Electronics Guide

Shunt and Series Faults

A shunt fault connects one or more phases to each other or to ground. Four types are calculated routinely:

• Three-phase fault: all three phases shorted together, with or without ground. It is the only balanced type.
• Single line-to-ground fault: one phase to ground, as when lightning flashes over one insulator string.
• Line-to-line fault: two phases shorted together without ground.
• Double line-to-ground fault: two phases shorted together and to ground.

A fault is bolted when the connection has negligible impedance. Real faults often include an arc, a tower-footing resistance, or a tree, which the calculation represents as a fault impedance Z_f. A series fault, by contrast, opens one or two phases, as a broken jumper or a single blown fuse does. Series faults, solved with differently connected sequence networks, are not treated here.

What the Results Are Used For

Single line-to-ground faults are the most common type on overhead lines. Three-phase faults are rare but often, though not always, produce the largest current. A study therefore calculates every type at every bus of interest, for the strongest and the weakest credible source conditions. Maximum currents set the duties of breakers, fuses, buses, and cables; minimum currents show whether protection can still detect the fault; branch currents set relay coordination; and bus voltages show which loads and inverter-based resources must ride through the sag.

The Bus Impedance Matrix

Programs build the bus admittance matrix Y_bus from the branch data and obtain the needed elements of its inverse, the bus impedance matrix Z_bus, by sparse factorization. The diagonal element Z_kk is the Thévenin impedance at bus k. An off-diagonal element Z_ik gives the voltage change at bus i when a current I_f is drawn from bus k:

ΔV_i = −Z_ik I_f

One factorization therefore yields the current for a fault at any bus and the voltage at every other bus during that fault. For unbalanced faults, a program builds one such matrix for each sequence network. With V_f = 1.0 per unit, the three-phase short-circuit level of a bus is S_base / |Z_th|, or 400 MVA for a Thévenin impedance of 0.25 per unit on a 100 MVA base.

Three Reactances and Their Time Constants

A fault study represents that decay with three direct-axis reactances:

• Subtransient reactance, X″_d, governs the first few cycles, while the damper windings oppose the flux change. Its extra current decays with the short-circuit subtransient time constant T″_d.
• Transient reactance, X′_d, governs the following period, while the field winding still opposes the change. Its extra current decays with the short-circuit transient time constant T′_d.
• Synchronous reactance, X_d, governs the sustained current once both effects have died away, with the excitation held constant.

For a three-phase short circuit at the terminals of an unloaded generator, the RMS symmetrical current is

I_ac(t) = E [(1/X″_d − 1/X′_d) exp(−t/T″_d) + (1/X′_d − 1/X_d) exp(−t/T′_d) + 1/X_d]

where E is the prefault RMS voltage. The current starts at E/X″_d and settles at E/X_d. A DC component, treated in the next section, adds to it and decays with the armature time constant T_a. Machine theory derives these reactances and time constants; a fault study needs only their values.

With assumed values of X″_d = 0.15, X′_d = 0.25, and X_d = 1.6 per unit on the machine rating, T″_d = 0.035 s, T′_d = 0.9 s, and E = 1.0 per unit, the symmetrical current starts at 6.67 per unit, falls to 3.80 per unit at 0.1 s and 1.74 per unit at 1 s, and approaches 0.625 per unit. For these values the sustained current at constant excitation is below rated current, one reason excitation systems force the field voltage up during faults.

Which Reactance for Which Question

The time of interest selects the reactance. The first-cycle peak and a breaker's closing duty depend on X″_d. Far from generators, the network impedance dominates and the symmetrical current barely decays. Near a generator, the current at contact parting is measurably lower, and time-delayed relays see a current governed by X′_d, X_d, and the excitation system.

Motors also feed faults. A synchronous motor behaves like a generator. An induction motor contributes current for only a few cycles, because its rotor flux has no separate excitation to sustain it once the supply voltage collapses. Studies therefore include induction motors in first-cycle calculations and reduce or drop them for later times.

The X/R Ratio Sets the Decay

Expressed in cycles, the time constant is (X/R)/(2π), so the ratio X/R alone sets how many cycles the offset lasts. At 60 Hz, X/R = 17 gives τ = 45 ms. The ratio is high near large generators and transformers, whose windings have little resistance compared with their leakage reactance, and lower on distribution circuits and cables. The worst-case peak occurs about half a cycle after the fault begins, which gives the approximation

i_p ≈ √2 I (1 + e^−πR/X)

where I is the RMS symmetrical current. IEC 60909-0 writes the peak as i_p = κ√2 I″_k, where I″_k is the initial symmetrical short-circuit current. For a circuit with a single R/X ratio, the standard's 2016 edition gives

κ = 1.02 + 0.98 e^−3R/X

The table compares the exact peak from the equation for i(t), maximized over the inception angle, with both approximations. The IEC factor stays within about 0.5 percent of the exact value across the range.

