Revision material: This lesson covers the KNEC Diploma in Electrical Engineering Module III June/July 2021 Electrical Power Systems and Electromagnetic Field Theory past paper. It reproduces all eight questions and provides worked revision answers. Candidates were instructed to answer three questions in Section A and two in Section B. The answers are independently prepared and are not an official KNEC marking scheme.
Section A: Electrical Power Systems
Question 1
(a) (3 marks) State three characteristics of an effective power system protective scheme.
(b) (7 marks) With the aid of a labelled diagram, describe the operation of a balanced beam relay.
(c) (8 marks) A one ampere overcurrent relay having a plug setting of 125% and a time multiplier setting of 0.6 is supplied by a current transformer rated 400:1 A. The fault current on the transformer is 4,000 A. Determine the (i) relay coil current for the fault; (ii) nominal relay coil current; (iii) plug setting multiplier; and (iv) relay time of response.
(d) (2 marks) Describe pilot protection scheme with reference to power systems.
Answer 1(a)
- Reliability: it operates when a fault occurs and remains inactive during normal conditions.
- Selectivity: it isolates only the faulty section, keeping healthy parts in service.
- Speed: it clears faults quickly to limit equipment damage and help maintain system stability.
Answer 1(b)
A balanced beam relay has a light beam on a pivot. The operating coil acts on one end, while a spring or adjustable counterweight provides restraint at the other end. The protected circuit current is supplied to the operating coil through a current transformer.
adjustable restraint / spring
↓
operating coil ─────┴──────────────┐
(from CT) movable beam │
─────●────────┘
pivot
moving contact ───┐
└── trip circuit → breaker trip coil
At normal current, the restraining torque keeps the beam balanced and the contacts open. A fault raises the CT secondary current. When operating torque exceeds restraining torque, the beam tilts, closes the trip contacts and energises the circuit-breaker trip coil, disconnecting the faulty section.
Answer 1(c)
The CT secondary current at the stated fault current is:
Irelay = 4,000 A × (1 A / 400 A) = 10 A
The relay is rated at 1 A. A 125% plug setting gives a pickup current of:
Ipickup = 1.25 × 1 A = 1.25 A
PSM = Irelay / Ipickup = 10 / 1.25 = 8
The operating time depends on the relay’s inverse-time characteristic, which is not specified in the question. Assuming the common IEC standard-inverse curve, t = 0.14 × TMS / (PSM0.02 − 1):
t = 0.14 × 0.6 / (80.02 − 1) ≈ 1.98 s
Answers: fault coil current = 10 A; nominal coil current = 1 A; PSM = 8; response time ≈ 1.98 s on the stated IEC standard-inverse assumption. A different relay curve gives a different time.
Answer 1(d)
Pilot protection compares electrical quantities measured at both ends of a protected line using a communication channel such as a pilot wire or fibre link. If the comparison indicates an internal fault, relays at the line ends trip their circuit breakers; external faults do not satisfy the trip condition.
Question 2
(a) (4 marks) Explain two factors that affect the sag of a conductor in transmission lines.
(b) (6 marks) Explain three factors that affect corona loss in overhead transmission lines.
(c) (10 marks) A transmission line of diameter 26.1 mm and unit weight 10.89 N/m is installed between two towers 300 m apart. The initial tension on the line is 15,450 N. During heavy loading, the ice thickness on the line is 12.5 mm and the wind load is 190 N/m². Taking the density of ice as 915 kg/m³, determine the (i) initial sag on the line; (ii) ice load; (iii) wind load; and (iv) total unit weight.
Answer 2(a)
- Span length: for level supports and small sag, sag is proportional to the square of the span, so a longer span increases sag.
- Conductor loading: the conductor’s own weight, ice and wind increase the vertical or resultant load and therefore increase sag.
- Tension and temperature: higher tension reduces sag, while heating lengthens the conductor and increases sag.
Answer 2(b)
- Operating voltage: higher voltage raises the electric stress at the conductor surface and increases corona loss.
- Conductor size and spacing: a larger conductor radius reduces surface electric stress; greater phase spacing also reduces it.
