GATE EE

Symmetrical components and fault analysis

Symmetrical components resolve any unbalanced three-phase set into three balanced sets (positive, negative and zero sequence), so an unsymmetrical fault becomes an interconnection of three single-phase sequence networks. Fault MVA and the symmetrical and unsymmetrical fault currents all follow from the sequence impedances at the fault bus. These notes cover the network connection for each fault, the effect of fault and neutral impedance, the zero-sequence path through transformers, the DC offset of a short circuit and fault studies with the bus impedance matrix.

8 previous-year questions · 7 solved · 5 practice questions

Symmetrical components and the a-operator

Convention used throughout: per-unit values on stated MVA and kV bases, phasors written with magnitude and angle, $j$ is the imaginary unit, $a = 1∠12 0^{\circ}$ , and quantities are per phase unless stated. Phase $a$ is the reference phase. The three sequence networks are named positive (1), negative (2) and zero (0).

Any unbalanced set of three phasors $V_{a}, V_{b}, V_{c}$ equals the sum of three sets. The positive-sequence set has equal magnitudes, displaced $12 0^{\circ}$ in the order $a$ - $b$ - $c$ . The negative-sequence set has equal magnitudes in the order $a$ - $c$ - $b$ . The zero-sequence set has three equal phasors in phase with each other.

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.

Facts worth remembering

A balanced set in the phase order $a$ - $b$ - $c$ has only a positive-sequence component; a balanced set in the order $a$ - $c$ - $b$ has only a negative-sequence component.
Line voltages carry no zero-sequence component, because $V_{ab} + V_{b c} + V_{c a} = 0$ .
The sequence components of phases $b$ and $c$ differ from those of $a$ only by factors of $a$ and $a^{2}$ , so one phase carries all the information.
With phase $a$ as reference, $V_{a 1}$ is the positive-sequence voltage of phase $a$ , the positive-sequence voltage of phase $b$ is $a^{2} V_{a 1}$ and the negative-sequence voltage of phase $b$ is $a V_{a 2}$ .

Sequence impedances and sequence networks

The impedance an element presents to a set of currents depends on the sequence. For static, balanced elements (transposed lines, transformers) reversing the phase order does not change the mutual coupling, so positive and negative sequence impedances are equal. Rotating machines respond differently to the forward and backward rotating fields, so $Z_{1} \neq = Z_{2}$ in general.

Element	Positive $Z_{1}$	Negative $Z_{2}$	Zero $Z_{0}$
Transmission line (transposed)	Series impedance per phase	$Z_{2} = Z_{1}$	Larger than $Z_{1}$ , because zero-sequence current returns through earth and any ground wires
Transformer	Leakage impedance	$Z_{2} = Z_{1}$	Equal to the leakage impedance for a bank of single-phase units, approximately equal for three-phase units; how it connects into the network depends on the winding connection and grounding
Synchronous generator	Sub-transient, transient or synchronous reactance, as the study requires	Different from $Z_{1}$ ; close to the sub-transient reactance	Small, set by winding leakage; the network value is $Z_{g 0} + 3 Z_{n}$ if grounded through $Z_{n}$ , and an open circuit if the neutral is ungrounded
Induction motor (as a source in the first cycles)	Sub-transient reactance	Close to $Z_{1}$	No zero-sequence path if the stator is delta or ungrounded star

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.

Transformer connection (HV - LV)	Zero-sequence representation
Yg - Yg (both neutrals solidly grounded)	Leakage impedance connects the two buses; zero-sequence current passes through
Yg - Yg grounded through $Z_{n 1}$ (HV) and $Z_{n 2}$ (LV)	Leakage impedance between the buses, with $3 Z_{n 1}$ and $3 Z_{n 2}$ in series
Yg - $Δ$	Leakage impedance from the Yg-side bus to reference; the delta-side bus is not connected to the Yg side (current circulates inside the delta)
Y (ungrounded) - any	Open circuit on the ungrounded side, so no zero-sequence path through the transformer
$Δ$ - $Δ$	Open circuit; no path to reference on either side
Yg through $Z_{n}$ - $Δ$	As for Yg - $Δ$ , with $3 Z_{n}$ added in series with the leakage impedance on the Yg side

