Question 1
Suppose $I_{A},\:I_{B}$ and $I_{C}$ are a set of unbalanced current phasors in a three-phase system. The phase-B zero-sequence current $I_{B0}=0.1\angle 0^{\circ}$ p.u. If phase-A current $I_{A}=1.1\angle 0^{\circ}$ p.u and phase-C current $I_{C}=\left ( 1\angle 120^{\circ}+0.1 \right )$ p.u, then $I_{B}$ in p.u is
 A $1\angle 240^{\circ}-0.1\angle 0^{\circ}$ B $1.1\angle 240^{\circ}-0.1\angle 0^{\circ}$ C $1.1\angle -120^{\circ}+0.1\angle 0^{\circ}$ D $1\angle -120^{\circ}+0.1\angle 0^{\circ}$
GATE EE 2021   Power Systems
Question 1 Explanation:
\begin{aligned} I_{B O} &=\frac{1}{3}\left(I_{A}+I_{B}+I_{C}\right) \\ 0.1 &=\frac{1}{3}\left(1.1 \angle 0+I_{B}+1 \angle 120^{\circ}+0.1\right) \\ I_{B} &=0.3-1.1 \angle 0-0.1-1 \angle 120^{\circ} \\ I_{B} &=-0.9-1 \angle 120^{\circ} \\ I_{B} &=0.1+1 \angle 240^{\circ} \\ I_{B} &=1 \angle-120^{\circ}+0.1 \angle 0^{\circ} \end{aligned}
 Question 2
A 3-Bus network is shown. Consider generators as ideal voltage sources. If rows $1,\:2$ and $3$ of the $Y_{Bus}$ matrix correspond to Bus $1,\:2$ and $3$, respectively, then $Y_{Bus}$ of the network is

 A $\begin{bmatrix} -4j & j & j\\ j& -4j & j\\ j& j & -4j \end{bmatrix}$ B $\begin{bmatrix} -4j & 2j & 2j\\ 2j& -4j & 2j\\ 2j& 2j & -4j \end{bmatrix}$ C $\begin{bmatrix} - \frac{3}{4}j& \frac{1}{4}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& -\frac{3}{4}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& \frac{1}{4}j & -\frac{3}{4}j \end{bmatrix}$ D $\begin{bmatrix} - \frac{1}{2}j& \frac{1}{4}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& -\frac{1}{2}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& \frac{1}{4}j & -\frac{1}{2}j \end{bmatrix}$
GATE EE 2021   Power Systems
Question 2 Explanation:

\begin{aligned} V_{1} &=j 1 i_{1}+j 1\left(i_{1}+i_{2}+i_{3}\right) \\ V_{1} &=2 j i_{1}+j i_{2}+j i_{3} \\ \text{and}\qquad \qquad V_{2} &=j i_{1}+2 j i_{2}+j i_{3} \\ V_{3} &=j i_{1}+j i_{2}+2 j i_{3} \\ \therefore \qquad \left[\begin{array}{l} V_{1} \\ V_{2} \\ V_{3} \end{array}\right]&=\left[\begin{array}{ccc} 2 j & j & j \\ j & 2 j & j \\ j & j & 2 j \end{array}\right]\left[\begin{array}{l} i_{1} \\ i_{2} \\ i_{3} \end{array}\right]\\ Z_{\text {bus }}&=\left[\begin{array}{ccc} 2 j & j & j \\ j & 2 j & j \\ j & j & 2 j \end{array}\right] \\ Y_{\text {bus }}&=\left[\begin{array}{ccc} \frac{-3 j}{4} & \frac{j}{4} & \frac{j}{4} \\ \frac{j}{4} & \frac{-3 j}{4} & \frac{j}{4} \\ \frac{j}{4} & \frac{j}{4} & \frac{-3 j}{4} \end{array}\right] \end{aligned}
 Question 3
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is $N_{L}$. The balanced Newton-Raphson method is used to carry out load flow study in polar form. $\text{H, S, M, and R}$ are sub-matrices of the Jacobian matrix J as shown below:
$\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}$
The dimension of the sub-matrix M is
 A $N_{L}\times \left ( N-1 \right )$ B $\left ( N-1 \right )\times \left ( N-1-N_{L} \right )$ C $N_{L}\times \left ( N-1+N_{L} \right )$ D $\left ( N-1 \right )\times \left ( N-1+N_{L} \right )$
GATE EE 2021   Power Systems
Question 3 Explanation:
$\left[\begin{array}{l} \Delta P \\ \Delta Q \end{array}\right]=J\left[\begin{array}{l} \Delta \delta \\ \Delta V \end{array}\right] \text { where } J=\left[\begin{array}{ll} H & S \\ M & R \end{array}\right]$
For size of M
Row = No. of unknown variables of $Q=N_{L}$
Column = No. of variable which has $\delta=N_{L}+\left(N-1-N_{L}\right)$
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
 Question 4
Bus 1 with voltage magnitude $V_1 = 1.1 p.u.$ is sending reactive power $Q_{12}$ towards bus 2 with voltage magnitude $V_2 = 1 p.u.$ through a lossless transmission line of reactance X. Keeping the voltage at bus 2 fixed at 1 p.u., magnitude of voltage at bus 1 is changed, so that the reactive power $Q_{12}$ sent from bus 1 is increased by 20%. Real power flow through the line under both the conditions is zero. The new value of the voltage magnitude, $V_1$, in p.u. (rounded off to 2 decimal places) at bus 1 is _______ .
 A 0.118 B 1.12 C 1 D 0.82
GATE EE 2020   Power Systems
Question 4 Explanation:
With real power zero, load angle $\delta =0$
with initial values, $V_{1}=1.1, \; \; \; V_{2}=1$
$Q_{12}=\frac{V_{1}^{2}}{X}-\frac{V_{1}V_{2}}{X}\sin \delta$
$\, \, =\frac{(1.1)^{2}}{X}-\frac{1.1\times 1}{X}\sin 0=\frac{0.11}{x}$

