV)

Motor Control Centre (0.400 kV)

Typical values of the constantsi

Voltage factor c (IEC 60909-0 Table 1)

Network feeder

Transformer

Generator

Induction motor

Cable

Typical values of the constants