Short-circuit current is the number that sizes breakers, sets relay pickups, fixes cable bracing, and decides whether a piece of equipment survives the worst day of its life. For a single fault location, the calculation reduces to something small and exact: take the positive, negative, and zero sequence Thévenin impedances seen at the fault point, connect the three sequence networks in the arrangement that the fault type dictates, and solve one series circuit. This calculator does that for the four shunt fault types and shows you the network interconnection it used.
Enter the prefault voltage and the three sequence impedances in per-unit, add a fault impedance if the fault is not bolted, and pick a fault type. The tool reports the sequence currents, all three phase currents and phase voltages at the fault point, the ground current, and the ratios that decide how the system is grounded. Give it a base MVA and base kV and it converts the per-unit answer to kA.
Fault Current Calculator
Three-phase, single-line-to-ground, line-to-line and double-line-to-ground faults from the sequence impedances
| Phase | |I| pu | ∠I deg | |V| pu | ∠V deg |
|---|---|---|---|---|
| A | — | — | — | — |
| B | — | — | — | — |
| C | — | — | — | — |
How sequence networks give the fault current
Fortescue’s theorem lets any three-phase set be written as the sum of a positive-sequence set, a negative-sequence set, and a zero-sequence set. In a network that is balanced everywhere except at the fault point, those three sequence networks are decoupled: each has its own Thévenin impedance seen from the fault, and only the boundary conditions at the fault itself tie them together. That is the whole trick. Every shunt fault type reduces to one specific way of wiring the three networks, and the prefault voltage \(V_f\) drives only the positive-sequence network.
Three-phase fault
A balanced fault produces balanced currents, so the negative and zero sequence currents are zero and only the positive-sequence network carries current:
\[ I_{a1}=\frac{V_f}{Z_1+Z_f},\qquad I_{a2}=I_{a0}=0,\qquad I_a=I_{a1} \]
Single-line-to-ground fault
With phase A faulted to earth through \(Z_f\), the boundary conditions \(I_b=I_c=0\) and \(V_a=Z_fI_a\) force the three sequence currents to be equal, which is the series connection of all three networks with \(3Z_f\) added:
\[ I_{a0}=I_{a1}=I_{a2}=\frac{V_f}{Z_1+Z_2+Z_0+3Z_f},\qquad I_a=3I_{a1}=\frac{3V_f}{Z_1+Z_2+Z_0+3Z_f} \]
The factor of three on \(Z_f\) is not a fudge. A single per-phase network carries \(I_{a1}\), but the physical fault carries \(I_a=3I_{a1}\), so representing the real drop \(Z_fI_a\) inside the sequence circuit requires an impedance of \(3Z_f\).
Line-to-line fault
For a fault between phases B and C through \(Z_f\), no current returns through earth, so \(I_{a0}=0\) and the positive and negative networks sit in series:
\[ I_{a1}=-I_{a2}=\frac{V_f}{Z_1+Z_2+Z_f},\qquad I_b=-I_c=-j\sqrt{3}\,I_{a1} \]
When \(Z_1=Z_2\) and the fault is bolted, this gives \(|I_b|=(\sqrt{3}/2)|I_{3\phi}|\approx 0.866\,|I_{3\phi}|\), the familiar result that a line-to-line fault draws about 87% of the three-phase current.
Double-line-to-ground fault
Phases B and C faulted to each other through \(Z_f\) and then to earth through \(Z_g\) puts the negative and zero branches in parallel behind the positive branch:
\[ I_{a1}=\frac{V_f}{Z_1+Z_f+\dfrac{(Z_2+Z_f)(Z_0+Z_f+3Z_g)}{Z_2+Z_0+2Z_f+3Z_g}} \]
and the current divider splits \(I_{a1}\) between the two parallel branches:
\[ I_{a2}=-I_{a1}\,\frac{Z_0+Z_f+3Z_g}{Z_2+Z_0+2Z_f+3Z_g},\qquad I_{a0}=-I_{a1}\,\frac{Z_2+Z_f}{Z_2+Z_0+2Z_f+3Z_g} \]
Back to phase quantities
Whatever the fault type, once the three sequence currents are known the phase currents follow from the inverse Fortescue transform with \(a=1\angle 120^\circ\):
\[ I_a=I_{a0}+I_{a1}+I_{a2},\qquad I_b=I_{a0}+a^{2}I_{a1}+aI_{a2},\qquad I_c=I_{a0}+aI_{a1}+a^{2}I_{a2} \]
The sequence voltages at the fault point come from the same three networks, and the same matrix converts them to phase voltages:
\[ V_1=V_f-Z_1I_{a1},\qquad V_2=-Z_2I_{a2},\qquad V_0=-Z_0I_{a0} \]
Those phase voltages are worth reading. During a bolted single-line-to-ground fault the faulted phase collapses to zero while the two healthy phases can rise well above their prefault value on a poorly grounded system, and that rise is what sets surge arrester ratings.
Worked example
Press Load example. It sets \(V_f=1.0\) pu, a purely reactive \(Z_1=Z_2=j0.10\) pu, \(Z_0=j0.05\) pu, a bolted fault (\(Z_f=Z_g=0\)), and a base of 100 MVA at 138 kV. The four cases can be checked by hand:
- Three-phase: \(1.0/0.10=10.0\) pu.
- Single-line-to-ground: \(3\times 1.0/(0.10+0.10+0.05)=12.0\) pu, which is 1.20 times the three-phase current.
