Question 1
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 1 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 2
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 2 Explanation:
Most relevant information needed to specify P at PV buses is solution of economic load dispatch.
 Question 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 Explanation:
\begin{aligned} [J]&=[2n-m-2] \\ 100&= [2n-20-2]\\ n&= 61 \end{aligned}
Total number of buses =n=61
 Question 7
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 7 Explanation:
Both $Y_{BUS}$ and $Z_{BUS}$ are unsymmetrical with transformer.
 Question 8
The bus admittance matrix for a power system network is
$\begin{bmatrix} -j39.9 & j20 & j20\\ j20 & -j39.9 & j20\\ j20 & j20 & -j39.9 \end{bmatrix}$pu
There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure. If this transmission line is removed from service, What is the modified bus admittance matrix?
 A $\begin{bmatrix} -j19.9 & j20 & 0\\ j20 & -j39.9 & j20\\ 0 & j20 & -j19.9\end{bmatrix}$ B $\begin{bmatrix} -j39.9 & j20 & 0\\ j20 & -j39.9 & j20\\ 0 & j20 & -j39.9 \end{bmatrix}$ C $\begin{bmatrix} -j19.9 & j20 & 0\\ j20 & -j39.9 & j20\\ 0 & j20 & -j19.95 \end{bmatrix}$ D $\begin{bmatrix} -j19.95 & j20 & 0\\ j20 & -j39.9 & j20\\ 0 & j20 & -j19.95 \end{bmatrix}$
GATE EE 2017-SET-1   Power Systems
Question 8 Explanation:
$Y_{bus}=\begin{bmatrix} -j39.9 & j20 & j20\\ j20 & -j39.9 &j20 \\ j20 & j20 & -j39.9 \end{bmatrix}$ Reactance is eliminated in between Bus 1 and 3.
\begin{aligned} Y_{11(new)} &=-j39.9-\frac{1}{j0.05}-j0.05 \\ &=-j19.95 \\ Y_{33(new)} &=-j39.9-\frac{1}{j0.05}-j0.05 \\ &=-j19.95 \\ Y_{13(new)} &=Y_{31}=-y_{13}-\frac{1}{j0.05} \\ Y_{31(new)} &=-j20-\frac{1}{j0.05}=0 \end{aligned}
$=\begin{bmatrix} -j19.95 & j20 &0 \\ j20 & -j39.9 & j20\\ 0 & j20 & -j19.95 \end{bmatrix}\;pu$
 Question 9
A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is _______.
 A 18 B 20 C 14 D 10
GATE EE 2017-SET-1   Power Systems
Question 9 Explanation:
Total number of buses $\equiv 10$
$G_1 \equiv$ slack bus
$G_2 \equiv$ PQ bus (reactive power limit is reached) Minimum number of non linear equation to be sloved = Number of unknown bus voltage variables $=(2 \times 10 -4-2)=14$
 Question 10
A 3-bus power system is shown in the figure below, where the diagonal elements of Y-bus matrix are $Y_{11}=-j12pu$, $\; Y_{22}=-j15pu$ and $Y_{33}=-j7pu$ The per unit values of the line reactances p, q and r shown in the figure are
 A p=-0.2, q=-0.1, r=-0.5 B p=0.2, q=0.1, r=0.5 C p=-5, q=-10, r=-2 D p=5, q=10, r=2
GATE EE 2017-SET-1   Power Systems
Question 10 Explanation:
Given,
\begin{aligned} Y_{11} &=-j12 \; p.u. \\ Y_{22} &=-j15 \; p.u.\\ Y_{33} &=-j7 \; p.u. \\ &\text{We know that,} \\ Y_{11} &=y_{12}+y_{13}=-j12 \; p.u. \;\;...(i) \\ Y_{22} &=y_{12}+y_{23}=-j15 \; p.u. \;\;...(ii) \\ Y_{33} &=y_{13}+y_{23}=-j7 \; p.u. \;\;...(iii) \\ &\text{From eq. (i) and (ii)} \\ y_{13}-y_{23} &=j3\; p.u. \\ y_{13}+y_{23} &=-j7 \; p.u. \\ y_{13}&= -j2\; p.u.\\ y_{23}&= -j5\; p.u.\\ y_{12}&= -j10\; p.u. \end{aligned}
The p.u. values of line reactances p, q and r are
\begin{aligned} jr&=\frac{1}{-j2}=j0.5\; p.u.\\ jp&=\frac{1}{-j5}=j0.2\; p.u.\\ jq&=\frac{1}{-j10}=j0.1\; p.u.\\ \therefore \; p&=0.2, q=0.1, r=0.5 \end{aligned}
There are 10 questions to complete. 