- Atmospheric and surface conditions: rain, fog, dust, rough joints and damaged surfaces promote ionisation and increase loss. Lower air density also lowers the disruptive voltage.
Answer 2(c)
The initial sag is calculated for the bare conductor, using the parabolic approximation:
sag = wL² / (8T) = 10.89 × 300² / (8 × 15,450) = 7.93 m
The ice-coated diameter is 26.1 + 2(12.5) = 51.1 mm = 0.0511 m. The ice annulus area per metre of conductor is:
Aice = (π/4)(0.0511² − 0.0261²) = 0.001516 m²
wice = 915 × 9.81 × 0.001516 = 13.61 N/m
Wind acts on the projected ice-coated diameter:
wwind = 190 × 0.0511 = 9.71 N/m
The vertical load is the conductor weight plus ice load. Combine it with the horizontal wind load as a resultant:
wvertical = 10.89 + 13.61 = 24.50 N/m
wtotal = √(24.50² + 9.71²) = 26.35 N/m
Answers: initial sag = 7.93 m; ice load = 13.61 N/m; wind load = 9.71 N/m; resultant total unit load = 26.35 N/m.
Question 3
(a) (3 marks) State three merits of zinc oxide surge diverter with reference to transmission line protection.
(b) (7 marks) With the aid of a labelled diagram, describe the working principle of a valve type surge diverter in power transmission line.
(c) (10 marks) A 3-phase, 50 Hz, 160 km overhead transmission line delivers a load of 10 MVA and 132 kV at 0.8 power factor lagging to a balanced load. The resistance, inductance and capacitance per km are 0.16 Ω, 1.2 mH and 0.0082 μF respectively. Using the nominal π method, determine the (i) total series impedance in the line; (ii) total shunt admittance in the line; (iii) voltage at the receiving end; (iv) current at the receiving end; and (v) voltage at the sending end.
Answer 3(a)
- Zinc oxide has a strongly nonlinear voltage-current characteristic, so it conducts heavily during a surge while drawing very little current at normal voltage.
- Gapless designs respond quickly and avoid the maintenance and spark-gap coordination required by older diverters.
- It clamps surges to a comparatively low residual voltage and can absorb repeated surge energy when correctly rated.
Answer 3(b)
Transmission line ── spark-gap assembly ── nonlinear resistor stack ── earth
│ │
blocks normal voltage conducts surge current and
until breakdown limits the follow current
In a valve-type diverter, series spark gaps normally insulate the line from earth. A surge raises the voltage until the gaps break down and arc. Surge current then flows through nonlinear resistor elements to earth. Their resistance falls at high surge current, limiting the voltage across protected equipment; as the surge subsides, resistance rises and limits the power-frequency follow current. The arc extinguishes at a current zero and the gaps recover their insulation.
Answer 3(c)
Use per-phase quantities. The total series resistance, inductance and capacitance are:
R = 0.16 × 160 = 25.6 Ω
L = 1.2 mH/km × 160 = 0.192 H
C = 0.0082 μF/km × 160 = 1.312 μF
Z = R + jωL = 25.6 + j(2π × 50 × 0.192) = 25.6 + j60.32 Ω per phase
Y = jωC = j(2π × 50 × 1.312 × 10⁻⁶) = j4.1218 × 10⁻⁴ S per phase
The receiving-end line voltage is given as 132 kV, hence:
VR = 132/√3 = 76.210 kV per phase; VR,LL = 132 kV
At 10 MVA and 0.8 lagging power factor, the receiving-end current is:
IR = 10 × 10⁶ / (√3 × 132 × 10³) = 43.74∠−36.87° A
For the nominal π model, A = D = 1 + YZ/2 and C = Y(1 + YZ/4). Using VS = AVR + ZIR:
A = D = 0.98757 + j0.005276
C = −1.0873 × 10⁻⁶ + j4.0962 × 10⁻⁴ S
VS,phase = 77.742 + j1.841 kV = 77.763∠1.356° kV
VS,LL = √3 × 77.763 = 134.69 kV
Answers: Z = 25.6 + j60.32 Ω/phase; Y = j0.0004122 S/phase; receiving voltage = 132 kV line-to-line; receiving current = 43.74∠−36.87° A; sending voltage = 134.69 kV line-to-line (approximately 1.36° ahead of the receiving-end phase reference).
