GATE EE · Power Systems

Fault Analysis – GATE EE Power SystemsPractice Questions & PYQs

Generate GATE-level questions on Symmetrical and Unsymmetrical faults. Focus on: 1. Symmetrical components and Sequence networks (Positive, Negative, Zero). 2. Symmetrical Three-phase faults. 3. Unsymmetrical faults: Single Line-to-Ground (LG), Line-to-Line (LL), and Double Line-to-Ground (LLG).

41 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2025NAT

The bus impedance matrix of a 3-bus system (in $pu$ ) is

Z_{\text {bus }}=\left[\begin{array}{lll} j 0.059 & j 0.061 & j 0.038 \\ j 0.061 & j 0.093 & j 0.066 \\ j 0.038 & j 0.066 & j 0.110 \end{array}\right]

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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📖 Explanation

\begin{aligned} & Z_{\text {Bus }}=\left[\begin{array}{lll} \mathrm{j} 0.059 & \mathrm{j} 0.061 & \mathrm{j} 0.038 \\ \mathrm{j} 0.061 & \mathrm{j} 0.093 & \mathrm{j} 0.066 \\ \mathrm{j} 0.038 & \mathrm{j} 0.066 & \mathrm{j} 0.110 \end{array}\right] \\ & \mathrm{Z}_{\mathrm{f}}=\mathrm{j} 0.007 \mathrm{pu} \end{aligned}

When the fault is at bus-2, $\mathrm{I}_{\mathrm{f}}=\frac{\mathrm{V}_{\mathrm{f} 2}}{\mathrm{Z}_{22}+\mathrm{Z}_{\mathrm{f}}}=\frac{1 \mathrm{pu}}{\mathrm{j} 0.093+0.007}=-\mathrm{j} 10 \mathrm{pu}$ $\therefore$ Voltage at bus-1,

\begin{aligned} \mathrm{V}_{1(\mathrm{~F})} & =\mathrm{V}_{1}^{0}-\mathrm{I}_{\mathrm{F}} \mathrm{Z}_{12} \\ & =1-(-\mathrm{j} 10)(\mathrm{j} 0.061) \\ & =1-0.61 \\ & =0.39 \mathrm{pu} \end{aligned}

Q2GATE 2024NAT

The single line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being $P_{\max } \sin \delta$ . where $\delta$ is the load angle and the mechanical power input is $0.5 P_{\max }$ . A fault occurs on line 2 such that the power output of the generator is less than 0.5. $P_{\max }$ during the fault. After the fault is cleared by opening line 2. The power output of the generator is $\left\{P_{\max } / \sqrt{2}\right\} \sin \delta$ . If the critical fault clearing angle is $\pi / 2$ radians, the accelerating area on the power angle carve is ______ times $P_{\max }$ (rounded off to 2 decimal places)

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📖 Explanation

\begin{aligned} A_1 & =A_2 \\ \int_{90^{\circ}}^{135^{\circ}}\left(\frac{P_{\max }}{\sqrt{2}} \sin \delta-0.5 P_{\max }\right) d \delta & =0.1 P_{\max } \end{aligned}

Explanation Figure 2

Q3GATE 2023NAT

The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a $Y-\Delta$ 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).

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📖 Explanation

From positive sequence network :

\begin{aligned} X_{1} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}

From negative sequence network :

\begin{aligned} \mathrm{X}_{2} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}

For $\mathrm{L}-\mathrm{L}$ fault, Fault current, $I_{f}=\frac{\sqrt{3} E}{\left(X_{1}+X_{2}\right)}$ $=\frac{\sqrt{3} \times 1}{0.12+0.12} =7.217 \mathrm{pu}$

Q4GATE 2022MCQ

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

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📖 Explanation

We have, $X_0 \gt X_1=X_2$ (for $3-\phi$ transposed transmission line)

Q5GATE 2019MCQ

In the circuit shown below, the switch is closed at t=0. The value of $\theta$ in degrees which will give the maximum value of DC offset of the current at the time of switching is

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📖 Explanation

If the switch is closed at t=0 in series R-L circuit. Then the circuit current i(t) expression is

\begin{aligned} i(t)&=\left \{ \frac{-V_m}{\sqrt{R^{2+}X_L^2}}\sin (\theta -\phi ) \right \}e^{-t/\tau }\\ &+\frac{V_m}{\sqrt{R^{2+}X_L^2}}\sin (\omega t -\phi ) \end{aligned}

