GATE EE · Power Systems

Compensation Techniques And Voltage Profile Control – GATE EE Power SystemsPractice Questions & PYQs

Generate GATE-level questions on Voltage control. Focus on: 1. Reactive power compensation: Shunt capacitors/reactors and Series capacitors. 2. Tap-changing transformers and Synchronous condensers. 3. Voltage profile and Power factor improvement.

18 questions · 18 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2025NAT

Consider a distribution feeder, with $\mathrm{R} / \mathrm{X}$ ratio of 5 . At the receiving end, a 350 kVA load is connected. The maximum voltage drop will occur from the sending end to the receiving end, when the power factor of the load is ___________ (round off to three decimal places).

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📖 Explanation

$\therefore \frac{\mathrm{X}}{\mathrm{R}}=\frac{1}{5}$ To have maximum voltage drop,

\begin{aligned} & \begin{aligned} \phi=\theta= & \tan ^{-1}\left(\frac{\mathrm{X}}{\mathrm{R}}\right) \\ & =\tan ^{-1}\left(\frac{1}{5}\right)=11.31 \\ \therefore \cos \phi & =\cos (11.31)=0.981 \end{aligned} \end{aligned}

Q2GATE 2018MCQ

Consider the two bus power system network with given loads as shown in the figure. All the values shown in the figure are in per unit. The reactive power supplied by generator $G_1$ and $G_2$ are $Q_{G1}$ and $Q_{G2}$ respectively. The per unit values of $Q_{G1}, Q_{G2}$ , and line reactive power loss ( $Q_{loss}$ ) respectively are

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📖 Explanation

At $G_2$ load demand is 20 pu, $\begin{aligned} P_S&=\left | \frac{V_SV_R}{X_L} \right | \sin \delta \\ 5&=\frac{1 \times 1}{0.1} \sin \delta \\ \delta &=30^{\circ}\\ Q_S&=\frac{V_S^2}{X_L}-\frac{V_SV_R}{X_L} \cos \delta \\ &=\frac{1^2}{0.1}-\frac{1 \times 1 }{0.1} \cos 30^{\circ}\\ &=1.34 p.u.\\ Q_R&=\left | \frac{V_SV_R}{X_L} \right | \cos \delta -\left | \frac{V_R^2}{X_L} \right | \\ 5&=\frac{1 \times 1}{0.1} \cos 30^{\circ}-\frac{1^2}{0.1} \\ &=-1.34p.u.\\ Q_{loss}&+Q_S-Q_R\\ &=1.34-(-1.34)=2.68p.u.\\ \text{At }G_1: \; Q_{G1}&=Q_{load}+Q_S\\ &=5+1.34=6.34p.u.\\ \text{At }G_2: \; Q_{G2}&=Q_{load}-(-Q_R)\\ &=10-(-1.34)=11.34p.u. \end{aligned}$ supplies only 15 pu. Remaining supplied by $G_1$ through transmission line.

\begin{aligned} P_S&=\left | \frac{V_SV_R}{X_L} \right | \sin \delta \\ 5&=\frac{1 \times 1}{0.1} \sin \delta \\ \delta &=30^{\circ}\\ Q_S&=\frac{V_S^2}{X_L}-\frac{V_SV_R}{X_L} \cos \delta \\ &=\frac{1^2}{0.1}-\frac{1 \times 1 }{0.1} \cos 30^{\circ}\\ &=1.34 p.u.\\ Q_R&=\left | \frac{V_SV_R}{X_L} \right | \cos \delta -\left | \frac{V_R^2}{X_L} \right | \\ 5&=\frac{1 \times 1}{0.1} \cos 30^{\circ}-\frac{1^2}{0.1} \\ &=-1.34p.u.\\ Q_{loss}&+Q_S-Q_R\\ &=1.34-(-1.34)=2.68p.u.\\ \text{At }G_1: \; Q_{G1}&=Q_{load}+Q_S\\ &=5+1.34=6.34p.u.\\ \text{At }G_2: \; Q_{G2}&=Q_{load}-(-Q_R)\\ &=10-(-1.34)=11.34p.u. \end{aligned}

Q3GATE 2017MCQ

A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that $1 kW \leq P \leq 2kW$ and $1{kVAR} \leq Q \leq {kVAR}$ . A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is

