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GATE EE · Power Systems

Switch Gear And Protection Practice Questions

Generate GATE-level questions on Protection systems. Focus on: 1. Relays: Overcurrent, Directional, Distance (Mho/Reactance), and Differential protection. 2. Circuit Breakers: Arc interruption theory, Restriking voltage, and RRRV. 3. Protection of Generators, Transformers, and Transmission lines.

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

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Q1GATE 2024MCQ

The figure shows the single line diagram of a 4-bus power network, Branches $b_{1}, b_{2}$ , $b_{3}$ and $b_{4}$ have impedances $4 z, z, 2 z$ and $4 z$ per-unit (pu), respectively, where $z=r+j x$ , with $r>0$ and $x>0$ . The current drawn from each load bus (marked as arrows) is equal to $I$ pu, where $I \neq 0$ . If the network is to operate with minimum loss, the branch that should be opened is

🚀 Solve in Practice Mode

📖 Explanation

Assume,

\begin{aligned} |z| & =|r+j X|=1 \text { p.u., } \\ I & =1 \text { p.u. } \end{aligned}

By opening branches ring becomes radial open $b_{1}$ Total power loss,

\begin{aligned} P_{L 1} & =1^{2} \times 2+2^{2} \times 4+3^{2} \times 1 \\ & =27 \text { p.u. } \end{aligned}

Open $b_{2}$

\begin{aligned} P_{L 2} & =\left(3^{2} \times 4+2^{2} \times 2+1^{2} \times 4\right) \\ & =48 \text { p.u. } \end{aligned}

Open $b_{3}$ $P_{L 3}=\left(2^{2} \times 1+1^{2} \times 4+1^{2} \times 4\right)$ $=12 \text { p.u. } \leftarrow \text { mininum power loss }$ Open $b_{4}$ By opening branch $b_{3}$ Power loss is minimum.

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

Explanation Figure 5

Explanation Figure 6

Q2GATE 2023MCQ

For the three-bus power system shown in the figure, the trip signals to the circuit breakers $B_1$ to $B_9$ are provided by overcurrent relays $R_1$ to $R_9$ , respectively, some of which have directional properties also. The necessary condition for the system to be protected for short circuit fault at any part of the system between bus $1$ and the $R-L$ loads with isolation of minimum portion of the network using minimum number of directional relays is

🚀 Solve in Practice Mode

📖 Explanation

$B_3,B_4$ and $B_5,B_7$ are directional over current relays. All remaining are $( B_1,B_2,B_6,B_8,B_9)$ Now directional over current relays.

Explanation Figure 2

Q3GATE 2023NAT

A balanced delta connected load consisting of the series connection of one resistor $(R=15 \Omega)$ and a capacitor $(C=212.21 \mu \mathrm{F})$ in each phase is connected to three-phase, $50 \mathrm{~Hz}, 415 \mathrm{~V}$ supply terminals through a line having an inductance of $\mathrm{L}=31.83 \mathrm{mH}$ per phase, as shown in the figure. Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals $V_{AB}$ in Volts is ____ (Round off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

Given circuit : where,

\begin{aligned} X_{L} & =2 \pi \times 50 \times 31.83 \times 10^{-3}=10 \Omega \\ R & =15 \Omega \\ X_{C} & =\frac{1}{2 \pi \times 50 \times 212.21 \times 10^{-6}}=15 \Omega \end{aligned}

Convert $\Delta$ to star :

\begin{aligned} Z_{Y} & =\frac{Z_{\Delta}}{3} \\ & =\frac{15-j 15}{3}=5-j 5 \end{aligned}

Now, per phase equivalent circuit : Using voltage division,

\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =\frac{5-\mathrm{j} 5}{5-\mathrm{j} 5+10 \mathrm{j}} \times \frac{415 \angle 0^{\circ}}{\sqrt{3}} \\ & =\frac{415}{\sqrt{3}} \angle-90^{\circ} \mathrm{V} \\ \therefore \quad \quad\left|\mathrm{V}_{\mathrm{AB}}\right| & =\sqrt{3} \times \frac{415}{\sqrt{3}}=415 \mathrm{~V} \end{aligned}

Explanation Figure 2

Explanation Figure 3

Q4GATE 2022MCQ

The most commonly used relay, for the protection of an alternator against loss of excitation, is