Asymmetrical RMS Current

Breakers and fuses respond to the RMS value of the total current. With full offset, the DC component is √2 I e^−t/τ, and treating it as constant over one cycle gives an RMS value of approximately

I_asym(t) = I √(1 + 2e^−2t/τ)

The ratio I_asym/I is the asymmetry factor. It starts near √3, about 1.73, and falls toward 1 as the offset decays. The same offset drives current transformers toward saturation, a measurement problem discussed in Power System Protection.

The Operator a

The method uses the operator

a = e^j120° = −1/2 + j√3/2 ≈ −0.5 + j0.866

Multiplying a phasor by a advances it by 120°. It follows that a² = e^−j120°, that a³ = 1, and that

1 + a + a² = 0

Synthesis and Analysis Equations

Take phase a as the reference and call the sequence components of its voltage V₀, V₁, and V₂. In abc sequence, the positive-sequence component of phase b lags that of phase a by 120°, so it is a²V₁; its negative-sequence component leads by 120°, so it is aV₂. The phase voltages are therefore

V_a = V₀ + V₁ + V₂

V_b = V₀ + a²V₁ + aV₂

V_c = V₀ + aV₁ + a²V₂

Solving these equations with 1 + a + a² = 0 gives the analysis equations, in the one-third convention used throughout this article:

V₀ = (V_a + V_b + V_c) / 3

V₁ = (V_a + aV_b + a²V_c) / 3

V₂ = (V_a + a²V_b + aV_c) / 3

The same equations hold for currents, with two consequences used constantly. The neutral or ground current is I_a + I_b + I_c = 3I₀. And line-to-line voltages contain no zero-sequence component, because V₀ cancels in V_a − V_b, so a delta winding or a line-to-line measurement cannot see it.

A Decomposition Example

A four-wire wye circuit has lost phase c. The remaining phases carry I_a = 100 A and I_b = 100 e^−j120° A, and I_c = 0. The analysis equations give I₀ = 33.3 e^−j60° A, I₁ = 66.7 A, and I₂ = 33.3 e^j60° A. As a check, the neutral carries 3I₀ = 100 e^−j60° A, which equals I_a + I_b, and the three components add back to I_a = 100 A.

Generators and Motors

• Positive sequence: the internal voltage E behind X″_d, or behind X′_d or X_d for later times.
• Negative sequence: a reactance X₂ with no source. The backward-rotating field sweeps the rotor at nearly twice synchronous speed, so the rotor circuits respond much as they do in the subtransient period, and X₂ is close to the subtransient reactance. For a salient-pole machine it is approximately the average of the direct-axis and quadrature-axis subtransient reactances.
• Zero sequence: a reactance X₀, usually smaller than X″_d, because in-phase currents in the three windings produce almost no air-gap field. It reaches the reference bus only through a grounded neutral, in series with 3Z_n if the neutral has a grounding impedance. An ungrounded or delta-connected machine is an open circuit in this network.

An induction motor appears in the positive-sequence network through its locked-rotor reactance for the first few cycles. In the negative-sequence network it presents a low impedance, close to its locked-rotor impedance, because its slip relative to the backward field is nearly 2.

Transformers

In the positive- and negative-sequence networks, a transformer is its series leakage impedance Z_T, with the magnetizing branch neglected. Wye–delta and delta–wye units also shift the phase by 30°, in opposite directions for the two sequences. This article follows IEEE C57.12.00, under which positive-sequence voltages and currents on the high-voltage side lead those on the low-voltage side by 30°. Crossing from the high-voltage side to the low-voltage side then multiplies positive-sequence voltages and currents by e^−j30° and negative-sequence quantities by e^j30°, leaving per-unit magnitudes unchanged. Units built to other conventions, such as the IEC vector group Dyn11, shift the other way. The shift does not change the current at the fault, but it changes how currents divide among the phases on the far side of the transformer.

The zero-sequence network depends on the winding connections and grounding. Zero-sequence current can flow in a wye winding only if its neutral is grounded, and it can circulate inside a delta but cannot leave it through the lines. Those two rules give the equivalent circuits in the table, in which Z_n is a neutral grounding impedance referred to the per-unit base of its winding.