EXAM TIP

A transformer winding gives zero-sequence current a path on its own side when it is star-connected with its neutral grounded. A delta or ungrounded-star winding gives no zero-sequence path into its own side's network. A delta winding opposite a grounded star lets the current circulate but keeps it out of the delta-side network. A star-delta bank shifts the positive and negative sequences in opposite directions: if positive-sequence quantities shift by $+ 3 0^{\circ}$ from one side to the other, negative-sequence quantities shift by $- 3 0^{\circ}$ . Zero sequence is not shifted. The shift does not change the current at the fault point. It must be applied when finding phase currents or voltages on the other side of the transformer, because the positive and negative components then recombine at different angles.

Symmetrical three-phase fault and fault MVA

A balanced three-phase fault involves only the positive-sequence network, because the system stays balanced. Take the pre-fault voltage at the fault bus as $E$ (1 pu if not given) and neglect pre-fault load current unless the question includes it.

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.

Machine reactance	Meaning	Where it is used
Sub-transient $X_{d}^{''}$	Smallest; the effective reactance in the first few cycles	Initial symmetrical short-circuit current, momentary current
Transient $X_{d}^{'}$	Larger; applies after the sub-transient effect decays	Current in the transient period, a few cycles after the fault
Synchronous $X_{d}$	Largest; the steady-state value	Steady-state fault current after all decay

EXAM TIP

If a question gives all three reactances, read the wording to see which instant is needed: 'initial' or 'sub-transient' means $X_{d}^{''}$ , 'steady state' means $X_{d}$ . A larger reactance gives a smaller current, so the sub-transient reactance gives the largest of the three currents.

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.

Pre-fault load current. With load, the pre-fault voltage $V_{f}$ at the fault bus is not exactly 1 pu. By Thevenin's theorem, the current into the fault is $V_{f} / (Z_{t h} + Z_{f})$ , using the actual pre-fault voltage. The current in each branch (a generator or line, say) after the fault is its pre-fault load current plus the change caused by the fault, found from the Thevenin circuit. When the question says to neglect load current, use $E = 1$ pu at no load.

Unsymmetrical faults

Each fault is placed symmetrically about the reference phase $a$ : the SLG fault on phase $a$ , and the LL and DLG faults on phases $b$ and $c$ . The system is balanced before the fault and pre-fault load current is neglected, so $E$ is the pre-fault (Thevenin) voltage of phase $a$ at the fault bus. $Z_{1}, Z_{2}, Z_{0}$ are the Thevenin sequence impedances at that bus and $Z_{f}$ is the fault impedance. The sequence networks are then connected as below.

Fault	Phases	Network connection	Zero-sequence current?
Single line-to-ground (SLG)	$a$ to ground through $Z_{f}$	All three networks in series, with $3 Z_{f}$ added	Yes
Line-to-line (LL)	$b$ to $c$ through $Z_{f}$	Positive and negative networks connected terminal to terminal, with $Z_{f}$ in series between them; zero network unused	No
Double line-to-ground (DLG)	$b$ and $c$ to ground through $Z_{f}$	Positive network in series with the parallel combination of the negative network and (zero network + $3 Z_{f}$ )	Yes
Three-phase	$a$ , $b$ , $c$ shorted together	Positive network only	No

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 $I_{b} + I_{c} = 3 I_{a 0}$ .

CHECK FOR LL AND DLG FAULTS

I_{a 1} + I_{a 2} + I_{a 0} = I_{a} = 0

Phase $a$ is healthy in these faults, so its sequence currents add to zero. Use it to check an answer.

WORKED EXAMPLE

A synchronous generator rated 20 MVA, 11 kV has $X_{1} = 0.20$ , $X_{2} = 0.20$ and $X_{0} = 0.10$ pu on its own rating, with a solidly grounded neutral. It is on no load at 1 pu voltage when a bolted single line-to-ground fault occurs on phase $a$ at its terminals. Find the fault current in per unit and in amperes.

