KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III
Paper codes: 2521/305 and 2601/305 · Session: March/April 2025 · Time: 3 hours
Revision note: Attempt the questions before reading the solutions. These independently prepared answers are for revision and are not an official KNEC marking scheme. The paper has eight 20-mark questions; answer any three from Section A and any two from Section B.
SECTION A: ELECTRICAL POWER SYSTEMS
Question 1 (20 marks)
(a)(i) Explain conductor swinging as applied to overhead-line conductor vibrations.
Answer (a)(i). Conductor swinging is the lateral movement of a suspended overhead conductor under wind pressure. Wind pushes the conductor away from its vertical hanging position; the span and adjacent phases may oscillate, reducing electrical clearance and potentially causing phase clashing or mechanical damage.
(a)(ii) With the aid of a diagram, describe the operation of a Stockbridge damper. (8 marks total for part (a))
Answer (a)(ii). A Stockbridge damper consists of two metal counterweights attached to the ends of a short flexible messenger cable, which is clamped to the conductor near a support. Conductor vibration makes the weights move out of phase with the conductor. The messenger cable bends repeatedly, dissipating vibration energy as heat through internal friction and strand flexing.
Overhead conductor ──────────────── support
│ clamp
messenger cable
╱ ╲
counterweight counterweight
Vibration bends the messenger cable; the moving weights
dissipate energy and reduce conductor vibration.
(b) State two demerits of corona in overhead-line transmission. (2 marks)
Answer (b). Corona causes (1) power loss and reduced transmission efficiency, and (2) radio/communication interference. It can also produce audible noise, ozone and gradual conductor-surface deterioration.
(c) A 150 kV overhead transmission line has conductors of 2 cm diameter. Corona occurs when line voltage exceeds 250 kV rms. At 50 °C and 76 cm Hg, ionisation occurs at a potential gradient of 30 kV/cm. With irregularity factor 1, determine (i) air-density factor and (ii) conductor spacing. (10 marks)
Answer (c). For temperature t = 50 °C and pressure b=76 cm Hg, the air-density factor is:
δ = 3.92b / (273 + t) = (3.92 × 76) / (273 + 50) = 0.9224
Use the stated corona-onset line voltage of 250 kV rms to calculate the phase voltage. Conductor radius r=1 cm. Using Peek’s disruptive-voltage equation, where 30 kV/cm is the standard peak air-gradient represented by the 21.1 kV rms coefficient:
Vph = 250 / √3 = 144.34 kV rms
Vd = 21.1 m0 δ r ln(D/r), m0 = 1
ln(D/r) = Vph / (21.1 δ r) = 144.34 / (21.1 × 0.9224 × 1) = 7.417
D = r e7.417 ≈ 1,664 cm ≈ 16.6 m
Answers from the stated 250 kV threshold: δ = 0.9224; conductor spacing D ≈ 16.6 m. The question describes a 150 kV line but gives a 250 kV corona threshold; these values conflict. The spacing above follows the specified 250 kV threshold and is unusually large, so the source figure should be checked rather than silently changed.
Question 2 (20 marks)
(a)(i) Explain the effect of wind and ice loading on sag in overhead lines. (a)(ii) A 132 kV line has equal-level supports, span 180 m, conductor diameter 0.95 cm, conductor weight 0.8 kg/m, ultimate stress 5000 kg/cm², wind pressure 50 kg/m² of projected area and safety factor 5. Find (I) total effective weight and (II) sag. (10 marks total for part (a))
Answer (a)(i). Ice adds vertical weight and increases the conductor’s exposed diameter. Wind produces a transverse load. The conductor is therefore loaded by the resultant of its vertical and horizontal loads; this generally increases resultant sag and tension and deflects the conductor downwind.
