Nepal Engineering Council · Electrical Engineering · Chapter 9
Power System Analysis
Tap an option to check it. Wrong picks show the right answer and the hint.
184 questions in 6 syllabus topics.
9.1 Transmission line parameters
32 questions · AElE0901
1. The geometric mean radius (GMR) of a solid round conductor of radius r, used in inductance calculations, is:
Show hintHide hint
Think of e raised to −1/4.
Show answerHide answer
Answer: A. 0.7788 r
Accounting for internal flux linkage, the solid conductor is replaced by a fictitious hollow conductor of radius r' = r·e^(−1/4) = 0.7788 r.
2. A three-phase line has conductors of radius 1 cm placed at the corners of an equilateral triangle of side 2 m. The inductance per phase is approximately:
Show hintHide hint
Use the GMR, not the physical radius.
Show answerHide answer
Answer: B. 1.11 mH/km
L = 0.2 ln(D/r') mH/km with r' = 0.7788 × 1 cm = 0.7788 cm, so L = 0.2 ln(200/0.7788) ≈ 1.11 mH/km.
3. For the same line (radius 1 cm, equilateral spacing 2 m), the capacitance per phase to neutral is approximately:
Show hintHide hint
Capacitance uses the physical radius.
Show answerHide answer
Answer: C. 0.0105 µF/km
C = 2πε0 / ln(D/r) = 55.6×10⁻¹²/ln(200) F/m ≈ 0.0105 µF/km. Capacitance uses the actual radius, not the GMR.
4. A transposed three-phase line has phase spacings of 3 m, 4 m and 5 m. Its equivalent spacing (GMD) is:
Show hintHide hint
It is a geometric, not arithmetic, mean of three distances.
Show answerHide answer
Answer: A. 3.91 m
GMD = (D12·D23·D31)^(1/3) = (3×4×5)^(1/3) = 60^(1/3) ≈ 3.91 m.
5. A two-conductor bundle uses sub-conductors of GMR 1.2 cm spaced 40 cm apart. The GMR of the bundle is about:
Show hintHide hint
Take the geometric mean of the sub-conductor GMR and the bundle spacing.
Show answerHide answer
Answer: C. 6.93 cm
For a 2-conductor bundle, Ds_b = √(Ds × d) = √(1.2 × 40) ≈ 6.93 cm.
6. Skin effect in an AC conductor becomes more pronounced when:
Show hintHide hint
Recall how skin depth depends on ω and ρ.
Show answerHide answer
Answer: D. Frequency and conductor diameter increase
Skin depth δ = √(2ρ/ωµ) shrinks with frequency, so thick conductors at higher frequency crowd current to the surface; higher resistivity actually weakens skin effect, and DC has none.
7. The non-uniform current distribution in a conductor caused by the alternating magnetic field of a nearby current-carrying conductor is called:
Show hintHide hint
The cause lies in a neighbouring conductor.
Show answerHide answer
Answer: B. Proximity effect
Proximity effect is the crowding of current due to the field of adjacent conductors; skin effect is due to the conductor's own field.
8. The main purpose of transposing the conductors of a three-phase overhead line is to:
Show hintHide hint
Think about unequal spacing between phases.
Show answerHide answer
Answer: A. Equalise the inductance and capacitance of the three phases
With unsymmetrical spacing each phase has a different inductance and capacitance; transposition gives each phase every position over a cycle, balancing the parameters.
9. Which of the following is NOT a result of using bundled conductors on an EHV line?
Show hintHide hint
Bundling raises the effective GMR.
Show answerHide answer
Answer: C. Increase in series inductive reactance
Bundling increases the effective GMR, which lowers inductance, raises capacitance, lowers surge impedance (raising SIL) and lowers surface gradient (less corona).
10. An aluminium conductor (ρ = 2.83 × 10⁻⁸ Ω·m) of cross-section 100 mm² is 10 km long. Its DC resistance is:
Show hintHide hint
Convert mm² to m² carefully.
Show answerHide answer
Answer: C. 2.83 Ω
R = ρl/A = 2.83×10⁻⁸ × 10 000 / (100×10⁻⁶) = 2.83 Ω.
11. Taking the effect of earth into account, the calculated capacitance of an overhead line:
Show hintHide hint
Use the method of images.
Show answerHide answer
Answer: B. Increases slightly
The earth acts like image conductors that bring an equivalent opposite charge closer, so the line capacitance increases slightly.
12. A single-phase two-wire line has conductors of radius 0.5 cm spaced 1 m apart. The loop inductance is approximately:
Show hintHide hint
The loop has two conductors, each contributing 0.2 ln(D/r').
Show answerHide answer
Answer: B. 2.22 mH/km
Loop inductance = 0.4 ln(D/r') mH/km = 0.4 ln(100/0.3894) ≈ 2.22 mH/km.
13. If the spacing between the conductors of an overhead line is increased while the conductor size is kept the same:
Show hintHide hint
Look at where D appears in each formula.
Show answerHide answer
Answer: A. Inductance increases and capacitance decreases
L ∝ ln(D/r') rises with D, while C ∝ 1/ln(D/r) falls with D.
14. The disruptive critical voltage of an overhead line is the:
Show hintHide hint
It is the starting point of corona, before it can be seen.
Show answerHide answer
Answer: B. Minimum phase-to-neutral voltage at which ionisation of air around the conductor begins
Corona starts when the surface gradient reaches the breakdown strength of air; the corresponding voltage is the disruptive critical voltage. The visible glow appears later, at the visual critical voltage.
15. The breakdown strength (dielectric strength) of air at 25 °C and 76 cm Hg, as used in corona calculations, is about:
Show hintHide hint
The rms value is the peak divided by √2.
Show answerHide answer
Answer: D. 30 kV/cm peak (21.1 kV/cm rms)
Air at standard conditions breaks down at g0 ≈ 30 kV/cm peak, i.e. 30/√2 ≈ 21.1 kV/cm rms.
16. A three-phase line has conductors of radius 1 cm with equilateral spacing 2 m. With irregularity factor 0.9 and air density factor 1, the disruptive critical voltage (phase to neutral) is about:
Show hintHide hint
Use g0 = 21.1 kV/cm rms with r in cm.
Show answerHide answer
Answer: D. 100.6 kV rms
Vd = g0·m0·δ·r·ln(D/r) = 21.1 × 0.9 × 1 × 1 × ln(200/1) ≈ 100.6 kV rms per phase.
17. The air density correction factor δ at a barometric pressure of 72 cm Hg and temperature 35 °C is about:
Show hintHide hint
Temperature must be in kelvin.
Show answerHide answer
Answer: D. 0.916
δ = 3.92 b/(273 + t) = 3.92 × 72/308 ≈ 0.916.
18. According to Peek's formula, the corona power loss of a line is proportional to:
Show hintHide hint
The loss rises steeply once corona starts.
Show answerHide answer
Answer: C. (V − Vd)², where V is the operating and Vd the disruptive critical phase voltage
Peek's formula: P = (242.2/δ)(f + 25)√(r/D)(V − Vd)² × 10⁻⁵ kW/km/phase.
19. Which of the following is an advantage of corona on overhead lines?
Show hintHide hint
Think about what happens to a lightning surge travelling on the line.
Show answerHide answer
Answer: D. It reduces the magnitude and steepness of travelling-wave surges
Corona acts like a safety valve: the increased effective radius and losses attenuate surges. It reduces efficiency, causes radio interference, and the ozone produced corrodes conductors.
20. Corona loss on a given overhead line is generally greatest:
Show hintHide hint
What weather lowers the critical voltage?
Show answerHide answer
Answer: A. During rain or fog
Rain drops and fog reduce the critical voltage (by increasing surface irregularity and lowering air strength), so corona loss is much greater in foul weather.
21. Corona on a transmission line can be reduced by:
Show hintHide hint
Lower the surface field strength.
Show answerHide answer
Answer: D. Increasing the conductor diameter or using bundled conductors
A larger effective radius lowers the surface electric gradient, raising the critical voltage; reducing spacing or raising voltage worsens corona.
22. For a given line, the visual critical voltage compared with the disruptive critical voltage is:
Show hintHide hint
Glow appears only after ionisation is well established.
Show answerHide answer
Answer: C. Higher
Ionisation begins at the disruptive critical voltage but a visible glow requires a higher gradient, so Vv = 21.1 mv δ r(1 + 0.3/√(δr)) ln(D/r) > Vd.
23. The gas produced around conductors by corona discharge, which can corrode conductors and fittings, is:
Show hintHide hint
It is also responsible for the characteristic smell near corona.
