Faults formulas

Symmetrical components and fault analysis. 18 formulas for GATE EE Power Systems, each with when to use it. Each is explained in the topic notes.

All 18 Faults formulas

THE A-OPERATOR

a = 1∠12 0^{\circ} = - \frac{1}{2} + j \frac{3}{2}, a^{2} = 1∠24 0^{\circ} = - \frac{1}{2} - j \frac{3}{2}, a^{3} = 1, 1 + a + a^{2} = 0

Multiplying a phasor by $a$ rotates it by $12 0^{\circ}$ anticlockwise without changing its magnitude. Also $a^{2} - a = - j 3$ .

SYNTHESIS (PHASE FROM SEQUENCE)

V_{a} V_{b} V_{c} = 111 1 a^{2} a 1 a a^{2} V_{a 0} V_{a 1} V_{a 2}

Same form for currents. Here $V_{b} = V_{a 0} + a^{2} V_{a 1} + a V_{a 2}$ and $V_{c} = V_{a 0} + a V_{a 1} + a^{2} V_{a 2}$ .

ANALYSIS (SEQUENCE FROM PHASE)

V_{a 0} = \frac{1}{3} (V_{a} + V_{b} + V_{c}), V_{a 1} = \frac{1}{3} (V_{a} + a V_{b} + a^{2} V_{c}), V_{a 2} = \frac{1}{3} (V_{a} + a^{2} V_{b} + a V_{c})

Same form for currents. The inverse matrix is $\frac{1}{3}$ times the complex conjugate of the synthesis matrix.

NEUTRAL CURRENT

I_{n} = I_{a} + I_{b} + I_{c} = 3 I_{a 0}

Only the zero-sequence component flows in the neutral (or earth) return path. If there is no neutral path, $I_{a 0} = 0$ in the line currents. So the line currents of a three-wire system (delta or ungrounded star) contain no zero-sequence component, although zero-sequence current can still circulate inside a delta winding.

THREE-PHASE COMPLEX POWER IN SEQUENCE COMPONENTS

S = V_{a} I_{a}^{*} + V_{b} I_{b}^{*} + V_{c} I_{c}^{*} = 3 (V_{a 0} I_{a 0}^{*} + V_{a 1} I_{a 1}^{*} + V_{a 2} I_{a 2}^{*})

The factor 3 appears because the transformation used here (with $\frac{1}{3}$ in the analysis equations) is not power-invariant. There are no cross terms between different sequences.

SEQUENCE NETWORK EQUATIONS AT THE FAULT BUS

V_{a 1} = E_{a} - Z_{1} I_{a 1}, V_{a 2} = - Z_{2} I_{a 2}, V_{a 0} = - Z_{0} I_{a 0}

For a system that is balanced before the fault, with balanced sources, only the positive-sequence network has a source: $E_{a}$ , the pre-fault voltage of the reference phase $a$ at the fault bus. The negative and zero networks are passive. $Z_{1}, Z_{2}, Z_{0}$ are the Thevenin impedances seen from the fault point, and $I_{a 1}, I_{a 2}, I_{a 0}$ are the sequence currents flowing out of the system into the fault.

NEUTRAL IMPEDANCE IN THE ZERO-SEQUENCE NETWORK

Z_{0} = Z_{0, machine} + 3 Z_{n}

A neutral impedance $Z_{n}$ carries $3 I_{a 0}$ , so it appears as $3 Z_{n}$ in the single-phase zero-sequence network. It does not appear in the positive or negative networks.

THREE-PHASE FAULT CURRENT (PER UNIT)

I_{f} = \frac{E}{Z _{1} + Z _{f}}

$Z_{1}$ is the positive-sequence Thevenin impedance at the fault bus, $Z_{f}$ the impedance in each phase of the fault. For a bolted fault $Z_{f} = 0$ .

BASE CURRENT

I_{ba se} = \frac{S _{ba se}}{3 V _{ba se}}, I_{f} (amperes) = I_{f} (pu) \times I_{ba se}

$S_{ba se}$ in VA for the three-phase rating and $V_{ba se}$ the line-to-line voltage in volts, giving $I_{ba se}$ in amperes.

FAULT MVA (SHORT-CIRCUIT CAPACITY)

Fault MVA = \frac{S _{ba se} ( MVA )}{Z _{1} ( pu )} = 3 V_{LL} I_{f} \times 1 0^{- 6}

For a bolted three-phase fault with $E = 1$ pu. In per unit it equals $1/ Z_{1}$ . $S_{ba se}$ is in MVA, $V_{LL}$ is the base (rated) line-to-line voltage in volts and $I_{f}$ is the fault current in amperes.

CHANGE OF BASE

Z_{p u, n e w} = Z_{p u, o l d} \frac{S _{ba se, n e w}}{S _{ba se, o l d}} (\frac{V _{ba se, o l d}}{V _{ba se, n e w}})^{2}

Convert every element to the common MVA base, and to the common voltage base in each voltage zone, before combining.