Answer (a)(ii). Treat the kilogram force units in the scanned question as kgf. Cross-sectional area, breaking tension and working tension are:
A = πd²/4 = π(0.95)²/4 = 0.7088 cm²
Tbreak = (5000 kgf/cm²)(0.7088 cm²) = 3544 kgf
T = Tbreak/5 = 708.8 kgf
Wind load per metre is ww=p d, with the diameter converted to metres:
ww = 50 × 0.0095 = 0.475 kgf/m
we = √(w² + ww²) = √(0.8² + 0.475²) = 0.9304 kgf/m
For a shallow catenary, the sag in the direction of the resultant load is:
fresultant = weL²/(8T) = 0.9304 × 180²/(8 × 708.8) = 5.316 m
Its components are fv=wL²/(8T) vertically and fw=wwL²/(8T) horizontally:
fvertical = 4.571 m, fwind = 2.714 m
Answers: total effective weight we=0.9304 kgf/m; resultant sag =5.316 m (vertical sag =4.571 m, sideways deflection =2.714 m).
(b) Draw the connection of sequence networks for (i) line-to-ground and (ii) line-to-line-to-ground faults. (4 marks)
Answer (b). For a single line-to-ground fault, positive-, negative- and zero-sequence networks are connected in series. The sequence currents are equal. For a double line-to-ground fault, the positive-sequence source/network is in series with the parallel combination of the negative-sequence network and the zero-sequence network (including 3Zf, if the fault impedance is Zf).
(i) Single line-to-ground (LG)
E₁ ── Z₁ ── Z₂ ── Z₀ ── 3Zf ── ground
I₁ = I₂ = I₀
(ii) Double line-to-ground (LLG)
E₁ ── Z₁ ──●── Z₂ ───────── ground
└── Z₀ + 3Zf ─── ground
negative and zero sequence in parallel
For a bolted fault, Zf=0, so the zero-sequence branch is simply Z0.
(c) A three-phase, four-wire system under faulted loading has IR=200∠30° A, IY=80∠240° A, and IB=40∠180° A. Determine the sequence components of current in the red line. (6 marks)
Answer (c). Use the R-Y-B phase sequence and a=1∠120°. Fortescue’s transformation is:
IR0 = (IR + IY + IB)/3
IR1 = (IR + aIY + a²IB)/3
IR2 = (IR + a²IY + aIB)/3
Convert the given phasors to rectangular form, substitute, then convert each result back to polar form:
IR0 = 31.07 + j10.24 = 32.71∠18.24° A
IR1 = 91.07 + j44.88 = 101.53∠26.24° A
IR2 = 51.07 + j44.88 = 67.99∠41.31° A
Answers: zero-sequence red current 32.71∠18.24° A; positive-sequence red current 101.53∠26.24° A; negative-sequence red current 67.99∠41.31° A.
Question 3 (20 marks)
(a) With reference to overhead-line transmission, explain (i) voltage regulation and (ii) transmission efficiency. (4 marks)
Answer (a).
(i) Voltage regulation: The change in receiving-end voltage when the load changes from no-load to full-load while sending-end voltage is held constant, expressed as a percentage of full-load receiving voltage. A common approximate line expression is:
% regulation = (|VS| − |VR|) / |VR| × 100%
(ii) Transmission efficiency: The ratio of real power delivered to the load to real power supplied at the sending end:
η = (PR/PS) × 100%
(b) Draw and label the circuit and phasor diagram of a nominal-π medium transmission line. (4 marks)
Answer (b). The nominal-π model places half the total shunt admittance at each end and the total series impedance between them.
Series current I
┌────────────[ Z ]────────────┐
Vₛ o───● ●───o Vᵣ → load
│ │
Y/2 Y/2
│ │
⏚ ⏚
Vᵣ as reference: Iᵣ lags Vᵣ by φ (lagging load);
I꜀ᵣ = j(Y/2)Vᵣ leads Vᵣ by 90°;
I = Iᵣ + I꜀ᵣ; Vₛ = Vᵣ + IZ.