Show answerHide answer
Answer: A. Ozone
Corona ionises air and produces ozone (and nitrogen oxides which form nitrous acid in moisture), causing corrosion.
24. For a base of 100 MVA (three-phase) and 132 kV (line-to-line), the base impedance is:
Show hintHide hint
Zbase = kV²/MVA.
Show answerHide answer
Answer: D. 174.24 Ω
Zbase = (kV_LL)²/MVA_3φ = 132²/100 = 174.24 Ω.
25. A transformer has a reactance of 0.1 pu on its own rating of 50 MVA, 11 kV. On a common base of 100 MVA, 11 kV its reactance is:
Show hintHide hint
Per-unit impedance is proportional to the base MVA.
Show answerHide answer
Answer: C. 0.20 pu
Znew = Zold × (Snew/Sold) × (Vold/Vnew)² = 0.1 × (100/50) × 1 = 0.20 pu.
26. A generator rated 25 MVA, 11 kV has a reactance of 0.2 pu. Expressed on a base of 100 MVA, 12 kV, the reactance is about:
Show hintHide hint
Scale by the MVA ratio and the inverse square of the kV ratio.
Show answerHide answer
Answer: A. 0.672 pu
Znew = 0.2 × (100/25) × (11/12)² ≈ 0.672 pu.
27. A key advantage of the per-unit system in transformer representation is that:
Show hintHide hint
Bases on the two sides follow the turns ratio.
Show answerHide answer
Answer: B. The per-unit impedance is the same whether referred to the primary or the secondary side
With base voltages in the ratio of the turns ratio, the per-unit leakage impedance is identical on both sides, so the ideal transformer can be removed from the circuit.
28. The base current for a three-phase system with base 100 MVA and 33 kV (line) is about:
Show hintHide hint
Use √3 for a three-phase base.
Show answerHide answer
Answer: A. 1750 A
Ibase = MVA/(√3 kV) = 100×10⁶/(√3 × 33×10³) ≈ 1750 A.
29. In converting an impedance diagram into a reactance diagram for fault studies, which of the following is usually neglected?
Show hintHide hint
Keep only what dominates fault currents.
Show answerHide answer
Answer: B. Resistances, magnetising branches of transformers and static loads
For short-circuit studies the reactance diagram retains only series reactances; resistance, shunt magnetising branches, line charging and static loads are omitted.
30. A single-line diagram of a power system represents a balanced three-phase network by:
Show hintHide hint
Balanced phases are identical apart from 120° shifts.
Show answerHide answer
Answer: D. One line and standard symbols for each component, the neutral being implied
For balanced systems one phase with the neutral return implied is sufficient; components are shown by standard symbols on one line.
31. A line impedance of 20 Ω on a system with base 50 MVA and 66 kV has a per-unit value of about:
Show hintHide hint
First find the base impedance.
Show answerHide answer
Answer: C. 0.230 pu
Zbase = 66²/50 = 87.12 Ω, so Zpu = 20/87.12 ≈ 0.230 pu.
32. In a balanced three-phase system with properly chosen three-phase and line-voltage bases, the per-unit line voltage and per-unit phase voltage are:
Show hintHide hint
The √3 appears in both the actual value and the base.
Show answerHide answer
Answer: B. Equal
The line base is √3 times the phase base, just as the actual line voltage is √3 times the phase voltage, so both have the same per-unit value.
9.2 Performance of transmission line
30 questions · AElE0902
33. In the analysis of a short transmission line (typically below about 80 km), which parameter is neglected?
Show hintHide hint
Charging current is small on short lines.
Show answerHide answer
Answer: A. Shunt capacitance
For short lines the charging current is negligible, so only the series impedance R + jX is modelled.
34. For any passive, linear, bilateral two-port network such as a transmission line, the ABCD constants satisfy:
Show hintHide hint
It is the determinant of the transmission matrix.
Show answerHide answer
Answer: A. AD − BC = 1
Reciprocity of a passive bilateral network gives AD − BC = 1; for a symmetrical network A = D as well.
35. The ABCD parameters of a short transmission line with series impedance Z are:
Show hintHide hint
Sending and receiving currents are equal on a short line.
Show answerHide answer
Answer: D. A = 1, B = Z, C = 0, D = 1
Vs = Vr + Z·Ir and Is = Ir, giving A = D = 1, B = Z, C = 0.
36. For a medium line represented by the nominal-π circuit (series impedance Z, total shunt admittance Y), the constant C is:
Show hintHide hint
In the π model, B is simply Z; the correction factor moves to C.
Show answerHide answer
Answer: B. Y(1 + YZ/4)
Nominal π: A = D = 1 + YZ/2, B = Z, C = Y(1 + YZ/4). (In the nominal-T, C = Y and B = Z(1 + YZ/4).)
37. In the nominal-T representation of a medium transmission line, the total shunt admittance is:
Show hintHide hint
The letter T shows where the shunt branch sits.
Show answerHide answer
Answer: C. Concentrated at the middle of the line
Nominal T places the full Y at the midpoint with Z/2 on each side; nominal π splits Y/2 at each end.
38. For a long transmission line of length l, propagation constant γ and characteristic impedance Zc, the constant C is:
Show hintHide hint
C has the dimension of admittance.
Show answerHide answer
Answer: C. sinh(γl)/Zc
Exact (distributed) model: A = D = cosh γl, B = Zc sinh γl, C = sinh γl / Zc.
39. A lossless line has inductance 1 mH/km and capacitance 0.01 µF/km. Its surge impedance is about:
Show hintHide hint
Zc = √(L/C), with both per unit length.
Show answerHide answer
Answer: B. 316 Ω
Zc = √(L/C) = √(10⁻³/10⁻⁸) ≈ 316 Ω.
40. A 220 kV line has a surge impedance of 400 Ω. Its surge impedance loading (SIL) is about:
Show hintHide hint
Use line voltage squared over Zc.
Show answerHide answer
Answer: D. 121 MW
SIL = (kV_LL)²/Zc = 220²/400 = 121 MW.
41. A lossless overhead line has L = 1.0 mH/km and C = 0.0111 µF/km. The velocity of propagation of a wave on it is about:
Show hintHide hint
v = 1/√(LC), per unit length quantities.
Show answerHide answer
Answer: B. 3.0 × 10⁵ km/s
v = 1/√(LC) = 1/√(10⁻³ × 1.11×10⁻⁸) ≈ 3.0 × 10⁵ km/s, close to the speed of light.
42. The wavelength of a 50 Hz wave on a lossless overhead line (propagation velocity ≈ 3 × 10⁵ km/s) is about:
Show hintHide hint
λ = v/f.
Show answerHide answer
Answer: D. 6000 km
λ = v/f = 3×10⁵/50 = 6000 km.
43. The Ferranti effect in a long transmission line refers to:
Show hintHide hint
Charging current flows through series inductance.
Show answerHide answer
Answer: D. Rise of receiving-end voltage above sending-end voltage at no load or light load
Line charging current flowing through the series inductance produces a voltage rise along the line when the load is light.
44. A 300 km lossless 50 Hz line (wavelength 6000 km) is energised at 220 kV at the sending end with the receiving end open. The receiving-end voltage is about:
Show hintHide hint
At no load, Vs = A·Vr with A = cos βl.
Show answerHide answer
Answer: A. 231 kV
βl = 2π × 300/6000 = 0.314 rad; Vr = Vs/cos βl = 220/0.951 ≈ 231 kV (Ferranti rise).
45. The Ferranti effect becomes more pronounced with:
Show hintHide hint
The approximate rise is proportional to ω²l².
Show answerHide answer
Answer: B. Longer line length and higher supply frequency
The no-load rise is approximately ω²LC·l²/2, so it grows with the square of both frequency and length.
46. The percentage voltage regulation of a transmission line is defined as:
Show hintHide hint
The reference is the full-load receiving-end voltage.
Show answerHide answer
Answer: C. (Vr,no-load − Vr,full-load)/Vr,full-load × 100
Regulation compares the receiving-end voltage at no load with that at full load, expressed as a percentage of the full-load value.
47. A 3-phase, 33 kV short line has R = 5 Ω and X = 10 Ω per phase. It delivers 10 MW at 0.8 pf lagging at 33 kV. The voltage regulation is about:
Show hintHide hint
Per phase, drop ≈ I(R cosφ + X sinφ).
Show answerHide answer
Answer: A. 11.6%
I = 10×10⁶/(√3 × 33×10³ × 0.8) ≈ 219 A; Vs(phase) = |Vr + I(R + jX)| ≈ 21268 V vs Vr = 19053 V; regulation ≈ 11.6% (approx. I(R cosφ + X sinφ)/Vr ≈ 11.5%).
