# MCQs on Power Systems

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01․ The inertia constant of 100 MVA, 50 Hz, 4 pole generator is 10 MJ/MVA. If the mechanical input to the machine is suddenly raised from 50 MW to 75 MW, the rotor acceleration will be equal to
12.5 electrical degrees/s²
125 electrical degrees/s²
225 electrical degrees/s²
22.5 electrical degrees/s²

Rotor acceleration α = Pa/M Where, Pa = Accelerating power M = Angular momentum = HS/(180*f) H = Inertia constant S = Rating of the machine f = Frequency M =10*100/(180*50) Rotor acceleration α = 25*180*50/1000 = 225 electrical degrees/s²

02․ If a generator of 250 MVA rating has an inertia constant of 6 MJ/MVA, its inertia constant on 100 MVA base is
6 MJ/MVA
15 MJ/MVA
2.5 MJ/MVA
10.5 MJ/MVA

Inertia constant (H) in pu = H/Sb Where, Sb = Base MVA Hpun = Hpuo* (MVA)bo/(MVA)bn Here, "n" represents the new values "o" represents the old values Hpun = 6*250/100 = 15 MJ/MVA

03․ If the inertia constant H = 10 MJ/MVA for a 50 MVA generator, the stored energy is
5 MJ
50 MJ
500 MJ
10 MJ

Inertia constant H = Energy stored / MVAbasebase Energy stored = 10 * 50 = 500 MJ

04․ Equal area criterion gives the information regarding
stability region
absolute stability
relative stability
swing curves

Equal area criterion used to transient stability analysis of power system, is consistency with transient energy function method for single machine-infinite bus system or two machines system. Equal area criteria is used more efficiently for a single generator stability analysis or single motor stability analysis and is is used to find the absolute stability of the system.

05․ Steady state stability of a power system is improved by
single pole switching
decreasing generator inertia
using double circuit line instead of single circuit line
all of the above

Steady state stability of a power system can be increased by using 1. Parallel transmission lines 2. Bundled conductors 3. Series capacitors

06․ The sequence components of the fault current are as follows: IR1 = j1.5 pu, IR2 = -j0.5 pu, IR0 = -j1. The type of fault is
LG
LL
LLL
LLG

For LLG fault, IR1 + IR2 + IR0 = 0

07․ The pu parameters for a 500 MVA machine on its own base are: Inertia, H = 20 pu; reactance X = 2 pu. The pu values of inertia and reactance on 100 MVA common base, respectively, are
4, 0.4
100, 10
4, 10
100, 0.4

Inertia constant (H) in pu = H/Sb Where, Sb = Base MVA Hpun = Hpuo* (MVA)bo/(MVA)bn Here, "n" represents the new values "o" represents the old values Hpun = 20*500/100 = 100 pu Zpun = Zpuo * (MVA)bn/(MVA)bo Zpun = 2*100/500 = 0.4 pu

08․ An isolated synchronous generator with transient reactance equal to 0.1 pu on a 100 MVA base is connected to the high voltage bus through a step up transformer of reactance 0.1 pu on a 100 MVA base. The fault level at the bus is
100 MVA
200 MVA
500 MVA
400 MVA

The fault level at the bus = MVAbase/Xpu The fault level at the bus = 100/0.2 = 500 MVA

09․ An over current relay, having a current setting of 125% connected to a supply circuit through a current transformer of ratio 400/5. The pick up value is
15 A
10 A
6.25 A
12.5 A

Relay pick up current = % current setting * current transformer secondary current Relay pick up current = 1.25*5 = 6.25 A

10․ An over current relay, having a current setting of 125% connected to a supply circuit through a current transformer of ratio 500/5. The plug setting multiplier for a fault current of 5 kA is
6
5
2
8

Plug setting multiplier = Fault current/(Pick up current * CT ratio) Relay pick up current = % current setting * current transformer secondary current = 1.25*5 = 6.25 A Plug setting multiplier = 5*1000/(6.25*500/5) = 8

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