Question 4
(a) (5 marks) (i) State three causes of faults in electrical power systems. (ii) Distinguish between balanced and unbalanced faults.
(b) (8 marks) The three phase voltages across a 3-phase unbalanced load are as follows: Va = (176 − j132) V, Vb = (−128 − j96) V and Vc = (−160 + j100) V. Determine the (i) positive-sequence voltage; (ii) negative-sequence voltage; and (iii) zero-sequence voltage.
(c) (7 marks) A 3-phase system supplies 15 A to a delta-connected balanced load. A fault causes the fuse on one of the lines to melt. Determine the symmetrical components of the line currents.
Answer 4(a)
(i) Common causes include insulation deterioration or breakdown, lightning and switching surges, and mechanical or environmental damage such as wind, falling trees, contamination or animals.
(ii) A balanced fault affects all three phases equally and produces equal-magnitude phase currents displaced by 120°; it is usually a three-phase fault. An unbalanced fault affects phases unequally, so phase currents and voltages are no longer symmetrical; examples include line-to-ground and line-to-line faults.
Answer 4(b)
Let a = 1∠120° = −0.5 + j0.866. The sequence components are:
V₀ = (Va + Vb + Vc)/3
V₁ = (Va + aVb + a²Vc)/3
V₂ = (Va + a²Vb + aVc)/3
Substituting the given phasors:
V₀ = −37.33 − j42.67 V = 56.69∠−131.19° V
V₁ = 163.25 − j35.43 V = 167.05∠−12.24° V
V₂ = 50.09 − j53.90 V = 73.58∠−47.10° V
Answer 4(c)
Take the fused line as phase A. For a balanced delta load of branch impedance Z, the two remaining line terminals see one branch Z in parallel with the other two branches in series (2Z). Thus the surviving line current is 0.866 of the original balanced line current:
I = (√3/2) × 15 = 12.99 A
Choose the reference so that Ia = 0, Ib = 12.99∠0° A and Ic = 12.99∠180° A. Then:
I₀ = (Ia + Ib + Ic)/3 = 0
I₁ = (Ia + aIb + a²Ic)/3 = 7.50∠90° A
I₂ = (Ia + a²Ib + aIc)/3 = 7.50∠−90° A
The positive- and negative-sequence components have equal magnitudes of 7.50 A, and the zero-sequence component is zero. The stated angles use the selected reference for the surviving phase-B current.
Question 5
(a) (4 marks) (i) Define stability with respect to power systems. (ii) State three causes of voltage instability in power systems.
(b) (8 marks) A 60 Hz, 100 MVA generator has an inertia constant of 10 MJ/MVA and supplies an electrical load of 50 MW. The mechanical input to the generator is suddenly raised to 60 MW. Determine the (i) stored energy in the rotor at synchronous speed, E; (ii) accelerating power on the generator, Pa; (iii) moment of inertia, M; and (iv) rotor acceleration, α.
(c) (8 marks) With the aid of a labelled diagram, describe the dark lamp method of synchronizing two 3-phase generators.
Answer 5(a)
(i) Power-system stability is the ability of the system to regain an acceptable operating equilibrium after a disturbance, with its interconnected machines remaining in synchronism.
(ii) Voltage instability can be caused by inadequate reactive-power support, heavy loading or long weak transmission paths, and disturbances or equipment limits that reduce voltage support, such as generator excitation limits or loss of a line.