48. For the same 33 kV line (R = 5 Ω per phase) delivering 10 MW at 0.8 pf lagging, the transmission efficiency is about:
Show hintHide hint
Losses are 3I²R for three phases.
Show answerHide answer
Answer: C. 93.3%
Losses = 3I²R = 3 × 219² × 5 ≈ 717 kW; η = 10/(10 + 0.717) ≈ 93.3%.
49. For a short line with resistance R and reactance X, the voltage regulation is approximately zero when the load power factor is:
Show hintHide hint
Set R cosφ − X sinφ to zero.
Show answerHide answer
Answer: C. Leading, with tan φ = R/X
Regulation ≈ I(R cosφ − X sinφ)/Vr for a leading load, which is zero when tan φ = R/X.
50. The voltage regulation of a short line is maximum (approximately) when the load power-factor angle φ is lagging and equal to:
Show hintHide hint
Write R cosφ + X sinφ as |Z| cos(θ − φ).
Show answerHide answer
Answer: B. The line impedance angle, i.e. tan φ = X/R
R cosφ + X sinφ = |Z| cos(θ − φ) is maximum when φ = θ = tan⁻¹(X/R).
51. When a lossless line delivers exactly its surge impedance loading (SIL):
Show hintHide hint
Compare I²ωL with V²ωC at SIL.
Show answerHide answer
Answer: D. The voltage profile is flat and the line's reactive generation equals its reactive absorption
At SIL, I²X = V²B per unit length, so the line is reactively self-sufficient and |V| is constant along it.
52. When a line is loaded well above its SIL, it:
Show hintHide hint
Which grows faster with load: I²X or V²B?
Show answerHide answer
Answer: A. Absorbs reactive power, and voltage tends to sag along the line
Above SIL the I²X absorption exceeds the line charging, so the line is a net reactive load and voltages fall unless supported.
53. The propagation constant of a transmission line is γ = α + jβ = √(zy). The real part α is called the:
Show hintHide hint
It describes amplitude decay.
Show answerHide answer
Answer: A. Attenuation constant
α (nepers/km) gives the decrease in amplitude per unit length; β (rad/km) gives the phase shift.
54. A transmission line terminated in its characteristic impedance has:
Show hintHide hint
Evaluate (ZL − Zc)/(ZL + Zc).
Show answerHide answer
Answer: D. No reflected wave at the receiving end
The reflection coefficient (ZL − Zc)/(ZL + Zc) is zero when ZL = Zc, so the line appears infinitely long.
55. A medium line is modelled by a nominal-π with Z = 10 + j50 Ω and Y = j4 × 10⁻⁴ S. The magnitude of constant A is about:
Show hintHide hint
A = 1 + YZ/2 for the π model.
Show answerHide answer
Answer: B. 0.990
A = 1 + YZ/2 = 1 + (j4×10⁻⁴)(10 + j50)/2 = 0.990 + j0.002, |A| ≈ 0.990.
56. A line has |A| = 0.95. With the sending-end voltage held at 132 kV, the receiving-end voltage at full load is 125 kV. The voltage regulation is about:
Show hintHide hint
No-load receiving voltage is not simply Vs.
Show answerHide answer
Answer: A. 11.2%
At no load Vr = Vs/|A| = 132/0.95 ≈ 138.9 kV; regulation = (138.9 − 125)/125 ≈ 11.2%.
57. When two transmission networks with ABCD matrices T1 and T2 are connected in cascade (T1 first), the overall ABCD matrix is:
Show hintHide hint
Substitute one two-port equation into the other.
Show answerHide answer
Answer: D. T1 × T2 (matrix product in order of connection)
Since [Vs; Is] = T1[Vm; Im] and [Vm; Im] = T2[Vr; Ir], the overall matrix is T1·T2.
58. A 132 kV, 50 Hz, 100 km line has a capacitance to neutral of 0.01 µF/km per phase. The charging current per phase (receiving end open, voltage assumed uniform) is about:
Show hintHide hint
Use phase voltage and total capacitance.
Show answerHide answer
Answer: C. 23.9 A
Ic = Vph·ωC = (132 000/√3) × 314.16 × 1×10⁻⁶ ≈ 23.9 A.
59. The surge impedance of a typical single-circuit HV overhead transmission line lies in the range of about:
Show hintHide hint
Cables are much lower than overhead lines.
Show answerHide answer
Answer: B. 300–500 Ω
Overhead lines typically have Zc ≈ 400 Ω, while underground cables have much lower values of roughly 30–60 Ω.
60. The surge impedance loading of a line can be increased by:
Show hintHide hint
SIL = V²/Zc; think about what lowers Zc.
Show answerHide answer
Answer: C. Using bundled conductors
Bundling lowers L and raises C, reducing Zc = √(L/C) and raising SIL = V²/Zc.
61. A 400 kV line has a surge impedance of 320 Ω. Its surge impedance loading is:
Show hintHide hint
Square the line voltage in kV.
Show answerHide answer
Answer: A. 500 MW
SIL = 400²/320 = 160 000/320 = 500 MW.
62. For a long line, the receiving-end and sending-end quantities are related by the hyperbolic functions of γl. Which is the correct expression for the sending-end voltage?
Show hintHide hint
At no load (Ir = 0) the answer must reduce to Vs = Vr cosh γl.
Show answerHide answer
Answer: B. Vs = Vr cosh(γl) + Ir Zc sinh(γl)
From the distributed-parameter solution, Vs = A·Vr + B·Ir with A = cosh γl and B = Zc sinh γl.
9.3 Fault calculations
31 questions · AElE0903
63. With the symmetrical-component operator a = 1∠120°, the value of 1 + a + a² is:
Show hintHide hint
Add three equal phasors spaced 120° apart.
Show answerHide answer
Answer: A. 0
1, a and a² are three unit phasors 120° apart, so their sum is zero.
64. The operator a² (where a = 1∠120°) is equal to:
Show hintHide hint
a² is a rotation of 240°.
Show answerHide answer
Answer: B. −0.5 − j0.866
a² = 1∠240° = cos240° + j sin240° = −0.5 − j0.866.
65. In a three-phase system Ia = 100∠0° A while Ib = Ic = 0. The magnitude of each of the sequence currents I0, I1 and I2 is:
Show hintHide hint
Each sequence formula divides by 3.
Show answerHide answer
Answer: C. 33.3 A
I0 = (Ia + Ib + Ic)/3, I1 = (Ia + aIb + a²Ic)/3, I2 = (Ia + a²Ib + aIc)/3; with only Ia present each equals 100/3 ≈ 33.3 A.
66. The line currents in a three-phase system are Ia = 10∠0° A, Ib = 10∠180° A and Ic = 0. The magnitude of the positive-sequence current is about:
Show hintHide hint
Find |1 − a| first.
Show answerHide answer
Answer: C. 5.77 A
I1 = (Ia + aIb + a²Ic)/3 = 10(1 − a)/3; |1 − a| = √3, so |I1| = 10√3/3 ≈ 5.77 A. Here I0 = 0.
67. A perfectly balanced three-phase set of currents (equal magnitude, 120° apart, sequence a-b-c) contains:
Show hintHide hint
Compare the given set with the definition of each sequence.
Show answerHide answer
Answer: B. Only positive-sequence components
A balanced abc set is itself a positive-sequence set, so I2 = I0 = 0.
68. For a delta-connected winding, zero-sequence currents:
Show hintHide hint
A delta has no neutral, but it is a closed loop.
Show answerHide answer
Answer: D. Can circulate inside the delta but cannot flow in the connected lines
Zero-sequence currents are in phase in all three phases; there is no return path in the three-wire lines, but they can circulate round the closed delta.
69. A generator neutral is grounded through an impedance Zn. In the zero-sequence network this neutral impedance appears as:
Show hintHide hint
How much current flows in the neutral in terms of I0?
Show answerHide answer
Answer: D. 3Zn in series
The neutral carries 3I0, so the voltage drop is 3I0Zn; in the per-phase zero-sequence network it is represented as 3Zn.
70. A generator has Z1 = j0.2, Z2 = j0.2 and Z0 = j0.1 pu, with a solidly grounded neutral and E = 1.0 pu. The fault current for a bolted single line-to-ground fault at its terminals is:
Show hintHide hint
The sequence networks are in series and the phase current is 3I0.
Show answerHide answer
Answer: A. 6.0 pu
If = 3E/(Z1 + Z2 + Z0) = 3/0.5 = 6.0 pu.