With increased value of voltage,
new value of $\; Q_{12}=1.2Q_{12},\; \; V_{2}=1$
$1.2\, Q_{12}=\frac{V_{1}^{2}}{X}-\frac{V_{1}\times 1}{X}=1.2\times \frac{0.11}{X}$
$V_{1}^{2}-V_{1}-0.132=0$
$V_{1}=1.12,\; \; -0.118$
Hence the practical value in per unit,$V_{1}=1.12$ p.u.
 Question 5
Out of the following options, the most relevant information needed to specify the real power (P) at the PV buses in a load flow analysis is
 A solution of economic load dispatch B rated power output of the generator C rated voltage of the generator D base power of the generator
GATE EE 2020   Power Systems
Question 5 Explanation:
Most relevant information needed to specify P at PV buses is solution of economic load dispatch.
 Question 6
The $Y_{bus}$ matrix of a two-bus power system having two identical parallel lines connected between them in pu is given as
$Y_{bus}=\begin{bmatrix} -j8 &j20 \\ j20 & -j8 \end{bmatrix}$
The magnitude of the series reactance of each line in pu (round off up to one decimal place) is ___________
 A 0.5 B 0.1 C 0.3 D 0.2
GATE EE 2019   Power Systems
Question 6 Explanation:
$Y_{12}=-(y_{12})=-j20$
$=\frac{Y_{12}}{2}=\frac{-j20}{2}=-j10$
Series reaactance of each line $=\frac{1}{-j10}=j0.1 \; p.u.$
 Question 7
The per-unit power output of a salient-pole generator which is connected to an infinite bus, is given by the expression, $P= 1.4 \sin\delta + 0.15 \sin 2 \delta$, where $\delta$ is the load angle. Newton-Raphson method is used to calculate the value of $\delta$ for P = 0.8 pu. If the initial guess is 30$^{\circ}$, then its value (in degree) at the end of the first iteration is
 A 15$^{\circ}$ B 28.48$^{\circ}$ C 28.74$^{\circ}$ D 31.20$^{\circ}$
GATE EE 2018   Power Systems
Question 7 Explanation:
\begin{aligned} P(\delta )&=1.4 \sin \delta +0.15 \sin 2\delta =0.8\\ &=1.4 \sin \delta +0.15 \sin 2\delta -0.8=0\\ P'(\delta )&=\frac{d}{d\delta }(P(\delta ))\\ &=1.4 \cos \delta +0.30 \cos 2\delta \text{Given, } \delta _0&=30^{\circ} \end{aligned}
By using Newton Raphson method for single variable,
\begin{aligned} \Delta \delta &=\delta _1-\delta _0=-\frac{f(\delta _0)}{f'(\delta_0)}=-\frac{P(\delta _0)}{P'(\delta_0)}\\ &=\delta _1 -30^{\circ}=-\frac{1.4 \sin 30^{\circ}+0.15 \sin 60^{\circ} -0.8}{1.4 \cos 30^{\circ}+0.3 \cos 60^{\circ}}\\ &=\delta _1 -30^{\circ}=-0.0219 \; red =-1.26^{\circ}\\ &=\delta _1 =28.74^{\circ} \end{aligned}
 Question 8
A 1000 x 1000 bus admittance matrix for an electric power system has 8000 non-zero elements. The minimum number of branches (transmission lines and transformers) in this system are _____ (up to 2 decimal places).
 A 4000 B 6000 C 3500 D 7000
GATE EE 2018   Power Systems
Question 8 Explanation:
Number of buses =1000
No. of non-zero element = 8000
\begin{aligned} \% \text{ sparsity}&=\% x\\ &=\frac{\text{Number of zero element}}{\text{Total number of elements }}\\ &=\frac{10^6 -8000}{10^6}=0.992\\ \text{NO. of Tr-lines}&=\frac{N^2(1-x)-N}{2}\\ &=\frac{1000^2(1-0.992)-1000}{2}\\ &=3500 \end{aligned}
 Question 9
In a load flow problem solved by Newton-Raphson method with polar coordinates, the size of the Jacobian is 100x100. If there are 20 PV buses in addition to PQ buses and a slack bus, the total number of buses in the system is ________.
 A 63 B 122 C 61 D 59
GATE EE 2017-SET-2   Power Systems
Question 9 Explanation:
\begin{aligned} [J]&=[2n-m-2] \\ 100&= [2n-20-2]\\ n&= 61 \end{aligned}
Total number of buses =n=61
 Question 10
The figure show the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where $\alpha$ is a complex number with non-zero real and imaginary parts.

For the given circuit, $Y_{bus}\; and \; Z_{bus}$ are bus admittance matrix and bus impedance matrix, respectively, each of size 2x2. Which one of the following statements is true?
 A Both $Y_{bus}\; and \; Z_{bus}$ are symmetric B $Y_{bus}$ is symmetric and bus $Z_{bus}$ is unsymmetric C $Y_{bus}$ is unsymmetric and $Z_{bus}$ is symmetric D Both $Y_{bus}\; and \; Z_{bus}$ are unsymmetric
GATE EE 2017-SET-2   Power Systems
Question 10 Explanation:
Both $Y_{BUS}$ and $Z_{BUS}$ are unsymmetrical with transformer.
There are 10 questions to complete.