Phasor directions for a lagging load (schematic, not to scale):
I꜀ᵣ = j(Y/2)Vᵣ
↑
│ I = Iᵣ + I꜀ᵣ
Vᵣ reference ────────────●────────↗ I
╲
↘ Iᵣ (lags Vᵣ by φ)
This schematic assumes charging current exceeds the load’s lagging reactive component; the series-current angle changes if their relative magnitudes differ. Add I·R along I and jI·X 90° ahead of I to obtain Vₛ.
For a lagging load, the phasor construction starts with VR horizontal. Draw IR behind it by the load angle φ, and the receiving-end charging current IcR vertically upward. Their vector sum is the series current I. The resistive drop I·R is along I, the reactive drop jI·X leads I by 90°, and adding both drops to VR gives VS. The sending-end charging current is IcS=j(Y/2)VS; sending current is IS=I+IcS.
(c) A 66 kV single-phase, 100 km medium line delivers 2000 kW at 0.8 power factor lagging. Per kilometre, R=0.2 Ω, X=0.8 Ω, and susceptance =15×10−6 S. Using the end-condenser method, find (i) load current, (ii) sending-end current and (iii) voltage drop. (12 marks)
Answer (c). In the end-condenser approximation, place the total shunt susceptance at the receiving end. The total series impedance and shunt susceptance are:
Z = 100(0.2 + j0.8) = 20 + j80 Ω, B = 100(15 × 10⁻⁶) = 0.0015 S
Receiving-end load current:
IR = P/(VRpf)∠−cos⁻¹(0.8) = 2,000,000/(66,000 × 0.8)∠−36.87° = 37.88∠−36.87° A
The charging current and sending-end current are:
Ic = jBVR = j(0.0015)(66,000) = j99.0 A
IS = IR + Ic = (30.30 − j22.73) + j99.0 = 30.30 + j76.27 = 82.07∠68.33° A
Sending voltage and the series phasor drop are:
VS − VR = ZIS = −5.496 + j3.950 kV
VS = 66 + (−5.496 + j3.950) = 60.504 + j3.950 = 60.63∠3.73° kV
The magnitude of the series phasor drop is about 6.77 kV. Because the charging current is large, the calculated sending-end voltage magnitude is lower than the receiving-end magnitude; the line shows a 5.37 kV receiving-end voltage rise relative to the sending-end magnitude.
Answers: load current 37.88∠−36.87° A; sending-end current 82.07∠68.33° A; series drop −5.496+j3.950 kV (magnitude 6.77 kV), with a receiving-end voltage rise of about 5.37 kV by magnitude.
Question 4 (20 marks)
(a) With the aid of a diagram, describe the operation of an expulsion-type arrester. (6 marks)
Answer (a). An expulsion arrester is connected between line and earth. At normal voltage its series gaps are non-conducting. A surge breaks down the gaps and is diverted to earth. The resulting power-frequency follow-current arc decomposes the fibre lining of the tube and generates gas. The gas blast expels and cools the ionised gases, lengthens the arc and de-ionises the gap. At a current zero the arc extinguishes; the arrester then returns to its insulating state.
Line terminal
│
External series gap
│
┌───────────────┐
│ Fibre-lined │ ← insulating expulsion tube
│ arc tube │
│ Internal gap │
└───────┬───────┘
│
Earth terminal
↑ gas vent / arc products expelled
(b) An overhead line has inductance 2.5×10−7 H/km and capacitance 0.95×10−9 F/km. Determine (i) surge impedance and (ii) wave velocity. (4 marks)
Answer (b). Use the printed per-kilometre constants:
Z₀ = √(L/C) = √[(2.5 × 10⁻⁷)/(0.95 × 10⁻⁹)] = 16.22 Ω
v = 1/√(LC) = 1/√[(2.5 × 10⁻⁷)(0.95 × 10⁻⁹)] = 6.489 × 10⁷ km/s = 6.489 × 10¹⁰ m/s
Answers from the printed values: Z0=16.22 Ω; formal velocity =6.489×1010 m/s. This exceeds the speed of light and is not physically plausible for an ordinary overhead line. The scanned paper’s L/C exponents or units likely contain a typo; no corrected values are assumed here.