Answer 5(b)
Stored rotor energy is the inertia constant multiplied by the machine rating:
E = H × Srated = 10 × 100 = 1,000 MJ = 1.00 × 10⁹ J
Immediately after the mechanical input rises, electrical load remains 50 MW:
Pa = Pm − Pe = 60 − 50 = 10 MW = 0.10 pu
The question gives electrical frequency but not the generator pole count. If the intended synchronous mechanical speed is taken as ωs = 2π × 60 = 377 rad/s (a 2-pole assumption), the physical moment of inertia is:
J = 2E/ωs² = 2 × 1.00 × 10⁹ / 377² = 1.407 × 10⁴ kg·m²
The corresponding acceleration is:
α = Pa/(Jωs) = 10 × 10⁶/(1.407 × 10⁴ × 377) = 1.885 rad/s²
In the per-unit swing equation, the inertia coefficient is M = 2H/ωs:
M = 2 × 10 / 377 = 0.05305 pu·s; αelectrical = (ωs/(2H))Pa,pu = 1.885 electrical rad/s²
Qualification: if the machine has more than two poles, its mechanical synchronous speed and physical J differ; the question does not state the pole count. The per-unit swing-equation result uses the conventional 60 Hz electrical angular frequency.
Answer 5(c)
In the dark-lamp method, three lamps are connected across corresponding phases of the running busbars and the incoming generator:
Running busbars synchronising lamps incoming generator
A ─────────────────────( L1 )──────────────────── a
B ─────────────────────( L2 )──────────────────── b
C ─────────────────────( L3 )──────────────────── c
breaker connects a, b, c to A, B, C
First check that the incoming generator has the same phase sequence as the busbars. Adjust its excitation until its voltage matches the bus voltage, and adjust its speed until frequencies are nearly equal. With corresponding phases connected across each lamp, each lamp shows the difference between the two phase voltages. When the voltages are in phase and equal, all three lamps are dark. Close the breaker at that instant. If the frequency difference is small, the lamps brighten and darken slowly, making the synchronising instant easier to identify.
Section B: Electromagnetic Field Theory
Question 6
(a) (2 marks) State Faraday’s law and write its mathematical expression in integral form.
(b) (9 marks) A point charge Q of 300 nC is located at point P1(2, 0, 3) in free space. Determine the electric field intensity at point P2(0, 1, 2) due to charge Q.
(c) (9 marks) A lossy dielectric has a relative permeability μr = 1.5, relative permittivity εr = 1 and conductivity σ = 2 × 10⁻⁸ S/m. The electric field strength at the dielectric is E = 100 sin ωt az V/m. Determine the (i) frequency at which the conduction and displacement currents are equal; and (ii) instantaneous displacement current density at the frequency in (i).
Answer 6(a)
Faraday’s law states that the induced electromotive force around a closed path equals the negative time rate of change of magnetic flux through the path:
∮C E · dl = − d/dt ∫S B · dS = −dΦB/dt
Answer 6(b)
The displacement vector from the charge at P1 to P2 is:
R = (0−2)ax + (1−0)ay + (2−3)az = −2ax + ay − az m
|R| = √6 m; E = QR/(4πε₀|R|³)
E = (8.988 × 10⁹)(300 × 10⁻⁹)(−2ax + ay − az)/(√6)³
E = (−366.9ax + 183.5ay − 183.5az) V/m
The field magnitude is approximately 449.4 V/m, directed away from the positive charge toward P2.
Answer 6(c)
The conduction-current density amplitude is σE0; the displacement-current density amplitude is ωεE0. Equating them gives:
σE0 = ωε₀εrE0 ⇒ ω = σ/(ε₀εr)
f = ω/(2π) = (2 × 10⁻⁸)/(2π × 8.854 × 10⁻¹²) = 359.7 Hz
At this frequency, ωε = σ. Differentiate the specified electric field to obtain the instantaneous displacement current density:
Jd = ε₀εr ∂E/∂t = 100ε₀ω cos(ωt)az
Jd = 2.00 × 10⁻⁶ cos(2π × 359.7t)az A/m²
Question 7
(a) (3 marks) List three properties of electric field lines.
(b) (8 marks) A 10 GHz plane wave travelling in free space has an amplitude of 15 V/m. Determine the (i) velocity of propagation; (ii) wavelength; (iii) propagation constant; and (iv) characteristic impedance.
