Revision material: This lesson covers the KNEC Diploma in Electrical and Electronic Engineering Module III June/July 2022 Electrical Power Systems and Electromagnetic Field Theory past paper. The paper asks candidates to answer three questions from Section A and two from Section B. All eight questions are included here for revision. Answers are independently prepared study guides and are not an official KNEC marking scheme.
Section A: Electrical Power Systems
Question 1
(a) (4 marks) Explain two types of conductor vibrations in power systems.
Answer 1(a)
- Aeolian vibration: Wind flowing past a conductor sheds alternating vortices, producing relatively high-frequency, low-amplitude oscillation. Repeated motion near clamps can cause fatigue and strand damage.
- Galloping: Wind acts on an asymmetrically iced or wet conductor and produces low-frequency, large-amplitude movement. It can cause phase conductors to clash, flash over, or impose large mechanical loads on supports.
(b) (6 marks) With aid of a labelled diagram, derive an expression for sag for supports at unequal levels.
Answer 1(b)
Assume a span of horizontal length L with the right support B higher than the left support A by h. Let the conductor weight per unit length be w, and let H be the horizontal component of tension. For small sag, approximate the conductor by a parabola whose lowest point is O.
B (higher support)
●
/|
conductor _/ | h
/ |
A (lower support) ●···················· straight support chord
\__● O____/
x₁ x₂
O is x₁ from A and x₂ from B; x₁ + x₂ = L
Taking the lowest point as the origin, the parabolic equation is y = wx²/(2H). If the supports are at horizontal distances x₁ and x₂ from O, then x₁ + x₂ = L. Their height difference gives:
h = w(x₂² − x₁²)/(2H) = w(x₂ − x₁)L/(2H)
x₁ = L/2 − Hh/(wL) and x₂ = L/2 + Hh/(wL)
The sags measured vertically from the lower and higher support levels to the lowest point are therefore:
sA = wx₁²/(2H) = (w/2H)[L/2 − Hh/(wL)]²
sB = wx₂²/(2H) = (w/2H)[L/2 + Hh/(wL)]²
Here sB − sA = h. These equations use the usual parabolic, small-sag approximation.
(c) (6 marks) An overhead transmission line of length 600 m is supported at equal levels. The cross-sectional area of the conductor is 0.2 cm². The line is subjected to the following conditions: vertical permissible sag = 8 m; breaking stress = 7000 kg/cm²; wind pressure = 3 kg/m² of the projected area; specific gravity of conductor = 9 g/cm³. Determine the (i) effective weight of the line; (ii) safety factor.
Answer 1(c)
The conductor diameter is obtained from its circular cross-section. Its mass per metre follows from the stated density and cross-sectional area.
d = √(4A/π) = √[4(0.2)/π] = 0.5046 cm = 0.005046 m
wv = 9 g/cm³ × 0.2 cm² × 100 cm/m = 180 g/m = 0.18 kgf/m
The wind load acts on the projected conductor area per metre. Treating the paper’s kg load units as kgf:
ww = 3 kgf/m² × 0.005046 m = 0.01514 kgf/m
weffective = √(wv² + ww²) = √(0.18² + 0.01514²) = 0.18064 kgf/m ≈ 1.772 N/m
Use the stated vertical sag to find the horizontal tension component, then include the vertical and wind reactions when estimating the resultant support tension.
H = wvL²/(8sv) = 0.18 × 600²/(8 × 8) = 1012.5 kgf
Tsupport ≈ √[H² + (weffectiveL/2)²] = 1013.95 kgf
Breaking load = breaking stress × area = 7000 × 0.2 = 1400 kgf
Safety factor = breaking load/Tsupport = 1400/1013.95 ≈ 1.38
Answers: effective resultant line load ≈ 0.18064 kgf/m; safety factor ≈ 1.38.
(d) (4 marks) State four merits of constant voltage transmission using a synchronous phase modifier.
Answer 1(d)
- It supplies or absorbs reactive power to regulate the receiving-end voltage as the load changes.
- When over-excited, it improves lagging power factor and reduces line current for a given real-power transfer.
- Lower current reduces transmission losses and voltage drop.