Solution

Since $Z_{f} = 0$ and the neutral is solidly grounded: $I_{a 0} = \frac{1}{j ( 0.20 + 0.20 + 0.10 )} = - j 2.0$ pu.
Fault current $I_{a} = 3 I_{a 0}$ , so $∣ I_{a} ∣ = 6.0$ pu.
Base current: $I_{ba se} = \frac{20 \times 1 0 ^{6}}{3 \times 11 \times 1 0 ^{3}} \approx 1049.7$ A.
$∣ I_{a} ∣ = 6.0 \times 1049.7 \approx 6298$ A.

Answer: 6.0 pu, about 6300 A

WORKED EXAMPLE

For the same generator and the same terminal location, find the fault current magnitude for a bolted line-to-line fault between phases $b$ and $c$ , and the ground current for a bolted double line-to-ground fault. Compare with the three-phase fault current.

Solution

Three-phase: $I_{f} = 1/0.20 = 5.0$ pu, about $5.0 \times 1049.7 \approx 5249$ A.
LL: $I_{a 1} = 1/ (0.20 + 0.20) = 2.5$ pu, so $∣ I_{b} ∣ = 3 \times 2.5 \approx 4.33$ pu, about 4545 A. With $Z_{1} = Z_{2}$ the LL current is $3 /2$ times the three-phase current.
DLG: $X_{2} ∥ X_{0} = \frac{0.20 \times 0.10}{0.30} \approx 0.0667$ pu, so $I_{a 1} = 1/ (0.20 + 0.0667) = 3.75$ pu.
$I_{a 0} = - 3.75 \times \frac{0.20}{0.20 + 0.10} = - 2.5$ pu, so the ground current is $3∣ I_{a 0} ∣ = 7.5$ pu, about 7873 A.
The SLG current $3/ (X_{1} + X_{2} + X_{0})$ exceeds the three-phase current $1/ X_{1}$ only when $X_{2} + X_{0} < 2 X_{1}$ . Here $X_{1} = X_{2}$ and $X_{0} < X_{1}$ , so the SLG current (6.0 pu) exceeds the three-phase current (5.0 pu). This does not hold for every system.

Answer: LL: 4.33 pu (about 4545 A). DLG ground current: 7.5 pu (about 7873 A). Three-phase: 5.0 pu (about 5249 A).

Effect of fault and neutral impedance; bus impedance matrix

How impedances enter the sequence networks

Fault impedance $Z_{f}$ in an SLG fault: add $3 Z_{f}$ in series in the connection of the three networks.
Fault impedance $Z_{f}$ between phases in an LL fault: add $Z_{f}$ in series in the loop of the positive and negative networks.
Fault impedance $Z_{f}$ in the earth path of a DLG fault: add $3 Z_{f}$ in series with $Z_{0}$ .
Neutral impedance $Z_{n}$ of a generator or transformer: add $3 Z_{n}$ to that element's zero-sequence impedance.
If the only ground source is a star point grounded through a very large impedance, or not grounded at all, $Z_{0}$ is very large and the ground-fault current is very small.

The bus impedance matrix $Z_{b u s}$ of each sequence network gives the Thevenin impedance at every bus at once. For a fault at bus $k$ , the diagonal element $Z_{k k}$ of that sequence's matrix is the Thevenin impedance at bus $k$ . The off-diagonal element $Z_{ik}$ is the voltage at bus $i$ per unit of current injected at bus $k$ . The matrix is either given or built by adding branches one at a time.

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.

Building Z_bus by adding a branch of impedance $Z_b$

New branch from a new bus to the reference: the new diagonal element is $Z_{b}$ , and all new off-diagonal elements are zero.
New branch from a new bus $q$ to an existing bus $p$ : $Z_{q i} = Z_{p i}$ for every existing bus $i$ , and $Z_{q q} = Z_{pp} + Z_{b}$ .
A branch from an existing bus to the reference, or between two existing buses, modifies every element of the matrix, by the formulas below (the size of the matrix does not change).

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.

EXAM TIP

Check the size of any answer. With a large $Z_{n}$ , the SLG fault current falls well below the three-phase fault current. A solidly grounded system with small $Z_{0}$ can give an SLG current above the three-phase value. Drawing the sequence-network connection first, with every terminal labelled, prevents most sign and connection mistakes.