71. For the same generator (Z1 = j0.2, Z2 = j0.2, Z0 = j0.1 pu, E = 1.0 pu), the fault current for a symmetrical three-phase fault at its terminals is:
Show hintHide hint
Only one sequence network matters for a balanced fault.
Show answerHide answer
Answer: A. 5.0 pu
A three-phase fault involves only the positive-sequence network: If = E/Z1 = 1/0.2 = 5.0 pu.
72. For the same generator (Z1 = Z2 = j0.2 pu, E = 1.0 pu), the fault current for a line-to-line fault at its terminals is about:
Show hintHide hint
Positive and negative networks in parallel; phase current is √3 times I1.
Show answerHide answer
Answer: C. 4.33 pu
If = √3 E/(Z1 + Z2) = 1.732/0.4 ≈ 4.33 pu.
73. For the same generator (Z1 = Z2 = j0.2, Z0 = j0.1 pu, E = 1 pu, solidly grounded), a double line-to-ground fault occurs at its terminals. The current flowing into ground is about:
Show hintHide hint
Find I0 by current division between Z2 and Z0, then multiply by 3.
Show answerHide answer
Answer: A. 7.50 pu
I1 = E/(Z1 + Z2‖Z0) = 1/(0.2 + 0.0667) = 3.75 pu; I0 = −I1·Z2/(Z2 + Z0) = −2.50 pu; ground current = 3|I0| = 7.50 pu.
74. For a single line-to-ground fault, the positive-, negative- and zero-sequence networks are connected:
Show hintHide hint
All three sequence currents are equal for this fault.
Show answerHide answer
Answer: C. In series
The conditions Ia1 = Ia2 = Ia0 and Va1 + Va2 + Va0 = 3ZfIa1 are satisfied by a series connection.
75. For a line-to-line fault (not involving ground), the sequence network connection is:
Show hintHide hint
No ground connection means no zero-sequence current.
Show answerHide answer
Answer: B. Positive- and negative-sequence networks in parallel, zero-sequence absent
With no ground path I0 = 0, and Ia1 = −Ia2, Va1 = Va2 (bolted), giving the parallel connection of positive and negative networks.
76. In a double line-to-ground fault, the three sequence networks are connected:
Show hintHide hint
The sum of the three sequence currents of phase a is zero.
Show answerHide answer
Answer: B. All in parallel at the fault point
Va1 = Va2 = Va0 (bolted) and Ia1 + Ia2 + Ia0 = 0 lead to a parallel connection of all three networks.
77. On overhead power systems, the most frequently occurring type of fault is:
Show hintHide hint
Think of a single insulator flashing over.
Show answerHide answer
Answer: C. Single line-to-ground fault
Roughly 70–80% of faults are single line-to-ground (e.g. insulator flashover to tower); three-phase faults are the rarest though usually the most severe.
78. A three-phase fault current of 5 pu flows on an 11 kV bus. If the base is 100 MVA, the fault current in kiloamperes is about:
Show hintHide hint
Multiply pu current by the base current.
Show answerHide answer
Answer: C. 26.2 kA
Ibase = 100×10⁶/(√3 × 11×10³) ≈ 5249 A, so If = 5 × 5249 ≈ 26.2 kA.
79. The Thevenin (positive-sequence) impedance at a bus is 0.1 pu on a 100 MVA base, and the prefault voltage is 1.0 pu. The three-phase short-circuit level of the bus is:
Show hintHide hint
Fault level is inversely proportional to Zth.
Show answerHide answer
Answer: B. 1000 MVA
Fault MVA = Base MVA/Zth(pu) = 100/0.1 = 1000 MVA (with 1.0 pu prefault voltage).
80. When a single line-to-ground fault occurs on an ungrounded (isolated neutral) three-phase system, the voltage of the healthy phases to earth:
Show hintHide hint
The neutral shifts to the faulted phase's potential.
Show answerHide answer
Answer: A. Rises to about √3 times the normal phase voltage
The faulted phase is at earth potential, so the healthy phases rise to line voltage, i.e. √3 times their normal value.
81. Arcing grounds are a common problem in which type of system?
Show hintHide hint
The fault current there is purely capacitive.
Show answerHide answer
Answer: A. Ungrounded (isolated neutral) systems
In isolated-neutral systems the capacitive earth-fault current causes repeated arc extinction and restriking, producing high overvoltages (arcing ground).
82. A 50 Hz three-phase system has a capacitance to earth of 1 µF per phase. The inductance of a Petersen (arc-suppression) coil required for resonant grounding is about:
Show hintHide hint
The coil reactance must equal the reactance of 3C.
Show answerHide answer
Answer: A. 3.38 H
L = 1/(3ω²C) = 1/(3 × 314.16² × 1×10⁻⁶) ≈ 3.38 H.
83. Which of the following is an advantage of a solidly grounded neutral system?
Show hintHide hint
What happens to the neutral potential?
Show answerHide answer
Answer: B. Voltages of healthy phases do not rise appreciably during an earth fault, permitting reduced insulation
Solid grounding fixes the neutral potential, so healthy-phase voltages stay near normal, but fault currents (and communication interference) are high.
84. For a transposed transmission line, the relation among the sequence impedances is usually:
Show hintHide hint
A static element looks the same to either phase rotation.
Show answerHide answer
Answer: A. Z1 = Z2 and Z0 is larger than Z1
Static, symmetrical elements have Z1 = Z2; the zero-sequence path includes earth/ground wire return, typically making Z0 about 2–3.5 times Z1.
85. For a star(grounded)–delta transformer, the zero-sequence equivalent circuit:
Show hintHide hint
Zero-sequence current can circulate in the delta but not leave it.
Show answerHide answer
Answer: B. Connects the grounded-star side to the reference through the leakage impedance, while the delta side is open
Zero-sequence current can flow in the grounded star because it circulates in the delta, but no zero-sequence current flows in the delta-side lines.
86. A line-to-line fault on a power system does not produce:
Show hintHide hint
Is ground involved?
Show answerHide answer
Answer: C. Zero-sequence current
There is no path to ground, so Ia + Ib + Ic = 0 and I0 = 0.
87. At the fault point in a phase-a single line-to-ground fault, the sequence currents satisfy:
Show hintHide hint
Only phase a carries current.
Show answerHide answer
Answer: D. Ia1 = Ia2 = Ia0
With Ib = Ic = 0, the symmetrical-component transformation gives Ia0 = Ia1 = Ia2 = Ia/3.
88. A single line-to-ground fault occurs through a fault impedance Zf = j0.05 pu on a system with Z1 = Z2 = j0.2 and Z0 = j0.1 pu (E = 1 pu). The fault current is about:
Show hintHide hint
The fault impedance appears three times in the series loop.
Show answerHide answer
Answer: D. 4.62 pu
If = 3E/(Z1 + Z2 + Z0 + 3Zf) = 3/(0.5 + 0.15) ≈ 4.62 pu.
89. The initial symmetrical (momentary) short-circuit current of a synchronous generator is calculated using its:
Show hintHide hint
Use the reactance that applies in the first few cycles.
Show answerHide answer
Answer: D. Subtransient reactance X"d
Immediately after the fault the damper windings limit current to E/X"d; it then decays to transient (X'd) and steady (Xd) values.
90. For a bolted three-phase fault at a bus, the positive-sequence voltage at the fault point is:
Show hintHide hint
The fault impedance is zero.
Show answerHide answer
Answer: C. Zero
A bolted symmetrical fault shorts all three phases together at zero impedance, so the voltage there is zero.
91. The method of symmetrical components, which resolves an unbalanced set of three phasors into three balanced sets, was proposed by:
Show hintHide hint
The method dates from 1918.
Show answerHide answer
Answer: D. C. L. Fortescue
Fortescue presented the method of symmetrical coordinates in 1918.
92. In terms of symmetrical components, the total complex power in a three-phase system is:
Show hintHide hint
There is a factor coming from the transformation matrix.
Show answerHide answer
Answer: D. S = 3(V0I0* + V1I1* + V2I2*)
Because the transformation matrix satisfies AᵀA* = 3I, the power is three times the sum of the sequence powers, with no cross terms.
93. Negative-sequence currents in a synchronous generator are undesirable mainly because they:
Show hintHide hint
Consider the relative speed of the backward field and the rotor.
Show answerHide answer
Answer: B. Induce double-frequency currents in the rotor, causing overheating
The negative-sequence field rotates backward at synchronous speed, i.e. at twice synchronous speed relative to the rotor, inducing 2f currents that heat the rotor.
9.4 Load flow
32 questions · AElE0904
94. In load flow studies, the quantities specified at a PV (generator/voltage-controlled) bus are:
Show hintHide hint
The name of the bus tells you.