- Voltage support improves usable transmission capacity and system stability.
Question 2
(a) (i) (4 marks) State four effects of surges on power systems transmission lines.
(ii) (3 marks) Outline three features of an overhead earth wire with reference to surge protection in power systems.
Answer 2(a)
(i) Effects of surges:
- They can puncture or flash over line and equipment insulation.
- They can damage transformers, generators, switchgear, insulators and other connected equipment.
- They can cause interruption of supply when protection trips the affected circuit.
- They can create dangerous overvoltages, arcing and fire risk, and shorten insulation life.
(ii) Overhead earth-wire features:
- It is mounted above the phase conductors so it intercepts direct lightning strokes.
- It is bonded to the tower steelwork and grounded at each tower to provide a path for lightning current to earth.
- Its shielding angle and position are selected to protect all phase conductors; it also needs adequate electrical and mechanical strength.
(b) (5 marks) With aid of a labelled voltage characteristic curve, explain the operation of a surge diverter.
Answer 2(b)
Voltage V
↑
│ surge region:
│ steep current rise
│ /
│ /
│────────────────● clamping / knee voltage
│ normal voltage │ small leakage current
│_________________│________________________→ Current I
Isurge → earth
During normal operation, a surge diverter presents very high impedance, so only a small leakage current flows. When a lightning or switching surge raises the terminal voltage above its protective level, its nonlinear element conducts heavily and diverts surge current to earth. The diverter limits the voltage across the protected equipment. When the surge has passed, it returns to its high-impedance state; in a gapped design, the power-frequency follow current is extinguished and the gap recovers.
(c) (8 marks) The terminal ends of two long transmission lines A and C are connected by a cable B of length 1 km. The surge impedances of A, B and C are 300 Ω, 80 Ω and 600 Ω respectively. A rectangular voltage wave of 30 kV and infinite length is initiated in A and travels to C. (i) Sketch the diagram of the arrangement. (ii) Determine: (I) transmitted wave voltage (V₂) to the cable; (II) transmitted wave voltage (V₄) to transmission line C; (III) reflected wave voltage (V₃) back to the cable.
Answer 2(c)
Incident wave 30 kV →
Line A, Z₁ = 300 Ω ── junction 1 ── Cable B, Z₂ = 80 Ω, length 1 km ── junction 2 ── Line C, Z₃ = 600 Ω
│ V₂ → │ V₄ →
V₁ ← (back into A) V₃ ← (back into B)
At a junction from surge impedance Z1 to Z2, the voltage transmission and reflection coefficients are 2Z2/(Z1 + Z2) and (Z2 − Z1)/(Z1 + Z2).
V₂ = 30 × 2(80)/(300 + 80) = 12.632 kV
The wave incident on the second junction is V₂. The transmission into line C is:
V₄ = V₂ × 2(600)/(80 + 600) = 22.291 kV
The wave reflected back into cable B is:
V₃ = V₂ × (600 − 80)/(600 + 80) = 9.659 kV
Answers: V₂ = 12.632 kV; V₄ = 22.291 kV; V₃ = +9.659 kV. The positive sign means the second-junction reflected voltage has the same polarity as V₂. The 1 km cable length sets the travel delay but does not change these first-junction voltage coefficients.
Question 3
(a) (5 marks) With aid of a line equivalent circuit representation, explain the Ferranti effect on overhead transmission lines.
Answer 3(a)
Sending end series line impedance Receiving end
Vₛ ────────┬────────────── Z ──────────────┬────────── Vᵣ
│ │
C/2 C/2
│ │
earth earth
(light load / open circuit at receiving end)
The Ferranti effect is the rise of receiving-end voltage above the sending-end voltage on a long line under light-load or open-circuit conditions. The line’s distributed capacitance draws a leading charging current. That current flowing through the line’s series inductance produces a voltage rise toward the receiving end. The effect becomes more pronounced with line length, frequency and capacitance, and can be reduced with shunt reactors or suitable voltage control.