🚀 Solve in Practice Mode

📖 Explanation

Currently no explanation available

Q5GATE 2019NAT

In a 132 kV system, the series inductance up to the point of circuit breaker locationis 50 mH. The shunt capacitanceat the circuit breaker terminal is 0.05 $\mu F$ . The critical value of resistance in ohms required to be connected across the circuit breaker contacts which will give no transient oscillation is_____

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} L&=50mH\\ C&=0.05\mu F\\ R_{cr}&=\frac{1}{2}\sqrt{\frac{L}{C}}\\ &=\frac{1}{2}\sqrt{\frac{50 \times 10^{-3}}{0.05 \times 10^{-6}}}\\ &=500\Omega \end{aligned}

Q6GATE 2019NAT

The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 $\Omega$ . If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is ________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} I_{sec}&=5 \times 20=100A \\ V &=I_{sec}R =100 \times 0.01\\ &=1V \\ \text{VA } &\text{output of the CT} \\ &= VI_{sec}=100 \times 1\\ &= 100VA \end{aligned}

Q7GATE 2016MCQ

A power system with two generators is shown in the figure below. The system (generators, buses and transmission lines) is protected by six overcurrent relays $R_1 \; to \; R_6$ . Assuming a mix of directional and nondirectional relays at appropriate locations, the remote backup relays for $R_4$ are

🚀 Solve in Practice Mode

📖 Explanation

Currently no explanation available

Q8GATE 2015MCQ

A 3-phase transformer rated for 33 kV/11 kV is connected in delta/star as shown in figure. The current transformers (CTs) on low and high voltage sides have a ratio of 500/5. Find the currents $i_1 \; and \; i_2$ , if the fault current is 300 A as shown in figure.

🚀 Solve in Practice Mode

📖 Explanation

$i_2=0$ , since entire current flows through fault $i_1=\frac{1}{\sqrt{3}}A$

Q9GATE 2014NAT

The over current relays for the line protection and loads connected at the buses are shown in the figure. The relays are IDMT in nature having the characteristic $t_{op}=\frac{0.14\times Time \; Multipller \; Setting}{(Plug \; Setting \; Multiplier)^{0.02}-1}$ The maximum and minimum fault currents at bus B are 2000 A and 500 A respectively. Assuming the time multiplier setting and plug setting for relay $R_B$ to be 0.1 and 5 A respectively, the operating time of $R_B$ (in seconds) is_____.

🚀 Solve in Practice Mode

📖 Explanation

Minimum and maximum fault currents are $I_{fmin}=500A, I_{fmax}=2000A$ For relay $R_B,\;TMS=0.1$ , plug setting =5A

\begin{aligned} \text{PSM}&=\frac{I_f}{I_{pk} \times \text{C.T. ratio}} \\ \text{C.T. ratio} &=\frac{\text{Full load current}}{I_{pk}} \\ &= \frac{100}{5}\\ \text{At realy, }&R_B\\ \text{with }I_{fmin}\Rightarrow & \text{PSM}_1=\frac{500}{\frac{100}{5}\times 5}=5\\ \text{with }I_{fmax}\Rightarrow & \text{PSM}_2=\frac{2000}{\frac{100}{5}\times 5}=20 \end{aligned}

Operating time with TMS=0.1, and from IDMT characterisrics $t_{op_2}=\frac{0.14 \times 0.1}{(20)^{0.02}-1}=0.226sec$

Q10GATE 2011MCQ

A negative sequence relay is commonly used to protect

🚀 Solve in Practice Mode

📖 Explanation

Currently no explanation available

Q11GATE 2011MCQ

A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options

🚀 Solve in Practice Mode

📖 Explanation

For transmission of bulk power of very long distance high voltage (400 kV) is used. To increase reliability, double circuit is used.

Q12GATE 2010MCQ

Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coils is

🚀 Solve in Practice Mode

📖 Explanation

$I_1=200 \times \left ( \frac{5}{400} \right )=2.75A$ $I_2=250 \times \left ( \frac{5}{400} \right )=3.125A$ Current through operating coil, $I_{OC}=I_2-I_1=3.125-2.75=0.375A$

Explanation Figure 2

Q13GATE 2010MCQ

A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3s. The symmetrical breaking current is

🚀 Solve in Practice Mode

📖 Explanation

Rated MVA of circuit breaker = $200 MVA \;\sqrt{3} \times |V(line)|_{rated} \times$ Symmetrical breaking current =Rated MVA of CB Symmetrical breaking current $=\frac{200}{\sqrt{3} \times 33}=35kA$

Q14GATE 2009MCQ

Match the items in List-I (Type of transmission line) with the items in List-II (Type of distance relay preferred) and select the correct answer using the codes given below the lists.