Show answerHide answer
Answer: A. Real power P and voltage magnitude |V|
At a PV bus the generator's real output and terminal voltage magnitude are fixed; Q and δ are computed.
95. At the slack (swing) bus in a load flow study, the specified quantities are:
Show hintHide hint
It acts as the reference for angles.
Show answerHide answer
Answer: A. Voltage magnitude and phase angle
The slack bus is the angle reference with fixed |V| and δ (usually 0°); its P and Q are unknown and balance the system losses.
96. The main reason for having a slack bus in a load flow problem is that:
Show hintHide hint
What quantity cannot be known in advance?
Show answerHide answer
Answer: C. Transmission losses are unknown until the solution is obtained, so one generator must balance them
Total generation must equal load plus losses, but losses depend on the solution; the slack generator absorbs this mismatch.
97. In the bus admittance matrix Ybus, the off-diagonal element Yik (i ≠ k) is equal to:
Show hintHide hint
Off-diagonal terms carry a minus sign.
Show answerHide answer
Answer: C. The negative of the total admittance directly connecting buses i and k
Yii is the sum of admittances terminating at bus i (including shunts); Yik = −yik for the branch between i and k, and zero if no branch exists.
98. A three-bus network has line reactances j0.1 pu (1–2), j0.2 pu (1–3) and j0.25 pu (2–3), with no shunt elements. The diagonal element Y22 of Ybus is:
Show hintHide hint
Add the admittances (not impedances) of the lines meeting at bus 2.
Show answerHide answer
Answer: B. −j14 pu
Y22 = y12 + y23 = 1/j0.1 + 1/j0.25 = −j10 − j4 = −j14 pu.
99. A line between buses 1 and 2 has series impedance 0.05 + j0.15 pu. The element Y12 of the bus admittance matrix is:
Show hintHide hint
Invert the complex impedance, then change the sign.
Show answerHide answer
Answer: A. -2 + j6 pu
y12 = 1/(0.05 + j0.15) = 2 − j6 pu, so Y12 = −y12 = -2 + j6 pu.
100. A power network has 10 buses (excluding ground) and 13 transmission lines, no two lines in parallel. The number of non-zero elements in its Ybus is:
Show hintHide hint
Each line contributes two symmetric off-diagonal entries.
Show answerHide answer
Answer: C. 36
Non-zeros = diagonal elements + 2 × number of branches = 10 + 2 × 13 = 36.
101. Adding a new line between existing buses i and k of a network modifies which Ybus elements?
Show hintHide hint
The line touches only two buses.
Show answerHide answer
Answer: C. Only Yii, Ykk, Yik and Yki
The new branch admittance adds to the two diagonal terms and subtracts from the two mutual terms; nothing else changes, which makes Ybus easy to modify.
102. Which statement about Ybus for a large practical power system is correct?
Show hintHide hint
How many neighbours does a typical bus have?
Show answerHide answer
Answer: C. It is highly sparse and, without phase-shifting transformers, symmetric
Each bus connects to only a few others, so most off-diagonal terms are zero; reciprocity makes it symmetric unless phase shifters are present.
103. The Gauss–Seidel load-flow update for the voltage at PQ bus i is:
Show hintHide hint
Start from Pi − jQi = Vi* Σ Yik Vk.
Show answerHide answer
Answer: D. Vi = (1/Yii)[(Pi − jQi)/Vi* − Σ(k≠i) Yik Vk]
From Si* = Vi* Σ Yik Vk, solving for Vi gives Vi = (1/Yii)[(Pi − jQi)/Vi* − Σ(k≠i) Yik Vk], using the latest available values of Vk.
104. In a two-bus system, bus 1 is the slack with V1 = 1.0∠0° pu, and the line reactance is j0.1 pu. Bus 2 has a load of 0.5 + j0.2 pu. Starting from V2 = 1.0∠0°, the first Gauss–Seidel iterate of V2 is:
Show hintHide hint
A load means negative injected P and Q.
Show answerHide answer
Answer: D. 0.98 − j0.05 pu
Y22 = −j10, Y21 = j10, S2 = −0.5 − j0.2. V2 = (1/−j10)[(−0.5 + j0.2)/1 − j10 × 1] = (−0.5 − j9.8)/(−j10) = 0.98 − j0.05 pu.
105. An acceleration factor is used in the Gauss–Seidel load flow method in order to:
Show hintHide hint
Think of over-relaxation.
Show answerHide answer
Answer: B. Reduce the number of iterations needed for convergence
Over-relaxation (α typically 1.4–1.6) extrapolates each voltage correction, speeding up the slow convergence of Gauss–Seidel.
106. Compared with the Gauss–Seidel method, the Newton–Raphson load flow method:
Show hintHide hint
Recall the convergence order of Newton's method.
Show answerHide answer
Answer: C. Converges quadratically, in a few iterations nearly independent of system size
NR typically converges in 3–5 iterations regardless of size, though each iteration is costlier because the Jacobian must be formed and factorised.
107. In the polar-form Newton–Raphson load flow, the Jacobian matrix relates:
Show hintHide hint
The Jacobian holds partial derivatives of P and Q.
Show answerHide answer
Answer: C. Mismatches ΔP and ΔQ to corrections Δδ and Δ|V|
[ΔP; ΔQ] = [J1 J2; J3 J4][Δδ; Δ|V|], with J1 = ∂P/∂δ, J2 = ∂P/∂|V|, J3 = ∂Q/∂δ, J4 = ∂Q/∂|V|.
108. A 10-bus system has 1 slack bus and 3 PV buses; the rest are PQ buses. In polar-form Newton–Raphson load flow, the order of the Jacobian matrix is:
Show hintHide hint
PV buses contribute only an angle unknown.
Show answerHide answer
Answer: A. 15 × 15
Unknowns: δ at 9 non-slack buses and |V| at 6 PQ buses, i.e. 9 + 6 = 15; so the Jacobian is 15 × 15 (= 2(n − 1) − m).
109. The fast decoupled load flow method is based on the observation that, in transmission networks:
Show hintHide hint
Consider the typical X/R ratio of HV lines.
Show answerHide answer
Answer: A. Real power depends mainly on voltage angles and reactive power mainly on voltage magnitudes
Because X ≫ R, ∂P/∂|V| and ∂Q/∂δ are small and are neglected, decoupling the P–δ and Q–|V| problems with constant matrices B′ and B″.
110. During load flow, if the reactive power required at a PV bus to hold its specified voltage exceeds the generator's Q limit:
Show hintHide hint
The generator can no longer regulate the voltage.
Show answerHide answer
Answer: B. The bus is treated as a PQ bus with Q fixed at the violated limit
The generator cannot supply more than its limit, so Q is fixed at that limit and |V| is allowed to vary (PV→PQ switching).
111. Two buses with voltages 1.0∠0° pu and 1.0∠−10° pu are connected by a lossless line of reactance 0.2 pu. The real power flowing from bus 1 to bus 2 is about:
Show hintHide hint
P = V1V2 sin δ / X.
Show answerHide answer
Answer: B. 0.868 pu
P = V1V2 sin δ / X = sin 10°/0.2 ≈ 0.868 pu.
112. The primary outputs of a load flow solution are:
Show hintHide hint
Everything else is computed from these.
Show answerHide answer
Answer: A. Magnitude and phase angle of the voltage at every bus
Once all bus voltages are known, line flows, losses and slack-bus generation follow directly.
113. Load flow equations must be solved by iterative techniques because they are:
Show hintHide hint
Look at how P depends on |V| and δ.
Show answerHide answer
Answer: D. Non-linear algebraic equations
Bus powers involve products of voltage magnitudes and trigonometric functions of angles, giving non-linear algebraic equations.
114. Compared with the bus admittance matrix, the bus impedance matrix Zbus of a large network is:
Show hintHide hint
The inverse of a sparse matrix is rarely sparse.
Show answerHide answer
Answer: B. A full (dense) matrix
Zbus = Ybus⁻¹; the inverse of a sparse network matrix is generally full, which is why load flow uses Ybus.
115. In the Gauss–Seidel method, at a PV bus in each iteration:
Show hintHide hint
|V| is specified but Q is not.
Show answerHide answer
Answer: D. Q is calculated, a new voltage is computed, and its magnitude is reset to the specified value keeping the new angle
Q is first estimated from current voltages; the voltage update is then performed and scaled so |V| equals the specified magnitude.
116. If there is no transmission line between buses 3 and 5 of a network, the element Y35 of Ybus is:
Show hintHide hint
Off-diagonal terms come only from branches.