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📖 Explanation

Load p.f.

\begin{aligned} \cos \phi &=\cos \left[ \tan ^{-1}\frac{Q}{P} \right]\$\\ &=\cos \left[ \tan^{-1}\frac{1}{1} \right]\$\\ &=0.707 \text{ lag}\\ Q_c&=P[\tan \phi _1 -\tan \phi _2]\\ 1&=(\tan \cos ^{-1} 0.707- \tan \phi _2)\\ \phi _2&=45^{\circ}\\ \Rightarrow \; \cos \phi _2&=0.707 \text{ lag} \end{aligned}

Q4GATE 2016MCQ

The inductance and capacitance of a 400 kV, three-phase, 50 Hz lossless transmission line are 1.6 mH/km/phase and 10 nF/km/phase respectively. The sending end voltage is maintained at 400 kV. To maintain a voltage of 400 kV at the receiving end, when the line is delivering 300 MW load, the shunt compensation required is

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📖 Explanation

\begin{aligned} Z_n&=\sqrt{\frac{L}{C}} \\ &= \sqrt{\frac{1.6 \times10^{-3}}{10 \times 10^{-9}}}\\ &= 400\Omega \\ \text{SIL} &=\frac{400 \times 400}{400}=400 MW \end{aligned}

In the second case SIL decreases means $Z_n$ increasess. $Z_n$ increases with increase in inductance 'L'. So, it is inductive. Load $\lt$ SIL means, line behaves capacitive to compensate it inductor to be placed.

Q5GATE 2014NAT

The complex power consumed by a constant-voltage load is given by ( $1 kW \leq P1 \leq 1.5 kW$ ), where, $1 kW \leq P1 \leq 1.5 kW$ and $0.5 kVAR \leq Q_1 \leq 1 kVAR$ . A compensating shunt capacitor is chosen such that $|Q| \leq 0.25 kVAR$ , where Q is the net reactive power consumed by the capacitor load combination. The reactive power (in kVAR) supplied by the capacitor is ____.

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📖 Explanation

\begin{aligned} \text{Given, }S_{load}&=P_1+jQ_1 \\ 1kW \leq P_1 &\leq1.5 kW \\ 0.5 \text{ kVAR} \leq Q_1 &\leq 1\text{ kVAR} \end{aligned}

Shunt capacitor will only supply reactive power and the total reactive power consumed by the load and the capacitor together will be less than 0.25 kVAR i.e. $Q \leq 0.25\text{ kVAR}$ Therefore, reactive power supplied by the capacitor will be $Q _C=(1-0.25)=0.75 \text{ kVAR (lagging)}$

Q6GATE 2014MCQ

Shunt reactors are sometimes used in high voltage transmission systems to

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Q7GATE 2012MCQ

For the system below, $S_{D1} \; and \; S_{D2}$ are complex power demands at bus 1 and bus 2 respectively. If $|V_2|=1$ pu, the VAR rating of the capacitor ( $Q_{G2}$ ) connected at bus 2 is

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📖 Explanation

\begin{aligned} &\text{Real power,} \\ P_r&= \frac{|V_S||V_R|}{|X} \sin \delta \\ 1 &=\frac{1.0 \times 1.0}{0.5}\sin \delta \\ \delta &= \sin^{-1} (0.5)=30^{\circ}\\ &\text{Reactive power,} \\ Q_r&= \frac{|V_S||V_R|}{|X} \cos \delta -\frac{|V|^2}{|X|}\\ &= \frac{1.0 \times 1.0}{0.5}\cos 30-\frac{1^2}{0.5} \\ &= \frac{\left ( \frac{\sqrt{3}}{2} \right )}{\frac{1}{2}}-2\\ &= 1.732-2=-0.268\\ \text{But, } & Q_r+Q_C=0\\ Q_C&=-Q_r=0.268\;p.u. \end{aligned}

Q8GATE 2011MCQ

For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a

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📖 Explanation

Series capacitance compensation reduces reactance of the line. Power flow in line $\propto \frac{1}{\text{reactance of the line}}$ As reactance of line decreases, power flow in the line ${C_a}$ increases.