(c)(i) State three merits of solid-state relays. (c)(ii) A star-connected, three-phase, 20 MVA, 66 kV alternator uses circulating-current protection. Its star point is earthed through resistance r. With 90% of the stator winding protected against earth fault and an earth-fault relay setting of 10%, determine (I) the minimum current that will operate the relay and (II) r. (10 marks total for part (c))
Answer (c)(i). Solid-state relays have fast operating speed, no moving contacts, high reliability, low maintenance and low burden. Any three are valid merits.
Answer (c)(ii). Take the 10% relay setting as 10% of the rated primary phase current. The alternator rated line current is:
Irated = S/(√3VL) = 20 × 10⁶/(√3 × 66 × 10³) = 174.95 A
Thus the minimum primary current for relay operation is:
Ipickup = 0.10 Irated = 17.50 A
For 90% winding protection, the unprotected part near the neutral is 10% of the phase winding. A fault at its boundary applies 10% of phase voltage across the neutral resistor. Set this limiting fault current equal to relay pickup:
Vph = 66/√3 = 38.105 kV
r = 0.10Vph/Ipickup = (0.10 × 38,105)/17.50 = 217.8 Ω
Answers under the stated relay-setting interpretation: minimum operating current 17.50 A; neutral earthing resistance r≈218 Ω.
Question 5 (20 marks)
(a) State three assumptions used in transient-stability analysis. (3 marks)
Answer (a). Common classical assumptions are: (1) mechanical input power remains constant during the short disturbance; (2) generator internal emf behind transient reactance is treated as constant; and (3) damping, losses and governor response are neglected during the first-swing calculation. The system is represented by a lumped equivalent network and rotor speeds remain close to synchronous speed.
(b) Draw the swing curves at loss of stability for (i) steady-state stability and (ii) dynamic stability. (4 marks)
Answer (b). A swing curve plots rotor angle δ against time after a disturbance. At steady-state loss of stability, the rotor angle runs away and pole slipping follows. At dynamic instability, rotor-angle oscillations grow with time because damping is inadequate or negative.
(i) Steady-state instability (ii) Dynamic instability
Rotor angle δ Rotor angle δ
│ / │ /\ /‾\ /‾‾\
│ / │ / \ / \ / \
│ / │ / \__/ \____/ \__
└────────── time └──────────────────────── time
runaway angle growing oscillations
(c)(i) State two requirements of a protective relay. (c)(ii) With the aid of a diagram, describe the operation of a definite-distance relay. (c)(iii) An overcurrent relay has a 150% current setting and is connected through a 500/5 A CT. A fault current of 5000 A flows. Determine (I) pickup current, (II) fault current in the relay coil and (III) plug-setting multiplier. (13 marks total for part (c))
Answer (c)(i). Any two: selectivity/discrimination, sensitivity, reliability, speed, stability for external faults, or simplicity and economy.
Answer (c)(ii). A definite-distance relay measures apparent impedance Zmathrmapp=V/I. Its reach is set for a defined portion of the protected line. If a fault lies within reach, the measured impedance falls below the setting and the relay trips after its definite time delay.
Generator / bus ── CB ── CT ───────── protected line
│ current I
├───────────────┐
relay │
│ │
Bus VT voltage V ────────┘ │
Relay trips CB when V/I
is below the reach setting.
The current transformer supplies the current input and the voltage transformer supplies the voltage input. Under normal conditions or for faults beyond reach, the impedance criterion is not met and the relay does not trip. For a close-in fault, V/I decreases below the set impedance, the timer starts, and the circuit breaker trips after the selected fixed delay. A directional element may be added if reverse faults must be blocked.
