This set of Power Systems Multiple Choice Questions & Answers (MCQs) focuses on “Comparison of Angle and Voltage Stability – 2”.
1. For a given system consisting of two generators
Units Rating(MW) Speed Drop(%R)
1. 400 0.04
2. 800 0.05
The units share load of P1=200MW and P2=600MW. The units are operating in parallel to share a load of 700 MW at 50 HZ. The load is increased by 130MW with B=0. Then the value new steady state frequency variation is ______________
a) 49.75 Hz
b) 51.25 Hz
c) 49.25 Hz
d) 48.75 Hz
View Answer
Explanation:
\(\Delta f = \frac{-\Delta P}{\left( \frac{1}{R_1} + \frac{1}{R_2} + B \right)}\)
On the 1000 MVA base:
\(R_1 = \frac{0.04 \times 1000}{400} = 0.1, \quad R_2 = \frac{0.05 \times 1000}{800} = 0.0625\)
Substitute values into the equation:
\(\Delta f = \frac{-130}{1000} \div \left( \frac{1}{0.1} + \frac{1}{0.0625} + 0 \right) = -0.005 \, \text{p.u.}\)
Now, calculate the final frequency:
\(f = f_0 + \Delta f = 50 + (5 \times 10^{-3} \times 50) = 49.75 \, \text{Hz}\)
2. A 50 bus power system Ybus has 80% sparsity. The total number of transmission lines will be ___________
a) 225
b) 500
c) 475
d) 100
View Answer
Explanation: Number of non zero elements = 50*50*20/100 = 500 non zero
Number of TL = (500-50)/2 = 225 transmission lines.
3. The given graph is the depiction of ________ on a large power system network.
a) L-G fault
b) Three phase motor getting short
c) Fault in feeder
d) Any of the mentioned
View Answer
Explanation: A LG fault will create a sudden dip in the voltage profile and it will be recovered.
4. A single core cable is graded by using three dielectrics with relative permittivity 5,4,3 respectively. The diameter of the conductor is 2cm and the overall diameter is 8 cm. If the three dielectric work at the same maximum stress of 40kV/cm, then the safe working rms voltage of cable is ______ kV
a) 57.72
b) 81.63
c) 84.67
d) 71.63
View Answer
Explanation: d=2cm, d1 = ?, d2= ?, D = 8cm
Ԑ1 = 5; Ԑ2 = 4; Ԑ3 = 3
gmax =40 kV/cm
Ԑ1d1 = Ԑ2d2 = Ԑ3d3
d1 = 2.5cm, d2 = 3.33 cm
\(V_{\text{peak}} = \frac{g_{\text{max}}}{2} \left[ \frac{d \ln d_1}{d} + d_1 \ln \frac{d_2}{d_1} + d_2 \ln \frac{D}{d_2} \right] = 81.63 \, \text{kV}\)
Rms value of the peak voltage = 81.63/√2 = 57.72 kV.
5. A generator delivers power of 1 pu to an infinite bus through a purely reactive network. The maximum power that could be delivered by generator is 2 pu. A three phase fault occurs at the generator which reduces the generator output to zero. The fault is restored after ‘tc’ seconds. The maximum swing of rotor angle is found to be δmax = 110o electrical. The rotor angle at ‘tc’ is ________ electrical deg.
a) 69.14o
b) 159.14o
c) 63.08o
d) 65.7o
View Answer
Explanation: \(\cos \delta_{\text{cr}} = \frac{P_m}{P_{\text{max}}} \left( \delta_{\text{max}} – \delta_0 \right) + \cos (\delta_{\text{max}})\)
\(\delta = \sin^{-1} \left( \frac{P}{P_{\text{max}}} \right) = \sin^{-1} \left( \frac{1}{2} \right) = 30^\circ\)
\(\cos \delta_{\text{cr}} = \frac{1}{2} \left( 110 – 30 \right) \cdot \frac{\pi}{180} + \cos 110^\circ\)
\(\delta_{\text{cr}} = 69.14^\circ\)
6. A system consists of an alternator having reactance of 0.5pu connectedto an infinite bus through a series of reactance of 1 pu. The generator terminal voltage of IBB is 1 pu and that of 1.2 pu. The steady state power system limit (in pu) is ___________
a) 1.152
b) 1.167
c) 1.765
d) 1.729
View Answer
Explanation: \(I = \frac{V_t – V}{jX} = \frac{1.2 \ell \theta – 1}{j1}\)
\(E = V_t + X_d I = 1.2 \ell \theta + j \cdot 0.5 \cdot \frac{1.2 \ell \theta – 1}{j1}\)
=1.2ʟθ+0.6ʟ(θ)-1
= (1.8cosθ-1)+j1.8sinθ
SSL occurs at δ=90o
1.8cosθ-1 = 0
θ= 73.87
|E|=1.729
P=1.729/1.5 = 1.152 pu
7. A system consists of an alternator having reactance of 0.5pu connected to an infinite bus through a series of reactance of 1 pu. The generator terminal voltage of IBB is 1 pu and that of 1.2 pu. The Steady state occurs at power angle of ________degree.
a) 0
b) 90
c) 180
d) 45
View Answer
Explanation: It occurs at δ=90o.
8. A system consists of an alternator having reactance of 0.5pu connectedto an infinite bus through a series of reactance of 1 pu. The generator terminal voltage of IBB is 1 pu and that of 1.2 pu.The emf induced in the alternator for the maximum power transfer will be __________
a) 1.729
b) 1.152
c) 1.2
d) 1.6
View Answer
Explanation: I = \(\frac{(Vt-V)}{jX} = \frac{1.2ʟθ-1}{j1}\)
E=Vt+XdI = 1.2ʟθ+j0.5*\(\frac{1.2ʟθ-1}{j1}\)
=1.2ʟθ+0.6ʟ(θ)-1
= (1.8cosθ-1)+j1.8sinθ
SSL occurs at δ=90o
1.8cosθ-1 = 0
θ= 73.87
|E|=1.729
9. A system consists of an alternator having reactance of 0.5pu connected to an infinite bus through a series of reactance of 1 pu. The generator terminal voltage of IBB is 1 pu and that of 1.2 pu. The emf induced in the alternator will have the phase difference with respect to reference for the maximum power transfer is _______________
a) 90
b) 0
c) 73.87
d) 86.25
View Answer
Explanation: It occurs at δ=90o for the maximum power transfer.
10. A three phase transmission line is having a three unit suspension insulation string. The voltage at the insulator unit nearest to the line is 20kV and that across the adjacent unit is 15 kV. The ratio of mutual to ground capacitance is ________________
a) 0.18
b) 0.2
c) 0.333
d) 0.16
View Answer
Explanation: \(\frac{V_3}{V_2} = \frac{1 + 3k + k^2}{1 + k} = \frac{15}{20}\)
K=0.18
Sanfoundry Global Education & Learning Series – Power Systems.
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