Q9GATE 2009MCQ

Match the items in List-I (To) with the items in the List-II (Use) and select the correct answer using the codes given below the lists.

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📖 Explanation

Shunt capacitor are used to provide part of the reactive VAR's required by the load to keep the voltage within desirable limits and to improve factor. Series reactor reduce current ripple. Shunt reactors are used across capacitive loads or ligjtly loaded lines to absorb some of the leading VARs to control the voltage across the load to within certain desirable limit. Series capacitor compensation reduces the series impedance of the line. Power flow in line $\propto \frac{1}{X_L}$ , power flow in line increases, as ${X_L}$ decreases.

Q10GATE 2008MCQ

A loss less transmission line having Surge Impedance Loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be

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📖 Explanation

Surge impedance loading= SIL =2280 MW

\begin{aligned} Z_c&= \text{Surge impedance}\\ \text{SIL}_{2}&=\frac{\text{SIL}_{1}}{\sqrt{1-\gamma _{se}}}\\ \text{SIL}_{2}&=\frac{2280}{\sqrt{1-0.3}}\\ &=2725.12 \text{ MW} \end{aligned}

Q11GATE 2007MCQ

The figure below shows a three phase self-commutated voltage source converter connected to a power system. The converter's dc bus capacitor is marked as C in the figure. The circuit in initially operating in steady state with $\delta$ =0 and the capacitor dc voltage is equal to $V_{dc0}$ . You may neglect all losses and harmonics. What action should be taken to increase the capacitor dc voltage slowly to a new steady state value.

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📖 Explanation

To increase the capacitor dc voltage, energy has to be supplied by the generator to capacitor through converter. Power flow from higher angle to lower angle. As angle of induced emf of generator is 0, so $\delta$ should be made negative to make the power flow in required direction.

Q12GATE 2007MCQ

A 230 V (Phase), 50 Hz, three-phase, 4-wire, system has a phase sequence ABC. A unity power-factor load of 4 kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and a pure capacitor in the other two phases. The value of inductor and capacitor is

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📖 Explanation

Taking $V_{AN}$ as the reference

\begin{aligned} V_{AN}&=230\angle 0^{\circ} V\\ V_{BN}&=230\angle -120^{\circ}V \\ V_{CN}&=230\angle 120^{\circ}V \\ I_A&= \frac{P}{V_{AN} \cos \phi }\\ &= \frac{4 \times 10^3}{230 \times 1}\\ &=17.4\angle 0^{\circ} \end{aligned}

If an pure inductor is present in phase B, then $I_B$ lags $V_{BN}$ by $90^{\circ}$ . If a pure capacitor is present in phase C, then $I_C$ leads $V_{CN}$ by $90^{\circ}$ . If current through neutral is to be zero.

\begin{aligned} I_B \sin 30^{\circ} &=I_C \sin 30^{\circ} \\ \Rightarrow \; I_B &=I_C=I \\ I_A&=I_B \cos 30^{\circ} +I_C \cos 30^{\circ} \\ &=2I \cos 30^{\circ}=\sqrt{3}I \\ |I_B|&=|I_C|=|I|=\frac{|I_A|}{\sqrt{3}} \\ &= \frac{17.04}{\sqrt{3}}=10.05\\ |I_B|&=\frac{|V_{BN}|}{\omega L} \\ 10.05 &=\frac{230}{2 \pi \times 50 \times L} \\ L&=72.847 mH \\ |I_C| &=|V_{CN}| \cdot \omega C\\ 10.05&=230 \times 2 \times \pi \times 50 \times C \\ C&=139.087 \mu F \end{aligned}

Q13GATE 2006MCQ

A 400 V, 50 Hz, three phase balanced source supplies power to a star connected load whose rating is 12 $\sqrt{3}$ kVA, 0.8 pf (lag). The rating (in kVAR) of the delta connected (capacitive) reactive power bank necessary to bring the pf to unity is

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📖 Explanation

Rating of the load $12\sqrt{3}$ at 0.8 pf lag Active power demand of the load

\begin{aligned} P_L&=12\sqrt{3} \times 0.8=16.63kW\\ \cos \theta &=0.8\\ \Rightarrow \;\sin \theta &=0.6 \end{aligned}

Reactive power demand of the laod

\begin{aligned} Q_L&=12\sqrt{3} \sin \theta \\ &=12\sqrt{3} \times 0.6\\ Q_L&=12.47 \text{ kVAR} \end{aligned}

For unity pf, the total reactive power is zero. Assuming kVAR rating of capacitor bank $=Q_C$

\begin{aligned} &Q_C+Q_L=0\\ &Q_C=-Q_L=-12.47\text{ kVAR} \end{aligned}

So, capacitor banks supplies 12.47 KVAR reactive power to the load.