Answer (c)(iii). A 150% setting on a 5 A relay gives a secondary pickup of:
Ipickup,sec = 1.50 × 5 = 7.5 A
The CT ratio is 500/5 = 100:1. Therefore primary pickup and relay-coil fault current are:
Ipickup,pri = 7.5 × 100 = 750 A
Ifault,relay = 5000 × (5/500) = 50 A
PSM = Ifault,relay/Ipickup,sec = 50/7.5 = 6.67
Answers: primary pickup 750 A (relay secondary setting 7.5 A); fault current in relay coil 50 A; plug-setting multiplier 6.67.
SECTION B: ELECTROMAGNETIC FIELD THEORY
Question 6 (20 marks)
(a) With reference to electromagnetic waves, state two (i) methods of photometric detection of data and (ii) types of electromagnetic-wave detectors. (4 marks)
Answer (a).
(i) Photometric detection methods: (1) Bolometric/thermal detection: absorbed radiation heats a sensing element, changing its resistance or thermoelectric output. (2) Photoelectric detection: incident photons release or generate charge carriers, producing a measurable current or voltage in a photocell or photodiode.
(ii) Detector types: Thermal detectors, such as bolometers, thermopiles and pyroelectric sensors; and photon detectors, such as photoelectric cells, photodiodes and photoconductive sensors. These are broad physical-response classes; common waveband examples include antennas or loops for radio waves and bolometric or photoelectric detectors for infrared/optical waves. A detector is selected to suit the waveband and signal level.
(b)(i) State Coulomb’s law. (b)(ii) Charges −4 μC and +5 μC are at (2,−1,3) and (0,4,−2), respectively. With zero potential at infinity, find the electric potential at (1,0,1). (10 marks total for part (b))
Answer (b)(i). Coulomb’s law states that the force between two point charges is along the line joining them, directly proportional to the product of their charges and inversely proportional to the square of their separation. In a medium of permittivity ε:
F = (1/(4πε)) |q₁q₂|/R²
Answer (b)(ii). Assume the coordinates are in metres. Electric potential is a scalar sum. Distances from each charge to P=(1,0,1) are:
R₁ = √[(1−2)² + (0+1)² + (1−3)²] = √6 = 2.449 m
R₂ = √[(1−0)² + (0−4)² + (1+2)²] = √26 = 5.099 m
Taking free-space Coulomb constant k=8.988×109 V·m/C:
V(P) = k(q₁/R₁ + q₂/R₂)
V(P) = 8.988 × 10⁹ [(-4 × 10⁻⁶)/2.449 + (5 × 10⁻⁶)/5.099] = −5.864 × 10³ V
Answer: V(P)≈−5.86 kV.
(c) For an infinitely long conductor, H=1/(2π r) aphi. Show that the Ampère circuital-law integral of H· d l around a closed path is equivalent to direct current I. (6 marks)
Answer (c). The printed expression omits I. To show the general law, use the standard field of a long straight conductor carrying current I: H=I/(2π r) aphi. Choose a circular Amperian path of radius r, centred on the conductor. At every point H is tangential and constant, and d l=aphi r dφ:
∮ H·dl = ∫₀²π [I/(2πr)] r dφ = [I/(2π)](2π) = I
This is Ampère’s circuital law for the steady current enclosed by the path. If the printed expression is taken literally with no I, the integral equals 1 A; the missing factor is required for arbitrary current.
Question 7 (20 marks)
(a) Explain (i) paramagnetic and (ii) diamagnetic materials. (4 marks)
Answer (a).
(i) Paramagnetic material: Has a small positive magnetic susceptibility and relative permeability slightly greater than one. It is weakly attracted into an applied magnetic field; its induced magnetisation is in the field direction.
(ii) Diamagnetic material: Has a small negative susceptibility and relative permeability slightly less than one. Its induced magnetisation opposes the applied field, so it is weakly repelled.