Show answerHide answer
Answer: D. Zero
Off-diagonal terms are non-zero only for directly connected buses.
117. A line between buses 1 and 2 has series admittance −j10 pu and total line-charging susceptance of 0.04 pu (π model). If this is the only element connected to bus 1, Y11 is:
Show hintHide hint
Only half of the line charging appears at each end.
Show answerHide answer
Answer: D. −j9.98 pu
Half of the charging susceptance (j0.02) is placed at each end: Y11 = −j10 + j0.02 = −j9.98 pu.
118. A 'flat start' in a load flow study means:
Show hintHide hint
It concerns the starting guess.
Show answerHide answer
Answer: B. Initial voltages at all PQ buses are taken as 1.0∠0° pu
Unknown magnitudes start at 1.0 pu and unknown angles at 0°, while specified values are used where known.
119. In the polar Newton–Raphson method, the number of mismatch equations contributed by each PV bus is:
Show hintHide hint
Count the unknowns at a PV bus.
Show answerHide answer
Answer: A. One (ΔP only)
At a PV bus |V| is specified, so only δ is unknown and only the ΔP equation is written; a PQ bus contributes both ΔP and ΔQ.
120. The bus power mismatch used in Newton–Raphson iterations is defined as:
Show hintHide hint
It measures how far the current estimate is from satisfying the equations.
Show answerHide answer
Answer: B. Specified power minus power calculated from the current voltage estimates
ΔPi = Pi,sp − Pi,calc and ΔQi = Qi,sp − Qi,calc; iterations stop when all mismatches fall below tolerance.
121. The real power injected at bus i in terms of Ybus elements (Yik = |Yik|∠θik) and bus voltages is:
Show hintHide hint
Real power uses the cosine term.
Show answerHide answer
Answer: B. Pi = Σk |Vi||Vk||Yik| cos(θik − δi + δk)
Pi − jQi = Vi* Σ Yik Vk; the real part gives the cosine expression and the negative imaginary part gives Qi = −Σ |Vi||Vk||Yik| sin(θik − δi + δk).
122. If a network has no shunt elements (no connections to ground), its Ybus matrix is:
Show hintHide hint
Sum the elements in any row.
Show answerHide answer
Answer: A. Singular, because the elements of each row sum to zero
Without shunt branches, each diagonal equals the negative sum of its off-diagonals, so the rows are linearly dependent and Ybus cannot be inverted.
123. In a Gauss–Seidel iteration, a bus voltage changes from 1.000 pu to a newly computed 0.980 pu (real values for simplicity). With an acceleration factor of 1.6, the accelerated value is:
Show hintHide hint
The correction, not the voltage, is multiplied by α.
Show answerHide answer
Answer: C. 0.968 pu
V_acc = V_old + α(V_new − V_old) = 1.000 + 1.6(0.980 − 1.000) = 0.968 pu.
124. Two buses with voltages 1.0∠0° pu and 1.0∠−10° pu are joined by a lossless line of reactance 0.2 pu. The reactive power leaving bus 1 towards bus 2 is about:
Show hintHide hint
Q12 = (V1² − V1V2 cos δ)/X.
Show answerHide answer
Answer: D. 0.076 pu
Q12 = (V1² − V1V2 cos δ)/X = (1 − cos10°)/0.2 ≈ 0.076 pu; with equal voltage magnitudes both ends supply half of the line's I²X.
125. For a lossless two-bus system with |V1| = |V2| = 1.0 pu, X = 0.2 pu and angle difference δ = 10°, the Jacobian term ∂P/∂δ (power transfer versus angle) is about:
Show hintHide hint
Differentiate sin δ.
Show answerHide answer
Answer: D. 4.92 pu/rad
P = V1V2 sin δ/X, so ∂P/∂δ = V1V2 cos δ/X = cos10°/0.2 ≈ 4.92 pu/rad.
9.5 Stability analysis
30 questions · AElE0905
126. A generator with internal emf 1.2 pu is connected to an infinite bus of 1.0 pu through a total reactance of 0.5 pu. The steady-state stability limit (maximum power transfer) is:
Show hintHide hint
P = (EV/X) sin δ; take the maximum.
Show answerHide answer
Answer: D. 2.4 pu
Pmax = EV/X = 1.2 × 1.0/0.5 = 2.4 pu.
127. For the same system (Pmax = 2.4 pu), the electrical power output at a load angle of 30° is:
Show hintHide hint
Use the power–angle equation.
Show answerHide answer
Answer: D. 1.2 pu
Pe = Pmax sin δ = 2.4 × sin 30° = 1.2 pu.
128. For a round-rotor generator connected to an infinite bus through a pure reactance (resistance neglected), the theoretical steady-state stability limit is reached at a load angle of:
Show hintHide hint
Where does sin δ have its maximum?
Show answerHide answer
Answer: A. 90°
Pe = Pmax sin δ peaks at δ = 90°; beyond this, increasing δ reduces power and synchronism is lost.
129. The swing equation of a synchronous machine, with H in seconds and powers in per unit, is:
Show hintHide hint
Accelerating power equals inertia times angular acceleration.
Show answerHide answer
Answer: B. (2H/ωs) d²δ/dt² = Pm − Pe
Newton's law for the rotor, expressed with the inertia constant H, gives (2H/ωs) d²δ/dt² = Pa = Pm − Pe (damping neglected).
130. A 100 MVA generator stores 500 MJ of kinetic energy at synchronous speed. Its inertia constant H on a 200 MVA common base is:
Show hintHide hint
H scales inversely with the base MVA.
Show answerHide answer
Answer: C. 2.5 s
On its own rating H = 500/100 = 5 s; on 200 MVA, H = 5 × 100/200 = 2.5 s.
131. The equal area criterion for transient stability is directly applicable to:
Show hintHide hint
It relies on a single power–angle curve.
Show answerHide answer
Answer: C. A single machine connected to an infinite bus (or a two-machine system)
The criterion uses the P–δ curve of one machine against an infinite bus (a two-machine system can be reduced to this form); multi-machine systems need numerical solution of swing equations.
132. According to the equal area criterion, the system remains stable after a disturbance if:
Show hintHide hint
Energy gained must be given back.
Show answerHide answer
Answer: A. The decelerating area available equals or exceeds the accelerating area
The kinetic energy gained during acceleration (A1) must be fully returned during deceleration (A2) before δ reaches its maximum allowable value.
133. A generator delivers power at δ0 = 30° to an infinite bus. A three-phase fault at its terminals reduces electrical output to zero, and clearing restores the original network. The critical clearing angle is about:
Show hintHide hint
Apply equal areas with δmax = π − δ0.
Show answerHide answer
Answer: B. 79.6°
cos δcr = (π − 2δ0) sin δ0 − cos δ0 = (2.094)(0.5) − 0.866 = 0.181, so δcr ≈ 79.6°.
134. For the case above (δ0 = 30°, δcr ≈ 79.6°, Pe = 0 during fault), the machine has H = 5 s, f = 50 Hz and Pm = 1.0 pu. The critical clearing time is about:
Show hintHide hint
Integrate the swing equation twice with constant accelerating power; use radians.
Show answerHide answer
Answer: B. 0.235 s
With Pe = 0, δ = δ0 + (ωsPm/4H)t², so tcr = √[4H(δcr − δ0)/(ωsPm)] = √[20 × 0.866/314.16] ≈ 0.235 s (angles in radians).
135. A generator operating at no load on an infinite bus (lossless system) is suddenly loaded. By the equal area criterion, the maximum step of mechanical input it can accept without losing synchronism is about:
Show hintHide hint
The step cannot reach Pmax because of rotor overshoot.
Show answerHide answer
Answer: C. 0.725 Pmax
Equating accelerating and decelerating areas from δ = 0 with δmax = π − δ1 gives sin δ1 (π − δ1) = 1 + cos δ1, i.e. δ1 ≈ 46.4° and Pm ≈ 0.725 Pmax.
136. Which of the following does NOT help improve the transient stability of a power system?
Show hintHide hint
What happens to Pmax when X increases?
Show answerHide answer
Answer: D. Increasing the transfer reactance between generator and load
Higher reactance lowers Pmax and the decelerating area; fast clearing, quick excitation and series compensation all improve stability.
137. Increasing the inertia constant H of a generator (other things unchanged) generally:
Show hintHide hint
Inertia resists change in speed.
Show answerHide answer
Answer: C. Improves transient stability by slowing the rotor angle swing
Larger H means smaller angular acceleration for the same accelerating power, so δ rises more slowly and the critical clearing time increases.
138. A generator with Pmax = 2.0 pu operates at δ = 30°. The synchronising power coefficient is about:
Show hintHide hint
Differentiate Pmax sin δ.