The first term of expression indicates DC offset current. For maximum value of DC offset current, the angle should be $90^{\circ}$

\begin{aligned} -\left ( \theta -\tan ^{-1}\left ( \frac{\omega L}{R} \right ) \right )&=90^{\circ}\\ -\left ( \theta -\tan ^{-1}\left ( \frac{377 \times 10 \times 10^{-3}}{3.77} \right ) \right )&=90^{\circ}\\ (\theta -45^{\circ})&=-90^{\circ}\\ \theta &=-45^{\circ} \end{aligned}

Q6GATE 2019NAT

Five alternators each rated 5 MVA, 13.2 kV with 25% of reactance on its own base are connected in parallel to a busbar. The short-circuit level in MVA at the busbar is_________

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📖 Explanation

Net reactance of parallel connection,

\begin{aligned} X&=\frac{0.25}{5}=0.05\;p.u.\\ I_{SC}&=\frac{1}{X}=\frac{1}{0.05}=20\; p.u.\\ \text{SC MVA}&=20 \times 5\\ &=100 \text{ MVA} \end{aligned}

Q7GATE 2019NAT

A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactancesof 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance in ohms (round off to one decimal place) is ______

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📖 Explanation

\begin{aligned} X_1&=0.25 \; p.u.\\ X_2&=0.15 \; p.u.\\ X_0&=0.05 \; p.u.\\ I_{f(LG)}&=I_{f(3-\phi )}\\ \frac{3V_{pu}}{(X_1+X_2+X_0+3X_n)}&=\frac{V_{pu}}{X_1}\\ \frac{3 \times 1}{(0.25+0.15+0.05+3X_n)}&=\frac{1}{0.25}\\ \frac{3}{0.46+3X_n}&=\frac{1}{0.25}\\ \Rightarrow \;\; X_n&=0.1\; p.u.\\ X_n&=0.1 \times Z_B\\ &=0.1 \times \frac{30^2}{50}\\ &=1.8\Omega \end{aligned}

Q8GATE 2018NAT

The series impedance matrix of a short three-phase transmission line in phase coordinates is

\begin{bmatrix} Z_{s} & Z_{m} &Z_{m} \\ Z_{m}& Z_{s}& Z_{m}\\ Z_{m}& Z_{m}& Z_{s} \end{bmatrix}

. If the positive sequence impedance is $(1 + j10)\Omega$ , and the zero sequence is $(4 + j31)\Omega$ , then the imaginary part of $Z_{m}$ (in $\Omega$ ) is ______(up to 2 decimal places).

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📖 Explanation

\begin{aligned} Z_1&=(1+j10)\Omega \\ Z_0&=(4+j31)\Omega \\ \text{we know,}\\ Z_1&=Z_s-Z_m\\ Z_0&=Z_s+2Z_m\\ Z_1-Z_0&=-3Z_m\\ Z_m&=\frac{Z_0-Z_1}{3}\\ &=\frac{4+j31-1-j10}{3}\\ &=\frac{3+j21}{3}\\ &=1+j7 \end{aligned}

The imaginary part of $Z_m$ is 7.00.

Q9GATE 2018MCQ

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 $I_{f1}=kI_{f}$ , where $I_{f}$ is the fault current with zero fault impedance. The relation between $Z_{f}$ and k is

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📖 Explanation

For LL fault: Without $Z_f$

\begin{aligned} \text{Given, } I_{f1}&=k \times I_f \\ \frac{\sqrt{3} E_g}{Z_1+Z_2+Z_f} &=\left ( \frac{\sqrt{3} E_g}{Z_1+Z_2} \right )k \\ Z_1+Z_2&=k(Z_1+Z_2+Z_f) \\ Z_f &=\frac{(Z_1+Z_2)(1-k)}{k} \end{aligned}

Explanation Figure 1

Q10GATE 2018NAT

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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📖 Explanation

For LG fault,

\begin{aligned} I_{f\;p.u.}&=\frac{3V_{Th}}{Z_1+Z_2+Z_0+3Z_n}\\ &=\frac{3 \times 1 }{0.1+0.05+0.01+3(0.01)}\\ &=15.789\; p.u.\\ I_{base}&=\frac{P}{\sqrt{3}\cdot V_L}\\ &=\frac{125 \times 10^6}{\sqrt{3} \times 15.5 \times 10^3}\\ &=4656.05A\\ I_f&=I_{f\;p.u.}\cdot I_{base}\\ &=15.789 \times 4656.05\\ &=73516.538=73.52kA \end{aligned}

Q11GATE 2017NAT

The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).