(b)(i) Write two of Maxwell’s equations in integral form and state their significance. (b)(ii) In free space, E=30cos(ω t−50x) ay V/m. Given εr=1, μr=1, determine (I) displacement-current density Jd and (II) magnetic-field intensity H. (13 marks total for part (b))
Answer (b)(i). Any two Maxwell equations are acceptable. Faraday’s law states that a time-varying magnetic flux induces a circulating electric field:
∮C E·dl = −d/dt ∬S B·dS
The Ampère–Maxwell law states that conduction current and time-varying electric flux produce a circulating magnetic field:
∮C H·dl = Ienc + d/dt ∬S D·dS
Answer (b)(ii). The phase ω t−50x describes propagation in the +x direction, with β=50 rad/m. In free space, v=c and ω=β c:
ω = 50 × 2.998 × 10⁸ = 1.499 × 10¹⁰ rad/s
The displacement-current density is Jd=∂ D/∂ t=ε0 ∂ E/∂ t:
Jd = −30ε₀ω sin(ωt − 50x) ay ≈ −3.982 sin(ωt − 50x) ay A/m²
For propagation in +x, E along +y requires H along +z, since E× H points in the propagation direction. The free-space intrinsic impedance is η0≈376.73 Ω:
H = (E/η₀) az = (30/376.73) cos(ωt − 50x) az = 0.07963 cos(ωt − 50x) az A/m
Answers: displacement-current density amplitude 3.982 A/m2, with the negative sine time dependence shown; magnetic-field intensity H=0.07963cos(ω t−50x) az A/m. The printed phrase “electric field intensity H” is a terminology error: H denotes magnetic-field intensity.
(c) Explain the significance of Faraday’s law in a time-varying magnetic field. (3 marks)
Answer (c). Faraday’s law says that a changing magnetic flux induces an electromotive force and a non-conservative electric field around a closed path. It is the operating principle of generators and transformers and, together with Ampère–Maxwell law, explains electromagnetic-wave propagation.
Question 8 (20 marks)
(a) State three characteristics of transverse electromagnetic waves. (3 marks)
Answer (a). (1) The electric field is perpendicular to the magnetic field. (2) Both fields are perpendicular to the direction of propagation. (3) In a uniform lossless medium, the fields are in phase and carry energy in the direction of propagation.
(b) Assuming conductivity σ=0, show that the electromagnetic-wave equation is ∇2 E−με ∂2 E/∂ t2=0. (9 marks)
Answer (b). In a homogeneous, source-free, lossless medium, Maxwell’s curl equations are:
∇ × E = −μ ∂H/∂t, ∇ × H = ε ∂E/∂t
Take the curl of Faraday’s law and substitute the Ampère–Maxwell law:
∇ × (∇ × E) = −μ ∂/∂t (∇ × H) = −με ∂²E/∂t²
Use the vector identity ∇×(∇× E)=∇(∇· E)−∇2E. In a charge-free homogeneous medium, ∇· E=0, hence the left side is −∇2E:
−∇²E = −με ∂²E/∂t²
Rearranging gives the required wave equation:
∇²E − με ∂²E/∂t² = 0
(c) A 200 MHz wave travels in a medium with μr=1, εr=20, and σ=4 mS/m. (i) Identify the dielectric type. (ii) Find intrinsic impedance Ω and attenuation constant α. (8 marks)
Answer (c). Convert σ=4 mS/m=0.004 S/m, and calculate the loss tangent:
tan δ = σ/(ωε) = 0.004 / [2π(200 × 10⁶)(20ε₀)] = 0.01798
Since σ/(ωε)≪1, this is a low-loss dielectric. With μ=μ0, ε=20ε0, f=200 MHz, and ω=2π f, the intrinsic impedance is:
η = √[jωμ/(σ + jωε)] ≈ 84.229 + j0.757 Ω = 84.23∠0.515° Ω
The attenuation constant is:
α = ω√(με/2 [√(1 + (σ/(ωε))²) − 1]) ≈ 0.1685 Np/m
Answers: low-loss dielectric; intrinsic impedance η≈84.23∠0.515° Ω; attenuation constant α≈0.1685 Np/m.