In a three-limb core-form transformer, in-phase fluxes cannot return through the core and must close through the air and the tank, so its zero-sequence magnetizing impedance is much lower than that of a bank of single-phase units and may need to be modeled. A delta tertiary gives a grounded wye–wye transformer a zero-sequence path to the reference bus, and a zigzag or grounded wye–delta grounding transformer provides one deliberately on an otherwise ungrounded system. Winding connections and vector groups belong to the study of power transformers.

Overhead Lines and Cables

A static line has Z₁ = Z₂. Its zero-sequence impedance is considerably larger, with a larger resistive part, because zero-sequence current returns through the earth and any shield wires, a larger, more resistive loop than the phase conductors form among themselves. Parallel circuits on shared towers are also coupled in the zero-sequence network through a mutual impedance that the positive-sequence network can ignore; neglecting it causes errors in ground-fault currents. A cable's zero-sequence impedance depends on how its sheaths or screens are bonded and grounded. Line capacitance is neglected in most fault calculations, except on ungrounded systems, where it sets the ground-fault current.

Three-Phase Fault

A balanced fault, with an impedance Z_f in each phase, draws only positive-sequence current: I₁ = E / (Z₁ + Z_f), and I₂ = I₀ = 0. Only the positive-sequence network is used, whether or not the fault involves ground.

Single Line-to-Ground Fault

Phase a is connected to ground through Z_f. The boundary conditions are I_b = I_c = 0 and V_a = Z_fI_a. With I_b = I_c = 0, the analysis equations give I₀ = I₁ = I₂ = I_a/3. Substituting the network equations into V_a = V₀ + V₁ + V₂ = 3Z_fI₁ gives E − (Z₁ + Z₂ + Z₀)I₁ = 3Z_fI₁, so

I₀ = I₁ = I₂ = E / (Z₁ + Z₂ + Z₀ + 3Z_f)

I_a = 3E / (Z₁ + Z₂ + Z₀ + 3Z_f)

The three networks are connected in series, with 3Z_f in the loop. The fault impedance is multiplied by 3 because each sequence current is one third of the phase current.

Line-to-Line Fault

Phases b and c are connected through Z_f. The boundary conditions are I_a = 0, I_c = −I_b, and V_b − V_c = Z_fI_b. The analysis equations give I₀ = 0 and I₂ = −I₁, and the synthesis equations then give I_b = (a² − a)I₁ = −j√3 I₁. Because V_b − V_c = (a² − a)(V₁ − V₂), the voltage condition reduces to V₁ − V₂ = Z_fI₁, and substituting the network equations gives

I₁ = −I₂ = E / (Z₁ + Z₂ + Z_f)

I_b = −I_c = −j√3 E / (Z₁ + Z₂ + Z_f)

The positive- and negative-sequence networks are connected in parallel, with Z_f between them. The zero-sequence network is not used, because no current reaches ground. For a bolted fault with Z₁ = Z₂, the fault current is √3/2, or 0.866, times the three-phase fault current.

Double Line-to-Ground Fault

Phases b and c are connected to each other and to ground through Z_f. The boundary conditions are I_a = 0 and V_b = V_c = Z_f(I_b + I_c). The first gives I₀ + I₁ + I₂ = 0 and I_b + I_c = 3I₀. The condition V_b = V_c gives V₁ = V₂, and then V_b = V₀ + (a² + a)V₁ = V₀ − V₁ = 3Z_fI₀. These equations describe all three networks connected in parallel, with 3Z_f in series with the zero-sequence network:

I₁ = E / [Z₁ + Z₂(Z₀ + 3Z_f) / (Z₂ + Z₀ + 3Z_f)]

I₂ = −I₁(Z₀ + 3Z_f) / (Z₂ + Z₀ + 3Z_f)

I₀ = −I₁Z₂ / (Z₂ + Z₀ + 3Z_f)

The current into ground is 3I₀, and the phase currents follow from the synthesis equations.

Summary of Connections

Which Fault Gives the Largest Current?

A bolted single line-to-ground fault draws more current than a bolted three-phase fault whenever |Z₁ + Z₂ + Z₀| is less than 3|Z₁|. With Z₁ = Z₂ and all impedances at the same angle, the ratio of the two currents is 3 / (2 + Z₀/Z₁), so the ground fault is the larger whenever Z₀ is smaller than Z₁, as it is close to grounded-wye transformer windings and solidly grounded generators. The ground current 3I₀ of a double line-to-ground fault is then larger still. Farther out on an overhead line, where the line's large zero-sequence impedance dominates, the three-phase fault is larger.

Comparing Fault Types

The same networks give every fault type at both 138 kV buses. At B2, the transformer's Z₀ = j0.10 is smaller than Z₁ = j0.25, and the ground fault exceeds the three-phase fault by 21 percent. At B3, the line's zero-sequence reactance reverses the order.