(b) (9 marks) A 132 kV, 250 km long, 50 Hz transmission line delivers 30 MVA at 0.85 power factor lagging. The conductor resistance is 0.12 Ω/km, equilateral spacing is 2 m, and the effective conductor diameter is 1.8 cm. Using the nominal π method, determine the (i) impedance of the line, Z; (ii) sending-end voltage, Vₛ; (iii) voltage regulation.
Answer 3(b)
Work per phase. The conductor radius is r = 0.009 m, and the inductive GMR is approximately 0.7788r. For an equilateral three-phase line:
GMR = 0.7788(0.009) = 0.007009 m
L′ = 2 × 10⁻⁷ ln(2/0.007009) H/m = 1.13074 mH/km
C′ = 2πε₀/ln(2/0.009) = 0.0102953 μF/km
Z = 250[0.12 + j(2π × 50 × 1.13074 × 10⁻³)] = 30 + j88.808 Ω/phase
Y = j2π(50)(250 × 0.0102953 × 10⁻⁶) = j0.00080859 S/phase
The receiving-end phase voltage and current are:
Vᵣ = 132/√3 = 76.210 kV/phase
Iᵣ = 30 × 10⁶/(√3 × 132 × 10³) ∠−cos⁻¹(0.85) = 131.216∠−31.788° A
For the nominal π model:
Y/2 Z Y/2
Vₛ ────┬───────────────////───────────────┬──── Vᵣ
│ │
earth earth
A = D = 1 + YZ/2 = 0.964095 + j0.012129
C = Y(1 + YZ/4) = −4.904 × 10⁻⁶ + j0.00079408 S
Vₛ = AVᵣ + ZIᵣ = 82.959 + j8.756 kV/phase = 83.419∠6.025° kV/phase
Vₛ,line = √3 × 83.419 = 144.49 kV line-to-line
For standard voltage regulation, hold the sending voltage fixed and calculate the no-load receiving voltage Vᵣ₀ = Vₛ/A:
% regulation = (|Vᵣ₀| − |Vᵣ,full-load|)/|Vᵣ,full-load| × 100 ≈ 13.53%
Answers: Z = 30 + j88.808 Ω/phase; Vₛ ≈ 144.49 kV line-to-line; standard regulation ≈ 13.53%. Comparing |Vₛ| directly with |Vᵣ| gives about 9.46%, but that is not the no-load receiving-end definition of regulation.
(c) (i) (2 marks) State two limitations of a horn-gap arrester in power system protection.
(ii) (4 marks) A 66 kV, three-phase, 50 Hz overhead line, 50 km long, has a capacitance to earth of 0.016 μF/km per phase. Determine: (I) line inductance; (II) capacitance current; (III) kVA rating of the arc-suppression coil.
Answer 3(c)
(i) Limitations:
- The breakdown voltage is affected by gap spacing, weather and electrode condition, so protection is comparatively imprecise.
- After the surge, the power-frequency arc or follow current may persist; the horn gap may not extinguish it reliably without a circuit breaker.
(ii) Calculations: Total phase-to-earth capacitance is 0.8 μF. The Petersen (arc-suppression) coil is tuned so its inductive current balances the total three-phase capacitive earth-fault current.
C = 0.016 × 50 = 0.8 μF per phase; Vphase = 66/√3 = 38.105 kV
IC,phase = ωCVphase = 2π(50)(0.8 × 10⁻⁶)(38.105 × 10³) = 9.577 A
IC,total = 3IC,phase = 28.731 A
Lcoil = 1/[3ω²C] = 1/[3(2π × 50)²(0.8 × 10⁻⁶)] = 4.222 H
Coil rating = VphaseIC,total = 38.105 × 10³ × 28.731 ≈ 1.095 × 10⁶ VA = 1095 kVA
Answers: tuned coil inductance ≈ 4.22 H; line capacitive current = 9.58 A per phase (28.73 A total); coil rating ≈ 1095 kVA.
Question 4
(a) (4 marks) With reference to protection systems, explain each of the following: (i) unit system; (ii) biased differential bus zone reduction.
Answer 4(a)
- Unit system: A restricted protection scheme uses current transformers at the boundaries of a defined zone. Under normal operation or an external fault, the currents entering and leaving the zone balance. An internal fault creates differential current, which operates the relay and trips the breakers enclosing that zone.