PREVIOUS-YEAR QUESTIONS

8 questions from the EE papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2018

GATE 2026 EE Q.19

GATE 2026 · EEQ.19 · 1 mark · MCQ

The initial three-phase voltage phasors (V_A, V_B and V_C) at a bus of a power network are as shown in Case-1. Due to a disturbance, the bus voltage phasors changed in phase by a small angle Δθ, and the magnitudes remained the same as depicted in Case-2.

Which one of the following statements is correct about the zero sequence components?

GATE 2026 EE question 19: Two boxed phasor diagrams, Case-1 and Case-2, each with a horizontal dashed real axis and a vertical dashed imaginary axis… — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
The zero sequence components in Case-1 and Case-2 have the same phase angle and magnitude
(B)
The magnitude of the zero sequence component in Case-1 is greater than that in Case-2
(C)
The magnitude of the zero sequence component in Case-2 is greater than that in Case-1
(D)
The zero sequence components in Case-1 and Case-2 have the same magnitude but different phase angles

Answer and solution

GATE 2026 EE Q.19: answer and solution

GATE 2025 EE Q.33

GATE 2025 · EEQ.33 · 1 mark · NAT

The bus impedance matrix of a 3-bus system (in pu) is Z_bus = [[j0.059, j0.061, j0.038],[j0.061, j0.093, j0.066],[j0.038, j0.066, j0.110]]. A symmetrical fault (through a fault impedance of j0.007 pu) occurs at bus 2. Neglecting pre-fault loading conditions, the voltage at bus 1, during the fault is ________ pu (round off to three decimal places).

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GATE 2025 EE Q.33: answer and solution

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GATE 2025 EE Q.33: answer and solution

GATE 2024 EE Q.46

GATE 2024 · EEQ.46 · 2 marks · MSQ

For a two-phase network, the phase voltages $V_{p}$ and $V_{q}$ are to be expressed in terms of sequence voltages $V_{α}$ and $V_{β}$ as $[V_{p} V_{q}] = S [V_{α} V_{β}]$ . The possible option(s) for matrix $S$ is/are

(A)
$[11 1 - 1]$
(B)
$[1111]$
(C)
$[1110]$
(D)
$[- 1 1 11]$

Official answer

Official answer: A, DSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

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GATE 2024 EE Q.46: official answer

GATE 2023 EE Q.60

GATE 2023 · EEQ.60 · 2 marks · NAT

The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y-Δ transformer. The positive, negative and zero-sequence diagrams of the system are shown in figures (ii), (iii) and (iv), respectively. All reactances in the sequence diagrams are in p.u. For a bolted line-to-line fault (fault impedance = zero) between phases 'b' and 'c' at bus 1, neglecting all pre-fault currents, the magnitude of the fault current (from phase 'b' to 'c') in p.u. is ____________ (Round off to 2 decimal places).

GATE 2023 EE question 60: Figure (i): single-line diagram. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

GATE 2023 EE Q.60: answer and solution

GATE 2022 EE Q.15

GATE 2022 · EEQ.15 · 1 mark · MCQ

The valid positive, negative and zero sequence impedances (in p.u.), respectively, for a 220 kV, fully transposed three-phase transmission line, from the given choices are

(A)
1.1, 0.15 and 0.08
(B)
0.15, 0.15 and 0.35
(C)
0.2, 0.2 and 0.2
(D)
0.1, 0.3 and 0.1

Answer and solution

GATE 2022 EE Q.15: answer and solution

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GATE 2018 EE Q.20

GATE 2018 · EEQ.20 · 1 mark · NAT

The series impedance matrix of a short three-phase transmission line in phase coordinates is [[Z_s, Z_m, Z_m],[Z_m, Z_s, Z_m],[Z_m, Z_m, Z_s]]. If the positive sequence impedance is (1 + j10) Ω, and the zero sequence is (4 + j31) Ω, then the imaginary part of Z_m (in Ω) is ______ (up to 2 decimal places).

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GATE 2018 EE Q.20: answer and solution

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GATE 2018 EE Q.20: answer and solution

GATE 2018 EE Q.21

GATE 2018 · EEQ.21 · 1 mark · NAT

The positive, negative and zero sequence impedances of a 125 MVA, three-phase, 15.5 kV, star-grounded, 50 Hz generator are j0.1 pu, j0.05 pu and j0.01 pu respectively on the machine rating base. The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the generator is j0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is ______ (up to 2 decimal places).