The Effect of Fault Resistance

Suppose the B3 fault has a resistance of 10 Ω, or 10 / 190.44 = 0.0525 per unit. The loop impedance becomes 3 × 0.0525 + j1.62 = 0.1575 + j1.62 per unit, and I_a falls only from 775 A to 771 A, because the resistance adds in quadrature with a much larger reactance. A 50 Ω fault, by the same calculation, draws 697 A.

Effectively Grounded Systems

The coefficient of grounding measures that rise. It is the highest RMS power-frequency line-to-ground voltage on an unfaulted phase during a ground fault on one or more phases, divided by the line-to-line voltage at the same place without the fault. IEEE C62.92.1, the IEEE guide to neutral grounding in electrical utility systems, takes 80 percent as the limit for an effectively grounded system. It adds that effective grounding is obtained approximately when, for all system conditions, X₀/X₁ is positive and no greater than 3 and R₀/X₁ is positive and less than 1. Effective grounding lowers temporary overvoltages, easing the duty on insulation and surge arresters.

The table gives the coefficient calculated from the sequence networks for bolted single and double line-to-ground faults, with R₁ = 0 and Z₂ = Z₁. For X₀/X₁ from 1 to 3 and R₀/X₁ up to 1, it stays below 78 percent, but the ratio test is only an approximation: X₀/X₁ = 0.1 with R₀/X₁ = 0.9 passes it yet gives 81.6 percent.

Resistance Grounding

A resistor between the neutral and ground adds 3R_n to the zero-sequence impedance. When 3R_n dominates the loop, the ground-fault current is close to 3E/3R_n = E/R_n. On a 13.8 kV system, where E = 7,967 V, an 8 Ω resistor limits a ground fault to about 996 A wherever it occurs. The price is a larger k, and therefore a larger rise on the unfaulted phases.

Ungrounded and Resonant-Grounded Systems

An ungrounded system has no intended ground path. Its zero-sequence impedance is the capacitive reactance of each phase to ground, 1/(ωC₀), which is far larger than Z₁ or Z₂, so a ground fault draws

I_f ≈ 3E / Z₀ = j3ωC₀E

This is three times the normal charging current of one phase, leading the faulted phase's prefault voltage by 90°. With C₀ = 1.0 μF per phase on a 13.8 kV, 60 Hz system, each phase normally draws 3.00 A and a ground fault draws 9.01 A. Meanwhile the neutral shifts to −E, so the unfaulted phases rise to √3 E while the line-to-line voltages stay balanced. Line-to-line loads keep running, but a second ground fault on another phase becomes a line-to-line fault through ground.

A resonant-grounded system connects a reactor L_n from neutral to ground. In the zero-sequence network, its 3jωL_n appears in parallel with the phase capacitance, and choosing L_n = 1/(3ω²C₀) makes the pair resonate; for the system above, L_n = 2.345 H. Only a small residual current, set by losses and mistuning, then flows in the fault, which lets many arcing faults on overhead lines extinguish themselves. The unfaulted phases still rise to line-to-line voltage.

The Reference Condition

The rating standard for AC high-voltage circuit breakers above 1000 V, IEEE C37.04, and its IEC counterpart, IEC 62271-100, rate interruption against a reference condition: a DC time constant of 45 ms, which is X/R = 17 at 60 Hz and about 14 at 50 Hz, and a relay operating time of half a cycle. The contact parting time, from fault inception until the contacts part, is the relay time plus the breaker's opening time. If the DC fraction remaining at that instant is d = e^−t/τ, the asymmetry factor is

S = √(1 + 2d²)

A breaker whose relays are no faster than half a cycle, applied where X/R does not exceed 17, can use its nameplate rating directly. Higher ratios leave more offset at contact parting. At 60 Hz, with contacts parting three cycles after fault inception, the remaining DC fraction is 15.2 percent at X/R = 10, 33.0 percent at X/R = 17, and 53.3 percent at X/R = 30, giving asymmetry factors of 1.023, 1.104, and 1.253.

The IEEE Application Method

IEEE C37.010-2016, the application guide for AC high-voltage circuit breakers rated on a symmetrical current basis, turns this into a procedure. The study finds the symmetrical current as the prefault voltage divided by the Thévenin reactance, the E/X method, and finds the X/R ratio from separately reduced resistance and reactance networks. Where X/R exceeds the breaker's basis, it multiplies E/X by a factor from the guide's curves, then compares the result with the breaker's symmetrical interrupting rating. For industrial and commercial systems, IEEE 3002.3-2018, the recommended practice for short-circuit studies, covers the whole study, from data collection to device duties.