- Biased differential bus-zone scheme: Busbar CT currents are compared for each protected zone. The relay trips when the differential (operating) current exceeds a set proportion of the restraining or through-current. Bias restraint helps prevent false operation from CT errors or saturation during heavy external faults, while zone selection limits the outage to the faulty bus section. The paper prints “reduction”; this answer explains the usual biased differential bus-zone protection arrangement.
(b) (4 marks) Describe two essential characteristics of a protective relay system.
Answer 4(b)
- Reliability: the scheme should operate dependably for faults within its zone and remain secure for normal conditions or faults outside that zone.
- Selectivity: it should isolate only the faulty equipment or smallest practical section, leaving healthy sections in service.
(c) (5 marks) With aid of a diagram, describe the operation of an induction-type overcurrent relay.
Answer 4(c)
Line current → CT → relay operating coil / AC electromagnet
│ phase-displaced fluxes
▼
aluminium induction disc
spindle and moving contact ── fixed contact
▲ │
control spring circuit-breaker
trip coil
The line current is reduced by a current transformer and fed to the relay operating coil. The relay’s AC magnetic system produces phase-displaced fluxes, which induce eddy currents in an aluminium disc. Their interaction creates torque. At normal current the control spring holds the disc and contacts open. When current exceeds the pickup setting, operating torque overcomes the spring; the disc rotates and closes the trip contacts. The circuit-breaker opens. In an inverse-time relay, a larger fault current produces faster disc movement and a shorter operating time.
(d) (7 marks) A 6000 kVA, 3300 V, star-connected alternator has a synchronous reactance of 4 Ω and resistance of 0.6 Ω per phase. It is protected using the Merz-Price balanced-current system, which operates when out-of-balance current exceeds 40% of the load current. Determine: (i) out-of-balance current; (ii) proportion of the winding unprotected if the star point is earthed by a resistor of 8 Ω.
Answer 4(d)
The rated line current of a three-phase alternator is:
Irated = 6000 × 10³/(√3 × 3300) = 1049.73 A
The stated relay threshold is 40% of rated load current:
Ipickup = 0.40 × 1049.73 = 419.89 A
The phase-to-neutral voltage is 3300/√3 = 1905.3 V. Even ignoring the alternator winding impedance, the largest earth-fault current through the 8 Ω neutral resistor is:
Iearth,max ≤ 1905.3/8 = 238.16 A < 419.89 A
Including the alternator’s 0.6 + j4 Ω per-phase impedance would reduce the fault current further. Since even a fault at the line end cannot reach relay pickup, the relay cannot detect an earth fault anywhere along the winding under the stated settings.
Answers: out-of-balance pickup current ≈ 419.9 A; proportion of winding unprotected = 100% on the stated data.
Question 5
(a) (i) (2 marks) Distinguish between active and passive faults in protection systems.
(ii) (5 marks) A 200 kV, 50 Hz, three-phase line 180 km long has a conductor diameter of 2 cm, with delta spacing of 4 m. The surrounding air temperature is 30°C, while the altitude is 2500 m, corresponding to a barometric pressure of 73.15 cmHg. Take irregularity factor = 0.85 and surface irregularity factor = 0.7. Determine: (I) air-density factor, δ; (II) critical disruptive voltage, Vd; (III) visual critical voltage, Vv.
Answer 5(a)
(i) An active fault immediately creates an abnormal current path, such as a short circuit or earth fault, and usually requires rapid isolation. A passive fault is an abnormal condition such as an open circuit, broken conductor or poor connection that does not necessarily create a large short-circuit current, but can interrupt service or develop into an active fault.
(ii) Use r = 1 cm, D = 400 cm, barometric pressure b = 73.15 cmHg, and temperature t = 30°C. The air-density factor is:
δ = 3.92b/(273 + t) = 3.92(73.15)/(273 + 30) = 0.9464
For disruptive critical voltage, use surface irregularity factor m₀ = 0.7. The standard formula gives phase-to-neutral rms voltage in kV:
Vd,phase = 21.1m₀δr ln(D/r) = 21.1(0.7)(0.9464)(1)ln(400) = 83.75 kV
Vd,line = √3 × 83.75 = 145.05 kV line-to-line
For visual critical voltage, use irregularity factor mᵥ = 0.85:
Vv,phase = 21.1mᵥδr[1 + 0.3/√(δr)]ln(D/r) = 133.05 kV
Vv,line = √3 × 133.05 = 230.46 kV line-to-line
Answers: δ ≈ 0.9464; Vd ≈ 83.75 kV/phase (145.05 kV line-to-line); Vv ≈ 133.05 kV/phase (230.46 kV line-to-line).