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GATE 2018 EE Q.21: answer and solution

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GATE 2018 EE Q.28

GATE 2018 · EEQ.28 · 2 marks · MCQ

The positive, negative and zero sequence impedances of a three phase generator are Z_1, Z_2 and Z_0 respectively. For a line-to-line fault with fault impedance Z_f, the fault current is I_{f1} = k I_f, where I_f is the fault current with zero fault impedance. The relation between Z_f and k is

(A)
Z_f = (Z_1+Z_2)(1-k)/k
(B)
Z_f = (Z_1+Z_2)(1+k)/k
(C)
Z_f = (Z_1+Z_2)k/(1-k)
(D)
Z_f = (Z_1+Z_2)k/(1+k)

Answer and solution

GATE 2018 EE Q.28: answer and solution

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GATE 2018 EE Q.28: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 12 marks · NAT

A generator on no load at 1 pu voltage has $X_{1} = 0.25$ , $X_{2} = 0.15$ and $X_{0} = 0.05$ pu, with its neutral solidly grounded. A bolted single line-to-ground fault occurs at its terminals. Find the magnitude of the fault current in per unit.

Answer and solution

Answer: $6.67$ (accepted range $\pm 0.02$ )

For an SLG fault with $Z_{f} = 0$ : $∣ I_{a 0} ∣ = \frac{1}{0.25 + 0.15 + 0.05} = \frac{1}{0.45}$ pu.
$∣ I_{f} ∣ = 3∣ I_{a 0} ∣ = \frac{3}{0.45} \approx 6.67$ pu.

PRACTICE 21 mark · MSQ

A bolted fault occurs at a bus of a three-phase system with a solidly grounded neutral. Which of the following faults give zero zero-sequence current $I_{a 0}$ at the fault?

(A)
Single line-to-ground fault
(B)
Line-to-line fault
(C)
Double line-to-ground fault
(D)
Three-phase fault

Answer and solution

Answer: (B), (D)

SLG and DLG faults involve the ground, so the zero-sequence network is part of the connection and $I_{a 0} \neq = 0$ .
LL fault: $I_{a 0} = 0$ because the fault has no ground path. Three-phase fault: the system stays balanced and only positive sequence flows, so $I_{a 0} = 0$ .
The correct options are B and D.

PRACTICE 31 mark · MCQ

In the zero-sequence network of a delta-star transformer ( $Δ$ - Yg, neutral of the Yg side solidly grounded), the transformer is represented as:

(A)
Its leakage impedance connected between the two buses
(B)
Its leakage impedance connected from the Yg-side bus to reference, with the delta-side bus not connected to the Yg-side bus
(C)
An open circuit on both sides
(D)
A short circuit between the two buses

Answer and solution

Answer: (B)

Zero-sequence current can flow in the grounded star and circulates inside the delta, so it does not pass into the delta-side network.
The Yg side therefore sees the leakage impedance to reference, and the delta-side bus is isolated from it.
The correct option is B.

PRACTICE 42 marks · NAT

At a bus of a system on no load, the positive and negative sequence Thevenin reactances are $X_{1} = X_{2} = 0.2$ pu and the pre-fault voltage is 1 pu. A fault impedance of $j 0.1$ pu is connected between phases $b$ and $c$ (a line-to-line fault). Find the magnitude of the current in phase $b$ in per unit.

Answer and solution

Answer: $3.46$ (accepted range $\pm 0.02$ )

For an LL fault: $∣ I_{a 1} ∣ = \frac{1}{0.2 + 0.2 + 0.1} = 2.0$ pu.
$∣ I_{b} ∣ = 3 \times 2.0 \approx 3.46$ pu.

PRACTICE 51 mark · NAT

At a bus in a system with a 100 MVA base, the positive-sequence Thevenin reactance is 0.04 pu, with pre-fault voltage 1 pu. Find the fault MVA for a bolted three-phase fault at the bus.

Answer and solution

Answer: $2500$ (accepted range $\pm 1$ )

Fault MVA $= S_{ba se} / Z_{1} = 100/0.04$ .
$= 2500$ MVA.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EE

Symmetrical components, Symmetrical and unsymmetrical fault analysis

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