Show answerHide answer
Answer: B. 1.732 pu/rad
Ps = dPe/dδ = Pmax cos δ = 2 × cos 30° ≈ 1.732 pu per electrical radian.
139. A 50 Hz generator with H = 5 s has a synchronising power coefficient of 1.732 pu/rad. Neglecting damping, the natural frequency of small rotor oscillations is about:
Show hintHide hint
Linearise the swing equation; convert rad/s to Hz at the end.
Show answerHide answer
Answer: D. 1.17 Hz
ωn = √(ωsPs/2H) = √(314.16 × 1.732/10) ≈ 7.38 rad/s, so fn ≈ 1.17 Hz.
140. Transient stability of a power system refers to its ability to:
Show hintHide hint
Large or small disturbance?
Show answerHide answer
Answer: B. Remain in synchronism after a large, sudden disturbance such as a fault
Transient stability deals with large disturbances (faults, line switching) over the first swing or few seconds; steady-state stability deals with small, slow changes.
141. Voltage collapse in a power system is primarily caused by:
Show hintHide hint
Voltage is tightly linked to which kind of power?
Show answerHide answer
Answer: B. Inability of the system to meet the reactive power demand of heavily loaded areas
Voltage stability is mainly a reactive-power problem: when Q support is insufficient, voltages decline progressively, leading to collapse.
142. A bus is voltage stable if its voltage magnitude:
Show hintHide hint
Injecting Q should normally support the voltage.
Show answerHide answer
Answer: C. Increases when reactive power injection at that bus is increased
Positive V–Q sensitivity (dV/dQ > 0) at every bus indicates voltage stability; a negative sensitivity at any bus indicates instability.
143. On the P–V (nose) curve of a load bus, the tip of the nose represents:
Show hintHide hint
Beyond this point there is no solution.
Show answerHide answer
Answer: A. The maximum loadability point, i.e. the voltage stability limit
Beyond the nose point no load-flow solution exists for higher loading; it marks the critical voltage and maximum power transfer.
144. A power system stabilizer (PSS) improves stability by:
Show hintHide hint
It acts through the exciter.
Show answerHide answer
Answer: C. Adding a supplementary signal to the excitation system to damp rotor oscillations
A PSS uses speed, frequency or power deviation to modulate excitation, producing a damping torque component on electromechanical oscillations.
145. A generator (E = 1.2 pu, reactance 0.2 pu) feeds an infinite bus (1.0 pu) through two identical parallel lines of 0.4 pu each. If one line is tripped, the maximum power transfer changes from:
Show hintHide hint
Recompute the series-parallel reactance.
Show answerHide answer
Answer: D. 3.0 pu to 2.0 pu
Before: X = 0.2 + 0.4/2 = 0.4 pu, Pmax = 1.2/0.4 = 3.0 pu. After: X = 0.2 + 0.4 = 0.6 pu, Pmax = 1.2/0.6 = 2.0 pu.
146. Single-pole (independent-pole) switching and auto-reclosing improve transient stability mainly because:
Show hintHide hint
How many phases are lost in an LG fault?
Show answerHide answer
Answer: C. Healthy phases continue to transmit power during a single line-to-ground fault
Opening only the faulted phase lets the other two phases keep transferring synchronising power, enlarging the decelerating area.
147. The swing curve (δ versus t) of a machine in a multi-machine system is usually obtained by:
Show hintHide hint
The swing equation is a non-linear differential equation.
Show answerHide answer
Answer: B. Numerical step-by-step solution of the swing equations
Swing equations are non-linear differential equations, solved by point-by-point, modified Euler or Runge–Kutta methods.
148. An infinite bus is characterised by:
Show hintHide hint
Think of an extremely large system.
Show answerHide answer
Answer: A. Constant voltage and constant frequency irrespective of the power drawn
An infinite bus represents a very large system with zero internal impedance and infinite inertia, so its V and f are fixed.
149. Series capacitors compensate 40% of the reactance of a line connecting a generator to an infinite bus (line reactance is the only reactance). The maximum power transfer increases by about:
Show hintHide hint
Pmax is inversely proportional to the net reactance.
Show answerHide answer
Answer: D. 66.7%
Net X becomes 0.6X, so Pmax rises by the factor 1/0.6 ≈ 1.667, an increase of about 66.7%.
150. A 100 MVA, 50 Hz generator has H = 4 MJ/MVA. Its angular momentum M at synchronous speed is about:
Show hintHide hint
M = 2 × stored energy / ωs, with ωs = 2πf.
Show answerHide answer
Answer: A. 2.546 MJ·s/elec. rad
M = GH/(πf) = (100 × 4)/(π × 50) ≈ 2.546 MJ·s per electrical radian.
151. Compared with the steady-state stability limit, the transient stability limit of a system is generally:
Show hintHide hint
Think about rotor overshoot after a large disturbance.
Show answerHide answer
Answer: A. Lower
Under a large disturbance the rotor overshoots, so the maximum power that can be transferred while remaining stable is less than Pmax.
152. From the transient stability point of view, the most severe type of fault is usually:
Show hintHide hint
Which fault blocks power transfer most?
Show answerHide answer
Answer: D. Three-phase fault
A three-phase fault close to the generator reduces transferred power the most (to nearly zero), causing maximum acceleration.
153. Damper (amortisseur) windings on a synchronous machine rotor help stability mainly by:
Show hintHide hint
Induced currents oppose relative motion.
Show answerHide answer
Answer: A. Damping out rotor oscillations about the synchronous speed
When the rotor oscillates, currents induced in the damper bars produce a torque opposing the relative motion, damping the swings.
154. During a bolted three-phase fault at the terminals of a generator, the accelerating power on its rotor is:
Show hintHide hint
What is Pe when the terminal voltage is zero?
Show answerHide answer
Answer: B. Equal to the mechanical input power Pm
Electrical output falls to zero (voltage zero), so Pa = Pm − Pe = Pm and the rotor accelerates.
155. On-load tap changers on transformers feeding loads can aggravate voltage instability because:
Show hintHide hint
Think about what load does when its voltage is restored.
Show answerHide answer
Answer: A. They restore load-side voltage and hence load power, increasing stress on a weakened transmission system
After a disturbance, tap changers raise distribution voltages, restoring voltage-dependent load and increasing reactive demand on the network, which can drive further voltage decline.
9.6 Voltage control and VAR compensation
29 questions · AElE0906
156. For a lossless line of reactance X connecting two buses with voltages V1∠δ and V2∠0, the real power transferred depends mainly on:
Show hintHide hint
Look at the power–angle equation.
Show answerHide answer
Answer: B. The angle difference δ between the bus voltages
P = V1V2 sin δ / X; with voltages near 1 pu, real power is governed chiefly by the angle difference.
157. In a high-voltage network, reactive power tends to flow:
Show hintHide hint
Reactive flow is governed by voltage magnitudes.
Show answerHide answer
Answer: B. From the bus with higher voltage magnitude to the bus with lower voltage magnitude
Q12 ≈ V1(V1 − V2 cos δ)/X; for small δ, reactive power flows from higher |V| towards lower |V|.
158. Two buses at 1.05 pu and 1.00 pu (in phase, δ ≈ 0) are connected by a lossless line of reactance 0.1 pu. The reactive power sent from the 1.05 pu bus is about:
Show hintHide hint
Use Q12 = V1(V1 − V2 cos δ)/X with δ = 0.
Show answerHide answer
Answer: A. 0.525 pu
Q12 = V1(V1 − V2 cos δ)/X = 1.05 × (1.05 − 1.00)/0.1 = 0.525 pu.
159. Two buses, each held at 1.0 pu, are connected by a lossless line of reactance 0.25 pu with an angle difference of 15°. The real power transferred is about:
Show hintHide hint
P = V1V2 sin δ / X.
Show answerHide answer
Answer: C. 1.035 pu
P = V1V2 sin δ/X = sin 15°/0.25 ≈ 1.035 pu.
160. A shunt capacitor bank is rated 10 Mvar at rated voltage. If the bus voltage falls to 0.9 pu, its reactive output becomes:
Show hintHide hint
Capacitor reactive power depends on the square of voltage.
Show answerHide answer
Answer: C. 8.1 Mvar
Q = V²ωC ∝ V², so Q = 10 × 0.9² = 8.1 Mvar; this drop just when support is needed is a weakness of fixed capacitors.
161. Shunt reactors are connected at the ends of long EHV lines mainly to:
Show hintHide hint
What happens to an EHV line's voltage at no load?