(b) (i) (2 marks) Explain the term “dynamic stability” with reference to power system stability.
(ii) (2 marks) State two areas of application of the equal area criterion in power generation and transmission of voltages.
Answer 5(b)
(i) Dynamic stability is a power system’s ability to remain in synchronism after small, gradual disturbances while the dynamic responses of generators, controls and loads are taken into account. Stable oscillations should be damped rather than grow.
(ii) The equal-area criterion is used to:
- assess transient stability of a generator connected to a large system after a fault, switching event or sudden change in load;
- estimate a critical clearing angle or clearing time, or compare the stability margin before and after a change in network transfer capability.
(c) (9 marks) A 50 Hz synchronous generator supplies 600 MW of power and is connected to a power system delivering 100 MW. The maximum power angle is — 86°. A three-phase fault occurs at its terminals. Determine: (i) initial input power; (ii) pre-clearing angle; (iii) critical clearing time; (iv) critical clearing angle. Take H = 10 MJ/MVA on a 100 MVA base.
Answer 5(c)
Interpretation used: The scan’s “maximum power angle is — 86°” is taken to mean the maximum rotor swing angle δm = 86°. The calculation also interprets 600 MW as the maximum transferable power Pmax and 100 MW as the pre-fault steady power transfer. With those meanings, the initial mechanical input equals the pre-fault electrical output. A solid three-phase terminal fault is assumed to reduce electrical output to approximately zero, and the post-fault transfer curve is assumed unchanged.
At pre-fault steady state:
Pinput = Pm = 100 MW
Pm/Pmax = 100/600 = 1/6 = sinδ₀
δ₀ = sin⁻¹(1/6) = 9.594°
During the fault, Pe ≈ 0, so the accelerating area is Pm(δc − δ₀). Equating it to the post-fault decelerating area up to δm = 86° gives:
Pm(δc − δ₀) = Pmax(cosδc − cosδm) − Pm(δm − δc)
cosδc = cosδm + (Pm/Pmax)(δm − δ₀) = 0.2920
δc = cos⁻¹(0.2920) = 73.02°
On the 100 MVA base, Pm = 1 pu. During the fault the swing equation has constant accelerating power. With ωs = 2πf:
tc = √[4H(δc − δ₀)/(ωsPm,pu)] = √[4(10)(73.02° − 9.594°)(π/180)/(2π × 50 × 1)] = 0.375 s
Answers on the stated interpretation: initial input power = 100 MW; pre-clearing angle δ₀ ≈ 9.59°; critical clearing angle δc ≈ 73.02°; critical clearing time ≈ 0.375 s. If “maximum power angle” or the 600 MW / 100 MW values were intended to denote different quantities, the source question needs clarification and these conditional figures should be revised.
Section B: Electromagnetic Field Theory
Answer TWO questions from this section in the examination.
Question 6
(a) (2 marks) Distinguish between linearly and circularly polarized wave with reference to wave propagation.
Answer 6(a)
For a linearly polarized wave, the electric-field vector at a fixed point oscillates along a straight line. For a circularly polarized wave, two perpendicular field components of equal magnitude with a 90° phase difference make the electric-field vector rotate with constant magnitude, tracing a circle.
(b) (9 marks) A 60 MHz uniform plane wave propagates in a medium with εr = 20, μr = 1 and σ = 0.05 S/m. The amplitude of the electric-field intensity is 20 V/m. Determine: (i) angular frequency ω; (ii) complex permittivity ε; (iii) propagation constant γ; (iv) intrinsic impedance η.
Answer 6(b)
Using the ejωt phasor convention, the lossy-medium complex permittivity is εc = ε − jσ/ω.