Q14GATE 2005MCQ

At an industrial sub-station with a 4 MW load, a capacitor of 2 MVAR is installed to maintain the load power factor at 0.97 lagging. If the capacitor goes out of service, the load power factor becomes

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📖 Explanation

\begin{aligned} Q_C &=P_1(\tan \phi _1 - \tan \phi _2) \\ 2&=4(\tan \phi _1 - \tan (\cos ^{-1} 0.97))\\ \cos \phi &=0.8 \; lag \end{aligned}

Q15GATE 2003MCQ

A 3-phase 11 kV generator feeds power to a constant power unity power factor load of 100 MW through a 3-phase transmission line. The line-to line voltage at the terminals of the machine is maintained constant at 11 kV. The per unit positive sequence impedance of the line based on 100 MVA and 11 kV is j0.2. The line to line voltage at the load terminals is measured to be less than 11 kV. The total reactive power to be injected at the terminals of the load to increase the line-to-line voltage at the load terminals to 11 kV is

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📖 Explanation

$|V_s|=|V_R|=11kV$ p.u. impedance of the line = j0.2 p.u. Impedance of the line = p.u. impedance x Base impedance $x= j0.2 \times \frac{11^2}{100}=j0.242\Omega$ Active power at receiving end $=P_R=100 MW$

\begin{aligned} P_R&=\frac{|V_s||V_R|}{x}\sin \delta \\ 100 &= \frac{11 \times 11}{0.242} \sin \delta \\ \delta &=11.54^{\circ} \end{aligned}

Reactive power at receiving end

\begin{aligned} Q_R&=\frac{|V_s||V_R|}{x}\cos \delta -\frac{|V_R|^2}{x} \\ &= \frac{11 \times 11}{0.242} \cos 11.54 -\frac{11^2}{0.242} \\ &=-10.1 \text{ MVAR} \end{aligned}

Since poer factor =1 Load reactive power should be zero Therefore, reactive power to be injected at terminals $=-Q_R=-(-10.1)=10.1 \text{ MVAR}$

Q16GATE 2003MCQ

A balanced delta connected load of (8 + j6) $\Omega$ per phase is connected to a 400 V, 50 Hz, 3-phase supply lines. If the input power factor is to be improved to 0.9 by connecting a bank of star connected capacitor the required kVAR of the of the bank is

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📖 Explanation

\begin{aligned} Q_C &=\frac{P_1(\tan \phi _1 - \tan \phi _2)}{3(2 \pi f V_{ph}^2)}\;\;...(i) \\ P_1&=\frac{3V_L^2}{Z} \cos \phi _1 \text{ for } \Delta \text{ load}\\ &\text{Star capacitor bank,} \\ V_{ph}&=\frac{V_L}{\sqrt{3}} \\ \text{where, } V_L&=400V \\ Z&=8+j6,\; |Z|=10 \\ \phi _1&=\tan ^{-1}(6/8) \\ \cos \phi _2 &=0.9 \\ &\text{Substitute \in equation (i)} \\ Q_C&=10.2 \text{ kVAR} \end{aligned}

Q17GATE 2001MCQ

A lossless radial transmission line with surge impedance loading

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Q18GATE 2001MCQ

Consider the model shown in figure of a transmission line with a series capacitor at its mid-point. The maximum voltage on the line is at the location

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📖 Explanation

\begin{aligned} V_{P_2}&=1 pu-1 pu \times j 0.1pu \\ V_{P_3}&=1 pu+1 pu \times j 0.05pu \\ V_{P_4}&=1 pu-1 pu \times j 0.05pu \\ \Rightarrow \;& P_3 \text{ is maximum.} \end{aligned}

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