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PAPERElectrical EngineeringPOWER SYSTEMSCopyright By NODIA &
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- EEPOWER SYSTEMS www.nodia.co.inYEAR 2013TWO MARKSQ.
1Forapowersystemnetworkwithnnodes,Z33ofitsbusimpedancematrixis .
j05 per unit. The voltage at node 3 is. 13 10c -per unit. If a
capacitor having reactance of. j35 -per unit is now added to the
network between node 3 and the reference node, the current drawn by
the capacitor per unit is(A). 0325 100c - (B). 032580c(C). 0371
100c - (D). 043380cStatement for Linked Answer Questions: 2 and 3In
the following network, the voltage magnitudes at all buses are
equal to 1 pu, the voltage phase angles are very small, and the
line resistances are negligible. All the line reactances are equal
to1 j WQ. 2The voltage phase angles in rad at buses 2 and 3 are(A).
012q =- ,. 023q =- (B)02q = ,. 013q =-(C). 012q = ,. 013q = (D).
012q = ,. 023q =Q. 3If the base impedance and the line-to line base
voltage are 100ohms and 100kV respectively, then the real power in
MW delivered by the generator connected at the slack bus is(A)10 -
(B) 0(C) 10(D) 20YEAR 2012ONE MARKQ. 4The bus admittance matrix of
a three-bus three-line system isY
j1310510181051013=---RTSSSSVXWWWWIf each transmission line between
the two buses is represented by an equivalentpnetwork, the
magnitude of the shunt susceptance of the line connecting bus 1 and
2 is(A) 4(B) 2(C) 1(D) 0GATE SOLVED PAPER - EEPOWER SYSTEMS
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Free* *Maximum Discount* www.nodia.co.inQ.
5Atwo-phaseloaddrawsthefollowingphasecurrents:( ) ( ), sin i t I tm
1 1w f = -( ) ( ) cos i t I tm 2 2w f = - . These currents are
balanced if 1fis equal to.(A) 2f - (B) 2f(C) ( / ) 22p f - (D) ( /
) 22p f +Q. 6Thefigureshowsatwo-generatorsystemapplyingaloadof40MW
PD = , connected at bus 2.The fuel cost of generators G1 and G2 are
:( ) 10000 / Rs MWh C PG 1 1=and( ) 12500 / Rs MWh C PG 2 2=and the
loss in the line is. , P P 05( ) ( ) loss pu G pu 12=where the loss
coefficient is specified in pu on a 100MVA base. The most economic
power generation schedule in MW is(A), P P 20 22G G 1 2= = (B), P P
22 20G G 1 2= =(C), P P 20 20G G 1 2= = (D), P P 0 40G G 1 2= =Q.
7Thesequencecomponentsofthefaultcurrentareasfollows:1.5 , pu I
jpositive=0.5 , pu I jnegative=- 1pu I jzero=- . The type of fault
in the system is(A) LG(B) LL(C) LLG(D) LLLGYEAR 2012TWO MARKSQ.
8For the system below,and S SD D 1 2 are complex power demands at
bus 1 and bus 2 respectively. If1pu V2= , the VAR rating of the
capacitor ( ) QG2 connected at bus 2 is(A) 0.2pu(B) 0.268pu(C)
0.312pu(D) 0.4puQ. 9A cylinder rotor generator delivers 0.5pu power
in the steady-state to an infinite bus through a transmission line
of reactance 0.5pu. The generator no-load voltage is 1.5pu and the
infinite bus voltage is 1pu. The inertia constant of the generator
is 5MW- / s MVA and the generator reactance is 1pu. The critical
clearing angle, in degrees, for a three-phase dead short circuit
fault at the generator terminal is(A) 53.5(B) 60.2(C) 70.8(D)
79.6GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inBuy Online
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www.nodia.co.inYEAR 2011ONE MARKQ. 10A 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 optionsQ. 11A negative sequence relay is commonly
used to protect(A) an alternator(B) an transformer(C) a
transmission line(D) a bus barQ. 12For enhancing the power
transmission in along EHV transmission line, the most preferred
method is to connect a(A) Series inductive compensator in the
line(B) Shunt inductive compensator at the receiving end(C) Series
capacitive compensator in the line(D) Shunt capacitive compensator
at the sending endYEAR 2011TWO MARKSQ. 13A load center of 120 MW
derives power from two power stations connected by 220 kV
transmission lines of 25 km and 75 km as shown in the figure below.
The three generators, G G1 2 and G3 are of 100 MW capacity each and
have identical fuel cost characteristics. The minimum loss
generation schedule for supplying the 120 MW load is(A) MW
lossesMWMWPPP802020123= +==(B) MWMW lossesMWPPP603030123== += GATE
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www.nodia.co.in(C) MWMWMW lossesPPP404040123=== +(D) MW
lossesMWMWPPP304545123= +==Q.
14Thedirectaxisandquadratureaxisreactancesofasalientpolealternatorare
1.2p.uand1.0p.urespectively.Thearmatureresistanceisnegligible.Ifthis
alternatorisdeliveringratedkVAatupfandatratedvoltagethenitspower
angle is(A) 30c(B) 45c(C) 60c(D) 90cQ. 15A three bus network is
shown in the figure below indicating the p.u. impedance of each
element.The bus admittance matrix, Y -bus, of the network is(A)
..0.....j0302020120080008002--RTSSSSVXWWWW(B)....j1550575125012525---RTSSSSVXWWWW(C)
.......j01020020120080008010 --RTSSSSVXWWWW(D)...
j1050575125012510--RTSSSSVXWWWWStatement For Linked Answer
Questions : 13 & 14Q. 16Two generator units G1 and G2 are
connected by 15 kV line with a bus at the mid-point as shown
below250MVA G1= , 15 kV, positive sequence reactance25% XG1=on its
own base100MVA G2= , 15 kV, positive sequence reactance10% XG2=on
its own base L1 and10km L2= , positive sequence reactance0.225 / km
XLW =GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inBuy Online
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www.nodia.co.inQ. 17In the above system, the three-phase fault MVA
at the bus 3 is(A) 82.55 MVA(B) 85.11 MVA(C) 170.91 MVA(D) 181.82
MVAYEAR 2010ONE MARKQ. 18Power is transferred from system A to
system B by an HVDC link as shown in the figure. If the voltage VAB
and VCD are as indicated in figure, and I 0 2 , then(A)0, 0, V V V
V < >AB CD AB CD1 (B)0, 0, V V V VAB CD AB CD2 2 1(C)0, 0, V
V V V >AB CD AB CD2 2 (D)0, 0 V V