ω = 2πf = 2π(60 × 10⁶) = 3.7699 × 10⁸ rad/s
ε = ε₀εᵣ = 8.8542 × 10⁻¹²(20) = 1.7708 × 10⁻¹⁰ F/m
εc = ε − jσ/ω = (1.7708 − j1.3263) × 10⁻¹⁰ F/m = ε₀(20 − j14.979)
For μ = μ₀μᵣ, calculate:
γ = √[jωμ(σ + jωε)] = α + jβ = (1.9858 + j5.9641) m⁻¹
η = √[jωμ/(σ + jωε)] = 71.505 + j23.809 Ω = 75.365∠18.416° Ω
Answers: ω = 3.7699 × 10⁸ rad/s; εc = (1.7708 − j1.3263) × 10⁻¹⁰ F/m; γ = 1.9858 + j5.9641 m⁻¹; η = 71.505 + j23.809 Ω. The specified 20 V/m amplitude does not change these medium constants.
(c) (4 marks) With reference to electrodynamics, define each of the following terms and write their mathematical expressions: (i) electric-field intensity E; (ii) electric-flux density D.
Answer 6(c)
- Electric-field intensity E: electric force per unit positive test charge at a point. In a linear medium, Differential definition: E = F/q; its SI unit is V/m (or N/C).
- Electric-flux density D: electric flux per unit area normal to the field; it describes the electric displacement through a surface. In a linear, isotropic medium, D = εE; its SI unit is C/m². Gauss’s law in differential form is ∇·D = ρv.
(d) (5 marks) For free-space concentric spherical shells with voltages V₁ = 0 V at radius r₁ = 0.2 m and V₂ = 200 V at radius r₂ = 4 m, determine: (i) electric-field intensity E; (ii) electric-flux density D.
Answer 6(d)
Between concentric spherical electrodes, the potential has the form V(r) = A + B/r. Applying the two boundary values gives:
V(r) = 210.526 − 42.1053/r V, 0.2 < r < 4 m
E = −∇V = −42.1053/r² ar V/m
D = ε₀E = −3.7281 × 10⁻¹⁰/r² ar C/m²
The negative radial direction means the field points inward, from the outer shell at 200 V toward the inner shell at 0 V.
Question 7
(a) (3 marks) Describe the significance of boundary conditions in electromagnetics analysis.
Answer 7(a)
Boundary conditions specify how electric and magnetic fields behave on either side of an interface between materials. They follow from Maxwell’s equations and account for any surface charge or surface current. They connect field solutions across regions and make electromagnetic problems with multiple media solvable.
(b) (7 marks) A laminated cylindrical core of diameter 0.1 m is uniformly wound and generates a flux density of magnitude B(t) = (1/π)sin(300t). Determine: (i) voltage per turn V; (ii) flux density B(r,t) due to a circular path r > 0.2 m.
Answer 7(b)
The core radius is 0.05 m and its cross-sectional area is A = π(0.05)². The flux through one turn is:
Φ(t) = B(t)A = [sin(300t)/π]π(0.05)² = 0.0025 sin(300t) Wb
By Faraday’s law, the induced voltage per turn is:
v(t) = −dΦ/dt = −0.75 cos(300t) V/turn
For a circular path outside the core, the changing enclosed flux induces a circumferential electric field. Faraday’s law gives:
∮E·dl = Eφ(2πr) = −dΦ/dt = −0.75 cos(300t)
Eφ(r,t) = −0.11937/r cos(300t) V/m, r > 0.2 m
Source wording note: the scanned paper appears to ask for flux density B(r,t) in part (ii). A changing magnetic flux around a circular path determines the induced electric field above; it does not determine an external magnetic flux density without additional current or material information. The result is therefore given for the likely intended electric-field quantity.
(c) (4 marks) State Maxwell’s equations in differential form.
Answer 7(c)
Gauss’s law for electricity: ∇·D = ρv
Gauss’s law for magnetism: ∇·B = 0
Faraday’s law: ∇×E = −∂B/∂t
Ampère–Maxwell law: ∇×H = J + ∂D/∂t
(d) (6 marks) A perfect dielectric medium has an electric-field intensity given by E = E₀ cos(ωt − kz)aₓ V/m. Determine: (i) the magnetic-field intensity Hᵧ; (ii) an expression for current density J.