- Line-to-line: \(\sqrt{3}\times 1.0/0.20=8.660\) pu.
- Double-line-to-ground: the parallel combination gives \(I_{a1}=1.0/(0.10+0.10\|0.05)=7.5\) pu, a ground current of \(3I_{a0}=15.0\) pu, and a faulted-phase current of 11.456 pu.
The base current is \(100/(\sqrt{3}\times 138)=0.4184\) kA, so the three-phase fault is 4.18 kA and the ground fault is 5.02 kA. This is the classic situation near a solidly grounded delta-wye transformer: because \(Z_0\) is smaller than \(Z_1\), the ground fault, not the balanced fault, sets the interrupting duty. Now raise \(X_0\) past 0.10 and watch the single-line-to-ground bar in the comparison chart drop below the three-phase bar.
Reading the grounding classification
The chip beside the headline number answers the question that matters in practice: does the ground fault exceed the balanced fault, and is the system effectively grounded? The effectively grounded test used here is the conventional one from IEEE Std C62.92.1, which asks for \(X_0/X_1\le 3\) and \(R_0/X_1\le 1\) at the point of interest. Treat it as a screening indicator rather than a verdict. The criterion was written to bound the temporary overvoltage on unfaulted phases so that arresters can be rated, it assumes reasonably linear network behaviour, and a real grounding study also weighs the transformer winding connections, any neutral grounding impedance, and the operating configurations that can leave part of the system ungrounded.
- Three-phase governs: X0/X1 at most 3, R0/X1 at most 1, and the single-line-to-ground current is below the three-phase current. Effectively grounded by the usual test, and the balanced fault sets equipment duty.
- SLG exceeds three-phase: still effectively grounded, but the zero-sequence impedance is small enough that the ground fault is the larger current. Size the interrupting duty on the ground fault, and expect high ground relay pickup currents.
- Not effectively grounded: the zero-sequence path fails one or both ratio limits, so ground-fault current is limited and neutral displacement during a ground fault is larger. Arrester selection, ground-fault detection sensitivity, and the ability to run unfaulted for any length of time all need explicit attention.
Working in per-unit?
The free Per-Unit System Cheat Sheet covers base selection, change of base, and the base-current formula this calculator uses to convert per-unit fault current to kA, on one printable page.
Frequently asked questions
Is the three-phase fault always the worst case?
No. The single-line-to-ground current exceeds the three-phase current whenever the zero-sequence impedance is smaller than the positive-sequence impedance, which is common at the terminals of a solidly grounded delta-wye transformer and at generator terminals with a low zero-sequence reactance. The comparison chart in the calculator makes the crossover obvious: change \(X_0\) and watch the bars swap order.
Which machine reactance should I enter for Z1?
It depends on the instant you care about. Use the subtransient reactance for the first cycle, which is what breaker closing and latching duty and instantaneous relay elements see. Use the transient reactance for currents a few cycles later, which is closer to interrupting duty on a slower breaker. The calculator has one impedance per sequence, so it gives you one snapshot of an otherwise decaying current, and it does not model AC decrement.
Where does DC offset go?
It is not here. This is a symmetrical RMS calculation. The asymmetrical peak depends on the X/R ratio at the fault point and on the point on the wave at which the fault occurs, and standards handle it with multiplying factors applied to the symmetrical current. If you need momentary or close-and-latch duty, take the symmetrical value from this tool and apply the factor from the applicable standard.
Why is Z2 usually close to Z1?
Static plant such as lines, cables, and transformers presents the same impedance to a positive-sequence set and a negative-sequence set, because reversing the phase rotation does not change a passive symmetrical element. Rotating machines are the exception: the negative-sequence field rotates backwards relative to the rotor, so the negative-sequence reactance differs from the positive-sequence value and is usually close to the subtransient reactance. Zero sequence is different again, since it depends on the return path through earth and on transformer winding connections, which is why \(Z_0\) can be either much smaller or much larger than \(Z_1\).
What does a nonzero fault impedance change?
It lowers the current and it raises the voltage left at the fault point, which is exactly why high-impedance ground faults are hard to detect with plain overcurrent elements. Note the asymmetry in the equations: a single-line-to-ground fault sees \(3Z_f\) in the sequence circuit while a three-phase fault sees only \(Z_f\), so a given arc impedance suppresses the ground fault more than the balanced fault.
Related reading
- Symmetrical Components Calculator: the companion tool. Decompose any unbalanced phase set into the sequence quantities this calculator consumes, and see the phasor signature of each fault type.
- SCR Calculator: convert a fault level in MVA into the Thévenin impedance you need for \(Z_1\), and screen grid strength at an interconnection point.
- Two-Bus Load Flow & Voltage Stability: the same Thévenin picture used for steady-state voltage rather than fault current.
- All EE Power School calculators: the full set of browser-based power system tools.
References
- J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, Cengage, chapters on symmetrical components and unsymmetrical faults.
- P. M. Anderson, Analysis of Faulted Power Systems, IEEE Press, 1995.
- J. L. Blackburn and T. J. Domin, Protective Relaying: Principles and Applications, CRC Press.
- IEEE Std C62.92.1, IEEE Guide for the Application of Neutral Grounding in Electrical Utility Systems, Part I: Introduction, for the effectively grounded criterion.
- IEEE Std 141 (Red Book) and IEEE Std 551 (Violet Book) for industrial short-circuit calculation practice.