🚀 Solve in Practice Mode

📖 Explanation

Impedance relay is a voltage restrained over current relay. $T=k_1I^{2-}k_2V^2$ Reactance Relay is an over-current relay with directional restraint. $T=k_1I^{2-}k_3VI \sin \theta$ mho relay is a voltage restrained direction relay. $T=k_3VI \cos (\theta -t)-k_2V^2$ mho relay is inherently directional.

Q15GATE 2008MCQ

Voltages phasors at the two terminals of a transmission line of length 70 km have a magnitude of 1.0 per unit but are 180 degree out of phase. Assuming that the maximum load current in the line is $1/5^{th}$ of minimum 3-phase fault current, Which one of the following transmission line protection schemes will not pick up for this condition ?

🚀 Solve in Practice Mode

📖 Explanation

Distance protection using mho relays with zoze set to 80% of the line impedance will not provide protection.

Q16GATE 2008MCQ

A two machine power system is shown below. The Transmission line XY has positive sequence impedance of $Z_{1}\Omega$ and zero sequence impedance of $Z_{0}\Omega$ An 'a' phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase 'a', are given by $V_a$ Volts and $I_a$ amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by

🚀 Solve in Practice Mode

📖 Explanation

As the ground distance relay is located at terminals x Impedance measured by the relay $=\frac{\text{Bus voltage}}{\text{current}}=\frac{V_a}{I_a}$

Q17GATE 2007MCQ

Consider the transformer connections in a part of a power system shown in the figure. The nature of transformer connections and phase shifts are indicated for all but one transformer. Which of the following connections, and the corresponding phase shift $\theta$ , should be used for the transformer between A and B ?

🚀 Solve in Practice Mode

📖 Explanation

Taking $V_1$ as the reference $V_1=220\angle 0^{\circ}kV$ $\angle \theta _1=0^{\circ}$ Phase difference $V_2$ and $V_1$ is $0^{\circ}$ . So, $V_2=400\angle 0^{\circ}kV$ $V_2$ leads $V_3$ by $30^{\circ}$ . So, $V_3=15\angle -30^{\circ}kV$ $\angle \theta _3=-30^{\circ}$ $V_3$ lags $V_4$ by $30^{\circ}$ . $\angle \theta _3=\angle \theta _4-30^{\circ}$ $\angle \theta _4=\angle \theta _3+30^{\circ}$ $\angle \theta _4=-30^{\circ}+30^{\circ}=0^{\circ}$ Phase difference between $V_1$ and $V_2$ is $\theta$ $\theta =\angle \theta _1-\angle \theta _4=0-0=0^{\circ}$

Explanation Figure 2

Q18GATE 2007MCQ

Consider the protection system shown in the figure below. The circuit breakers numbered from 1 to 7 are of identical type. A single line to ground fault with zero fault impedance occurs at the midpoint of the line (at point F), but circuit breaker 4 fails to operate ("Stuck breaker"). If the relays are coordinated correctly, a valid sequence of circuit breaker operation is

🚀 Solve in Practice Mode

📖 Explanation

If the relays are co-ordinated correctly, due to fault ina particular section, relay in that section must operate first, then relays in near by section if first fails, it is back up protection. Thus, the sequence is [5,6,7,3,1,2].

Q19GATE 2006MCQ

Keeping in view the cost and overall effectiveness, the following circuit breaker is best suited for capacitor bank switching

🚀 Solve in Practice Mode

📖 Explanation

The arc time constant is the least (a few $\mu s$ ) in vaccum circuit breakers as compared to other breaker types. The rapid building up of dielectric strength after final arc extinction ( $20kV/\mu s$ ) is unique feature of VCBs. These arc, therefore, ideally suited for capacitor switching.

Q20GATE 2006MCQ

In a biased differential relay the bias is defined as a ratio of

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} N_0k|I_1-I_2|& \gt \frac{N_r}{2}k|I_1+I_2|\\ |I_1-I_2|&\gt \frac{N_r}{N_0}\left | \frac{I_1+I_2}{2} \right | \end{aligned}

where, $N_0=$ number of turns of the operating coil $N_r=$ number of turns of the restraining coil. The relay operates when the magnitude of the difference in the currents at the two ends of a protected element exceeds a certain percentage $((N_r/N_0)/100)$ of half the magnitude of their sum. Hence, the name percentage differentail relay.

Explanation Figure 1

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