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📖 Explanation

\begin{aligned} Z_1&=0.2\\ Z_2&=0.2\\ Z_0&=0.1\\ V_{Th}&=1\\ I_f&=3.75\\ &\text{For LG fault,}\\ I_f&=\frac{3V_{Th}}{Z_1+Z_2+Z_0+3Z_n}\\ 3.75&=\frac{3 \times 1}{0.2+0.2+0.1+3Z_n}\\ Z_n&=0.1\;p.u. \end{aligned}

Q12GATE 2016MCQ

If the star side of the star-delta transformer shown in the figure is excited by a negative sequence voltage, then

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Q13GATE 2016NAT

The magnitude of three-phase fault currents at buses A and B of a power system are 10 pu and 8 pu, respectively. Neglect all resistances in the system and consider the pre-fault system to be unloaded. The pre-fault voltage at all buses in the system is 1.0 pu. The voltage magnitude at bus B during a three-phase fault at bus A is 0.8 pu. The voltage magnitude at bus A during a three-phase fault at bus B, in pu, is ________.

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📖 Explanation

Voltage at bus B after 3-phase fault at A=0.8 p.u.

\begin{aligned} V_B &=V_{(prefault)}-Z_{12} \times I_{f(A)} \\ 0.8&= 1.0-Z_{12} \times 10 \\ Z_{12} &=0.02 p.u. \\ V_A &=V_{(prefault)}-Z_{12}I_{f(B)} \\ V_A&=1.0-(0.02 \times 8) \\ &= 0.84p.u. \end{aligned}

Q14GATE 2016NAT

A 50 MVA, 10 kV, 50 Hz, star-connected, unloaded three-phase alternator has a synchronous reactance of 1 p.u. and a sub-transient reactance of 0.2 p.u. If a 3-phase short circuit occurs close to the generator terminals, the ratio of initial and final values of the sinusoidal component of the short circuit current is ________.

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📖 Explanation

\begin{aligned} I''&=\frac{E_g}{X_d''};\; I=\frac{E_g}{X}\\ \frac{I''}{I}&=\frac{X}{X_d''}=\frac{1.0}{0.2}=5.0\; p.u. \end{aligned}

Q15GATE 2016NAT

Two identical unloaded generators are connected in parallel as shown in the figure. Both the generators are having positive, negative and zero sequence impedances of j0.4 p.u., j0.3 p.u. and j0.15 p.u., respectively. If the pre-fault voltage is 1 p.u., for a line-to-ground (L-G) fault at the terminals of the generators, the fault current, in p.u., is ___________.

🚀 Solve in Practice Mode

📖 Explanation

$I_f=3I_{a_1}=\frac{3}{0.2+0.15+0.15}=6 \; p.u.$

Explanation Figure 2

Q16GATE 2016NAT

A 30 MVA, 3-phase, 50 Hz, 13.8 kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%, 15% and 5% respectively. A reactance ( $X_n$ ) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases 'b' and 'c', with a fault impedance of j0.1 p.u. The value of $X_n$ (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.

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📖 Explanation

Base current,

\begin{aligned} I_B&=\frac{30 \times 10^3}{\sqrt{3} \times 13.8}=1255.109A\\ I_f&=4270A\\ I_{p.u.}&=\frac{4270}{1255.109}=3.402p.u.\\ I_{g1}&=\frac{E_a}{X_1+(X_2||X_0)}\\ &\text{where, }\\ X_0&=X_0+3(Z_n+Z_f)\\ X_0&=3Z_n+0.35\\ 3.402&=\frac{1.0}{0.15+\frac{0.15 \times (3Z_n+0.35)}{0.15+3Z_n+0.35}}\\ &\text{Solving the equation,}\\ Z_n&=1.07 \;p.u. \end{aligned}

Q17GATE 2016NAT

The single line diagram of a balanced power system is shown in the figure. The voltage magnitude at the generator internal bus is constant and 1.0 p.u. The p.u. reactances of different components in the system are also shown in the figure. The infinite bus voltage magnitude is 1.0 p.u. A three phase fault occurs at the middle of line 2. The ratio of the maximum real power that can be transferred during the pre-fault condition to the maximum real power that can be transferred under the faulted condition is _________.