Show answerHide answer
Answer: A. Absorb excess reactive power and limit voltage rise at light load
At light load the line charging produces surplus reactive power and a Ferranti rise; shunt reactors absorb it.
162. The main purpose of series capacitor compensation on a long transmission line is to:
Show hintHide hint
It acts in series with the line inductance.
Show answerHide answer
Answer: D. Reduce the effective series reactance and increase power transfer capability
Series capacitors cancel part of the inductive reactance, raising Pmax = V1V2/(X − Xc) and improving stability and voltage regulation.
163. A well-known adverse effect associated with series capacitor compensation of lines connected to turbine-generators is:
Show hintHide hint
The resonance frequency is below the system frequency.
Show answerHide answer
Answer: B. Subsynchronous resonance
The series LC circuit resonates at a frequency below 50 Hz which can interact with torsional modes of the turbine-generator shaft (SSR).
164. A line of reactance 100 Ω has a series capacitor of 40 Ω reactance inserted. The degree of series compensation is:
Show hintHide hint
Ratio of capacitive to line reactance.
Show answerHide answer
Answer: D. 40%
Degree of compensation k = Xc/XL = 40/100 = 40%.
165. A static VAR compensator (SVC) typically consists of:
Show hintHide hint
It is a thyristor-switched shunt susceptance.
Show answerHide answer
Answer: A. A thyristor-controlled reactor combined with fixed or thyristor-switched capacitors
SVCs use TCR/TSC/FC combinations for variable susceptance; a STATCOM uses a voltage-source converter, and a synchronous condenser is a rotating machine.
166. Compared with an SVC, a STATCOM is better at low system voltages because:
Show hintHide hint
Compare a current-source-like device with a susceptance.
Show answerHide answer
Answer: D. It can supply its rated current nearly independent of the system voltage
STATCOM output current is limited by converter rating, so Q falls only linearly with V; an SVC behaves as a susceptance with Q ∝ V².
167. A synchronous condenser is:
Show hintHide hint
It is a rotating machine.
Show answerHide answer
Answer: A. A synchronous motor running without mechanical load, whose excitation is varied to supply or absorb reactive power
Over-excited it supplies lagging vars (acts like a capacitor); under-excited it absorbs vars (acts like a reactor), with smooth control.
168. An on-load tap-changing transformer controls voltage by:
Show hintHide hint
A transformer is not a reactive power source.
Show answerHide answer
Answer: A. Changing the turns ratio, redistributing reactive flow without generating reactive power
Tap changers only alter the voltage ratio; they do not produce vars, so the reactive power must still come from elsewhere in the system.
169. An over-excited synchronous generator connected to a grid:
Show hintHide hint
Higher internal emf than terminal voltage.
Show answerHide answer
Answer: C. Delivers lagging reactive power to the system
Increasing field current raises E above V, and the generator supplies reactive power (lagging pf operation).
170. A load of 500 kW operates at 0.8 power factor lagging. The capacitor kvar required to raise the power factor to 0.95 lagging is about:
Show hintHide hint
Use P × (tan φ1 − tan φ2).
Show answerHide answer
Answer: D. 211 kvar
Qc = P(tan φ1 − tan φ2) = 500(0.750 − 0.329) ≈ 211 kvar.
171. A 30 kvar, 400 V, 50 Hz three-phase capacitor bank is delta-connected. The capacitance required per phase is about:
Show hintHide hint
In delta, each capacitor sees the line voltage and supplies one third of the kvar.
Show answerHide answer
Answer: A. 199 µF
Per phase Q = 10 kvar at 400 V (delta): C = Q/(ωV²) = 10 000/(314.16 × 400²) ≈ 199 µF.
172. A load of 1.0 + j0.5 pu is supplied through a line with R = 0.02 pu and X = 0.1 pu, receiving voltage 1.0 pu. The approximate voltage drop is:
Show hintHide hint
ΔV ≈ (PR + QX)/V.
Show answerHide answer
Answer: C. 0.07 pu
ΔV ≈ (PR + QX)/V = (1.0 × 0.02 + 0.5 × 0.1)/1.0 = 0.07 pu.
173. On high-voltage transmission lines where X ≫ R, the voltage drop along the line is governed mainly by:
Show hintHide hint
Compare PR with QX.
Show answerHide answer
Answer: C. The reactive power flow
ΔV ≈ (PR + QX)/V; with X ≫ R the QX term dominates, so voltage is controlled by managing reactive flow.
174. A 220 kV, 50 Hz line, 200 km long, has a capacitance to neutral of 0.009 µF/km per phase. The total three-phase reactive power generated by line charging (uniform voltage assumed) is about:
Show hintHide hint
Three-phase charging = V_LL² × ω × C per phase.
Show answerHide answer
Answer: B. 27.4 Mvar
Q = V_LL² ωC = (220×10³)² × 314.16 × (0.009×10⁻⁶ × 200) ≈ 27.4 Mvar.
175. For a lossless line of reactance X with both end voltages held at V, ideal shunt compensation that holds the midpoint voltage at V raises the maximum transferable power from V²/X to:
Show hintHide hint
The line is split into two halves of X/2 each.
Show answerHide answer
Answer: B. 2V²/X
Each half of the line (X/2) can transmit V²/(X/2) sin(δ/2), so Pmax = 2V²/X, reached at δ = 180°.
176. Which of the following devices can only generate (not absorb) reactive power?
Show hintHide hint
Which one has no controllable inductive part?
Show answerHide answer
Answer: A. Fixed shunt capacitor bank
A fixed capacitor always supplies vars; the others can both supply and absorb reactive power over a continuous range.
177. A FACTS controller that combines both series and shunt converters sharing a common DC link is the:
Show hintHide hint
It is the 'unified' device.
Show answerHide answer
Answer: B. Unified Power Flow Controller (UPFC)
The UPFC has a shunt converter (STATCOM-like) and a series converter (SSSC-like) linked through a DC capacitor, controlling P, Q and voltage.
178. A bus has a three-phase short-circuit level of 500 MVA. Switching in a 10 Mvar shunt capacitor bank will raise the bus voltage by approximately:
Show hintHide hint
ΔV/V ≈ ΔQ / short-circuit MVA.
Show answerHide answer
Answer: D. 2%
ΔV/V ≈ ΔQ/Ssc = 10/500 = 0.02, i.e. about 2%.
179. In power system operation, real power–frequency (P–f) and reactive power–voltage (Q–V) control loops can be treated separately because:
Show hintHide hint
Recall the same idea behind fast decoupled load flow.
Show answerHide answer
Answer: D. Real power is strongly coupled to angle/frequency and reactive power to voltage magnitude
With X ≫ R, ∂P/∂δ and ∂Q/∂|V| dominate, so the two control problems are loosely coupled.
180. For a short line represented by ABCD constants, the radius of the receiving-end power circle diagram is:
Show hintHide hint
The radius term contains both end voltages.
Show answerHide answer
Answer: C. |Vs||Vr|/|B|
Receiving-end complex power Sr = |Vs||Vr|/|B|∠(β − δ) − |A||Vr|²/|B|∠(β − α); the first term gives the radius and the second the centre.
181. A short line has R = 0.1 pu and X = 0.3 pu with |Vs| = |Vr| = 1.0 pu. The maximum real power that can be received is about:
Show hintHide hint
Resistance reduces the maximum receivable power below VsVr/|Z|.
Show answerHide answer
Answer: A. 2.16 pu
Pr,max = VsVr/|Z| − Vr²R/|Z|² = 1/0.3162 − 0.1/0.1 ≈ 2.16 pu.
182. A lossless line of reactance 0.5 pu between two 1.0 pu buses is compensated by 30% series capacitance. The new maximum power transfer is about:
Show hintHide hint
Subtract the capacitive reactance first.
Show answerHide answer
Answer: B. 2.86 pu
Net X = 0.5(1 − 0.3) = 0.35 pu, so Pmax = 1 × 1/0.35 ≈ 2.86 pu (from 2.00 pu without compensation).
183. Generator automatic voltage regulators (AVRs) control the terminal voltage by adjusting the:
Show hintHide hint
Q–V control acts through excitation.
Show answerHide answer
Answer: C. Field (excitation) current
Changing excitation changes the internal emf and hence the reactive power output and terminal voltage; turbine input controls real power.
184. Which is the most suitable location for a shunt capacitor bank used to relieve a heavily loaded distribution feeder supplying lagging loads?
Show hintHide hint
Supply vars where they are used.
Show answerHide answer
Answer: D. Close to the load end where reactive power is consumed
Supplying vars locally reduces reactive current through the whole feeder, cutting I²R losses and voltage drop.