Answer 7(d)
This is a uniform plane wave travelling in the +z direction. Since aₓ × aᵧ = a_z, the magnetic field is along +y. In a perfect dielectric, η = √(μ/ε).
Hᵧ = (E₀/η) cos(ωt − kz) A/m
A perfect dielectric has zero conductivity, so its conduction current density is zero. The time-varying electric field still produces displacement current density:
Jconduction = σE = 0
Jdisplacement = ∂D/∂t = ε∂E/∂t = −ωεE₀ sin(ωt − kz)aₓ A/m²
The Ampère–Maxwell current density in the field equation is the displacement-current density above when the medium is a perfect dielectric.
Question 8
(a) (2 marks) Distinguish between skin effect and skin depth with reference to electromagnetic waves.
Answer 8(a)
Skin effect is the tendency of alternating current and electromagnetic fields to concentrate near a conductor’s surface as frequency increases. Skin depth is the depth at which the field or current amplitude has fallen to 1/e (about 36.8%) of its surface value; for a good conductor it is approximately δs = √[2/(ωμσ)].
(b) (7 marks) A 5 GHz wave propagates in a medium characterized by permeability μᵣ = 2, permittivity εᵣ = 20 and conductivity σ = 4.0 S/m. The electric-field intensity is E = 0.2e−αzcos(2πft − βz)aₓ. Determine: (i) complex permittivity ε; (ii) propagation constant γ; (iii) attenuation constant α.
Answer 8(b)
Use the ejωt convention, ω = 2πf, ε = ε₀εᵣ, and μ = μ₀μᵣ.
ω = 2π(5 × 10⁹) = 3.1416 × 10¹⁰ rad/s
εc = ε − jσ/ω = ε₀(20 − j14.380) = (1.7708 − j1.2732) × 10⁻¹⁰ F/m
γ = √[jωμ(σ + jωε)] = α + jβ = (225.56 + j700.10) m⁻¹
α = 225.56 Np/m
Answers: εc = (1.7708 − j1.2732) × 10⁻¹⁰ F/m; γ = 225.56 + j700.10 m⁻¹; α = 225.56 Np/m. The supplied field expression shows the attenuation factor but is not needed to compute the medium constants.
(c) (6 marks) Table 1 shows the characteristics of different media. Complete the table.
| Property | Lossy dielectric | Lossless dielectric | Conductor |
|---|---|---|---|
| Conductivity | σ ≠ 0 (finite) | σ = 0 | σ ≫ ωε (ideal conductor: σ → ∞) |
| Permittivity | ε = ε₀εᵣ | ε = ε₀εᵣ | ε = ε₀εᵣ (finite-material model) |
| Permeability | μ = μ₀μᵣ | μ = μ₀μᵣ | μ = μ₀μᵣ |
For a lossy dielectric, conductivity is nonzero but not large compared with the displacement-current term. A lossless dielectric has zero conductivity. A good conductor has conductivity much larger than ωε; an ideal conductor is the limiting case σ → ∞. Permittivity and permeability remain material properties and can be specified for each medium.
(d) (5 marks) A charge q = 5 μC is placed at a distance a = 20 cm from an infinite grounded conducting plate. Determine: (i) force acting on the charge; (ii) work required to remove the charge to infinity.
Answer 8(d)
By the method of images, replace the grounded plane with an image charge −q the same distance behind the plane. The real and image charges are separated by 2a = 0.4 m. Thus the attractive force is:
F = (1/4πε₀)q²/(2a)² = (8.988 × 10⁹)(5 × 10⁻⁶)²/(0.4)² = 1.404 N
The force is directed toward the plate. The image-charge potential energy at distance a is U(a) = −q²/(16πε₀a); it approaches zero at infinity. The quasistatic external work required to take the charge to infinity is:
Wexternal = U(∞) − U(a) = q²/(16πε₀a) = 0.2809 J
Answers: force = 1.404 N toward the plate; external work required to remove the charge to infinity = 0.2809 J.