🚀 Solve in Practice Mode

📖 Explanation

The ratio $=\frac{P_1}{P_2}=\frac{X_1}{X_2}$

\begin{aligned} 1.\; \Rightarrow \; \;&\frac{j0.35 \times j0.6}{j1.2} &=j0.175 \\ 2.\; \Rightarrow \; \; &\frac{j0.6 \times j0.25}{j1.2} &=j0.125 \\ 3.\; \Rightarrow \; \; &\frac{j0.35 \times j0.25}{j1.2} &=j0.0729 \end{aligned}

\begin{aligned} X_{12}&=X_2\\ &=j0.375+j0.125+\frac{j0.375\times j0.125}{j0.0729}\\ &=j1.1430\;p.u.\\ \frac{P_1}{P_2}&=\frac{X_2}{X_1}=\frac{j1.1430}{j0.5}=2.286 \end{aligned}

Explanation Figure 2

Explanation Figure 3

Q18GATE 2015MCQ

A sustained three-phase fault occurs in the power system shown in the figure. The current and voltage phasors during the fault (on a common reference), after the natural transients have died down, are also shown. Where is the fault located?

🚀 Solve in Practice Mode

📖 Explanation

From the phasor $I_2$ and $I_4$ are having $180^{\circ}$ phase shift. $|I_1|$ is higher than all other. Hence fault occurred at point Q.

Q19GATE 2014MCQ

For a fully transposed transmission line

🚀 Solve in Practice Mode

📖 Explanation

For a transmission line (static device), $X_1=X_2 \text{ and } X_0 \gt\gt X_1$

Q20GATE 2014MCQ

In an unbalanced three phase system, phase current $I_a = 1\angle -90^{\circ}$ pu, negative sequence current $I_{b2}=4 \angle( -150^{\circ})$ , zero sequence current $I_{c0}=3 \angle -90^{\circ}$ pu. The magnitude of phase current $I_{b}$ in pu is

🚀 Solve in Practice Mode

📖 Explanation

Given,

\begin{aligned} I_a&= 1\angle (-90^{\circ})pu\\ I_b&= 4\angle (-150^{\circ})pu\\ I_c&= 3\angle (90^{\circ})pu \end{aligned}

As zero sequence current in all the three phases of unbalanced $3-\phi$ system are equal, therefore, $I_{a_0}=I_{b_0}=I_{c_0}=3\angle 90^{\circ}pu$ Also, $I_{b_2}=4\angle -150^{\circ}$ =Negative phase sequence current of phase b. Therefore, referring to the negative phase sequence phasor, $I_{b_2}=4\angle (-150^{\circ}-120^{\circ})=4\angle-270^{\circ}pu$ We know that, phase current vectors in matrix form are given by

\begin{aligned} \begin{bmatrix} I_a\\ I_b\\ I_c \end{bmatrix}&=\begin{bmatrix} 1 &1 &1 \\ 1& a^2 &a \\ 1 & a &a^2 \end{bmatrix}\begin{bmatrix} I_{a_0}\\ I_{a_1}\\ I_{a_2} \end{bmatrix} \\ \text{Here, } I_a&=I_{a_0}+I_{a_1}+I_{a_2} \\ \text{Given, } I_a&=1\angle -90^{\circ} \\ I_{a_0}&=3\angle 90^{\circ} \\ I_{a_2}&=4\angle -270^{\circ} \\ \text{Therefore,}& \\ I_{a_1}&=I_a-(I_{a_0}+I_{a_2}) \\ &=(1\angle -90^{\circ})-(3\angle 90^{\circ}+4\angle 270^{\circ}) \\ I_{a_1}&=8\angle -90^{\circ}pu \\ \text{Therefore,}& \\ I_b&=I_{a_0}+a^2I_{a_1}+aI_{a_2}\\ &=(3\angle 90^{\circ})+(1\angle 120^{\circ})^2 (8\angle -90^{\circ}) \\ &+(1\angle 120^{\circ})(4\angle -270^{\circ})\\ &=3\angle 90^{\circ}+8\angle 150^{\circ}+4\angle -150^{\circ}\\ I_b&=11.53\angle 154.3^{\circ}pu\\ &\text{Magnitude of phase current,}\\ I_b&=11.53\;pu \end{aligned}

Explanation Figure 1

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