(c) (9 marks) Two uniform line charges L1 and L2 of density 8.854 nC/m each are located in space along the plane z = 0 at y = +6 m and y = −6 m respectively. At point P(0, 0, 6), determine the (i) electric field strength due to line 1; (ii) electric field strength due to line 2; and (iii) total field strength.
Answer 7(a)
- At each point, the tangent to a field line gives the direction of the electric field.
- Field lines originate on positive charge and terminate on negative charge, or extend to infinity.
- Field lines never cross; closer spacing represents a stronger field.
Answer 7(b)
For a wave in free space, the propagation speed is c and the wavelength is c/f:
v = c = 3.00 × 10⁸ m/s
λ = c/f = (3.00 × 10⁸)/(10 × 10⁹) = 0.0300 m
β = 2π/λ = 209.44 rad/m; α = 0
γ = α + jβ = j209.44 m⁻¹
η₀ = √(μ₀/ε₀) ≈ 376.7 Ω
Answer 7(c)
Interpret each uniform line charge as an infinitely long line parallel to the x-axis, at the stated y and z coordinates. From either line to P, the perpendicular distance is:
ρ = √(6² + 6²) = √72 = 8.485 m
For an infinite line charge, E = λ/(2πε₀ρ) aρ. Since λ/ε₀ ≈ 1,000 V, the field vectors are:
E₁ = (−13.26ay + 13.26az) V/m
E₂ = (13.26ay + 13.26az) V/m
The y-components cancel and the z-components add:
Etotal = 26.53az V/m
Question 8
(a) (4 marks) Draw electric field lines between each of the following pairs of charges: (i) two equal and opposite charges; (ii) two equal and like positive charges.
(b) (8 marks) A potential is given by the expression V = 2(x + 1)²(y + 2)²(z + 3)² volts in free space. Determine each of the following at point P(1, 2, 3): (i) potential; (ii) electric field; and (iii) electric flux density.
(c) (2 marks) Define each of the following with respect to magnetic circuits: (i) susceptibility; and (ii) hysteresis.
(d) (6 marks) A magnetic material has a relative permeability of 3. The material is placed in a magnetic field strength of 8 A/m. Determine the (i) magnetic flux density in the material; (ii) magnetization; and (iii) susceptibility.
Answer 8(a)
(i) Equal and opposite charges: field lines leave the positive charge and terminate on the negative charge.
↗──────────────↘
↗ ↘
(+) ───────────────────→ (−)
↘ ↗
↘──────────────↗
(ii) Equal positive charges: lines leave both charges and curve away from the region between them; no line joins one positive charge to the other.
↖ ↑ ↗ ↖ ↑ ↗
\\ | ) ( | /
\\| ) ( |/
(+) (+)
/ | ) ( |\\
/ | ) ( | \\
↙ ↓ ↘ ↙ ↓ ↘
Answer 8(b)
At P(1, 2, 3), x + 1 = 2, y + 2 = 4 and z + 3 = 6:
V = 2(2²)(4²)(6²) = 4,608 V
The electric field is the negative gradient of potential:
E = −∇V = −[4(x+1)(y+2)²(z+3)²ax + 4(x+1)²(y+2)(z+3)²ay + 4(x+1)²(y+2)²(z+3)az]
E(P) = (−4,608ax − 2,304ay − 1,536az) V/m
In free space, D = ε₀E:
D(P) = (−4.080 × 10⁻⁸ax − 2.040 × 10⁻⁸ay − 1.360 × 10⁻⁸az) C/m²
Answer 8(c)
- Magnetic susceptibility: the dimensionless ratio χm = M/H that relates a material’s magnetization M to the applied magnetic field strength H in a linear, isotropic material.
- Hysteresis: the lag of magnetic flux density or magnetization behind the applied magnetic field during cyclic magnetisation.
Answer 8(d)
For a linear magnetic material, μr = 1 + χm:
χm = μr − 1 = 3 − 1 = 2
B = μ₀μrH = (4π × 10⁻⁷)(3)(8) = 3.016 × 10⁻⁵ T
M = χmH = 2 × 8 = 16 A/m
Answers: B = 30.16 μT; magnetization M = 16 A/m; susceptibility χm = 2.