SHORT CIRCUIT OF A SERIES R-L CIRCUIT (DC OFFSET)

i (t) = \frac{V _{m}}{∣ Z ∣} [sin (ω t + α - θ) - sin (α - θ) e^{- R t / L}], ∣ Z ∣ = R^{2} + ω^{2} L^{2}, θ = tan^{- 1} \frac{ω L}{R}

A voltage $v = V_{m} sin (ω t + α)$ drives a series $R$ - $L$ circuit that carries no current at $t = 0$ . The angle $α$ sets the instant of the fault on the voltage wave. There is no DC offset when $α = θ$ . The DC offset is largest when $α - θ = \pm 9 0^{\circ}$ . Then, if $R$ is small compared with $ω L$ , the first current peak approaches $2 V_{m} /∣ Z ∣$ , twice the steady-state peak. This model takes a constant reactance and ignores the machine's change from sub-transient to synchronous reactance.

SINGLE LINE-TO-GROUND FAULT (PHASE A)

I_{a 1} = I_{a 2} = I_{a 0} = \frac{E}{Z _{1} + Z _{2} + Z _{0} + 3 Z _{f}}, I_{f} = I_{a} = 3 I_{a 0}

Fault conditions: $I_{b} = I_{c} = 0$ and $V_{a} = Z_{f} I_{a}$ . The fault current through the ground is $I_{a}$ .

LINE-TO-LINE FAULT (PHASES B AND C)

I_{a 1} = - I_{a 2} = \frac{E}{Z _{1} + Z _{2} + Z _{f}}, I_{a 0} = 0, I_{b} = - I_{c} = - j 3 I_{a 1}

$Z_{f}$ is the impedance connected between $b$ and $c$ . $I_{a} = 0$ and $∣ I_{b} ∣ = ∣ I_{c} ∣ = 3 ∣ I_{a 1} ∣$ . No current flows to ground.

DOUBLE LINE-TO-GROUND FAULT (PHASES B AND C TO GROUND)

I_{a 1} = \frac{E}{Z _{1} + \frac{Z _{2} ( Z _{0} + 3 Z _{f} )}{Z _{2} + Z _{0} + 3 Z _{f}}}, I_{a 2} = - I_{a 1} \frac{Z _{0} + 3 Z _{f}}{Z _{2} + Z _{0} + 3 Z _{f}}, I_{a 0} = - I_{a 1} \frac{Z _{2}}{Z _{2} + Z _{0} + 3 Z _{f}}

$Z_{f}$ is the impedance in the common earth path of both phases. $I_{a} = 0$ and the current into the ground is $I_{b} + I_{c} = 3 I_{a 0}$ .

THREE-PHASE FAULT USING Z_BUS

I_{f} = \frac{V _{k} ( 0 )}{Z _{k k} + Z _{f}}, V_{i} (F) = V_{i} (0) - Z_{ik} I_{f}

$V_{k} (0)$ is the pre-fault voltage at bus $k$ , $V_{i} (F)$ the post-fault voltage at bus $i$ , and $Z_{b u s}$ the positive-sequence matrix.

SEQUENCE VOLTAGES AT BUS I FOR AN UNSYMMETRICAL FAULT AT BUS K

V_{i}^{(1)} = V_{i} (0) - Z_{ik}^{(1)} I_{k}^{(1)}, V_{i}^{(2)} = - Z_{ik}^{(2)} I_{k}^{(2)}, V_{i}^{(0)} = - Z_{ik}^{(0)} I_{k}^{(0)}

Find the sequence fault currents $I_{k}^{(1)}, I_{k}^{(2)}, I_{k}^{(0)}$ from the formulas of the previous section with $Z_{1} = Z_{k k}^{(1)}$ , $Z_{2} = Z_{k k}^{(2)}$ , $Z_{0} = Z_{k k}^{(0)}$ and $E = V_{k} (0)$ . Each sequence voltage is updated with its own matrix, then combined into phase voltages with the synthesis matrix. If a star-delta transformer lies between buses $i$ and $k$ , shift the positive and negative sequence values by its phase shift before combining.

ADDING A BRANCH TO EXISTING BUSES (KRON REDUCTION)

Z_{ij, new} Z_{ij, new} = Z_{ij} - \frac{Z _{i p} Z _{p j}}{Z _{pp} + Z _{b}} = Z_{ij} - \frac{( Z _{i p} - Z _{i q} ) ( Z _{p j} - Z _{q j} )}{Z _{pp} + Z _{q q} - 2 Z _{pq} + Z _{b}} (existing bus p to reference) (between existing buses p and q)

Apply to every element $Z_{ij}$ of the existing matrix. Both results come from adding a temporary extra row and column and eliminating it.

Where this was asked: 2018 (3), 2022 (1), 2023 (1), 2024 (1), 2025 (1), 2026 (1). See the 8 previous-year questions on Faults. This shows what was asked, not what will be asked.

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Faults: the notes and 8 solved previous-year questions