(a) It is given that E⃗ = Eₘ sin (ω t - α z) â_y in free space α > 0. (i) Determine D⃗, B⃗ and H⃗. Plot E⃗ and H⃗ at t = 0…

(a)(i) D, B and H, and the plot at t = 0

Assumptions. The medium is free space, which is lossless and source-free (σ = 0, ρ_v = 0, J = 0). The fields are time-harmonic, so no static field is added when integrating.

D. From the constitutive relation D = ε₀E:

D = ε₀Eₘ sin(ωt − αz) â_y C/m²

B from Faraday's law. Faraday's law is ∇ × E = −∂B/∂t. Only E_y exists and it varies only with z, so:

∇ × E = −(∂E_y/∂z) â_x = αEₘ cos(ωt − αz) â_x

Setting this equal to −∂B/∂t and integrating in time gives:

B = −(αEₘ/ω) sin(ωt − αz) â_x Wb/m²

H. From H = B/μ₀:

H = −(αEₘ/ωμ₀) sin(ωt − αz) â_x A/m

Ampère's law fixes α. Ampère's law is ∇ × H = ∂D/∂t. Here ∇ × H = (∂H_x/∂z) â_y = (α²Eₘ/ωμ₀) cos(ωt − αz) â_y, and ∂D/∂t = ωε₀Eₘ cos(ωt − αz) â_y. Equating the two gives α² = ω²μ₀ε₀, so α = ω√(μ₀ε₀). Then α/(ωμ₀) = √(ε₀/μ₀) = 1/120π, and

H = −(Eₘ/120π) sin(ωt − αz) â_x A/m

At t = 0:

• E = −Eₘ sin(αz) â_y
• H = (Eₘ/120π) sin(αz) â_x

E and H are perpendicular to each other and to z. They are in time and space phase, reaching their maxima together. At every z, E × H points along +â_z; for example, at αz = π/2, (−â_y) × (+â_x) = +â_z.

(a)(ii) Travelling wave in +z; speed and E/H set by free space

Travelling wave. The field depends on t and z only through the phase φ = ωt − αz. A point of constant phase satisfies ωt − αz = constant, so z = (ωt − constant)/α. Its velocity is dz/dt = ω/α, which is positive because α > 0. The pattern therefore moves bodily in the +z direction without change of shape. The same conclusion follows from the one-dimensional wave equation, ∂²E_y/∂z² = μ₀ε₀ ∂²E_y/∂t², which E_y satisfies once α² = ω²μ₀ε₀.

Speed.

u = ω/α = 1/√(μ₀ε₀)

With the given values, μ₀ε₀ = 4π × 10⁻⁷ × (1/36π) × 10⁻⁹ = (1/9) × 10⁻¹⁶ s²/m², so

u = 3 × 10⁸ m/s

Ratio E/H.

E/H = ωμ₀/α = √(μ₀/ε₀) = √(4π × 10⁻⁷ × 36π × 10⁹) = 120π ≈ 377 Ω

This is the intrinsic impedance η₀.

Both results contain only μ₀ and ε₀; neither Eₘ nor ω appears. So the wave speed and the ratio E/H depend solely on the properties of free space.

(b)(i) Slip-ring induction motor

Data: s_m = 0.125, T_max = 2T_fl, N_s = 120f/P = 120 × 50/4 = 1500 rpm.

With stator impedance neglected, s_m = R₂/X₂, and the torque ratio is

T/T_max = 2/(s/s_m + s_m/s)

I. Full-load slip and speed. Put T_fl/T_max = 1/2 and x = s/s_m:

x + 1/x = 4 → x² − 4x + 1 = 0 → x = 2 ± √3

The stable (low-slip) root is x = 0.268.

• s_fl = 0.125 × 0.268 = 0.0335 (3.35 %)
• N_fl = 1500(1 − 0.0335) = 1449.8 rpm

II. Starting current. The rotor current is I₂ = E₂/√((R₂/s)² + X₂²), which is proportional to 1/√((s_m/s)² + 1). Hence:

I_st/I_fl = √((s_m/s_fl)² + 1)/√(s_m² + 1) = √(3.732² + 1)/√(1.0156) = 3.864/1.008

I_st ≈ 3.83 × I_fl

Cross-check through torque. T_st/T_max = 2/(8 + 0.125) = 0.246, so T_st = 0.492 T_fl. Since T ∝ I²/s, (I_st/I_fl)² = 0.492/0.0335 = 14.7, which again gives 3.83.

(b)(ii) Synchronous motor for p.f. correction

Existing load. P₁ = 600 kW at 0.6 lagging, so Q₁ = 600 × tan(cos⁻¹0.6) = 600 × 4/3 = 800 kVAr (lagging).

Synchronous motor input. P₂ = 100/0.9 = 111.1 kW. The 90 % efficiency matters here, because the motor draws its losses from the supply.

Combined requirement.

• P = 600 + 111.1 = 711.1 kW
• At 0.9 lagging, Q = 711.1 × tan(cos⁻¹0.9) = 711.1 × 0.4843 = 344.4 kVAr

Motor reactive power. Q₂ = 344.4 − 800 = −455.6 kVAr. The motor must draw leading reactive power, i.e. run over-excited.

Results.

• kVA rating: S₂ = √(111.1² + 455.6²) = 468.9 kVA
• Operating p.f.: cos φ₂ = 111.1/468.9 = 0.237 leading (φ₂ ≈ 76.3°)

The motor must be over-excited so that it draws leading current and cancels the plant's lagging kVAr.

(c) Buck-boost converter

Operating point.

• V₀ = −V_s D/(1 − D) = −12 × 0.25/0.75 = −4 V. The output is inverted.
• R = |V₀|/I₀ = 4/1.25 = 3.2 Ω.
• Average inductor current: I_L = I₀/(1 − D) = 1.667 A.

(i) Inductor ripple. During the on-time, V_s appears across L:

Δi_L = V_s D/(Lf) = (12 × 0.25)/(150 × 10⁻⁶ × 20 × 10³) = 3/3 = 1.0 A (peak-to-peak)

The inductor current therefore swings between I_min = 1.667 − 0.5 = 1.167 A and I_max = 2.167 A. Since I_min is above zero, the converter runs in CCM.

(ii) Critical values.

Critical inductance, at the CCM/DCM boundary where I_min = 0 (Δi_L = 2I_L):

L_c = (1 − D)²R/(2f) = 0.75² × 3.2/(2 × 20 × 10³) = 45 µH

Critical capacitance, for continuous capacitor voltage where ΔV_c = 2V₀:

C_c = D/(2fR) = 0.25/(2 × 20 × 10³ × 3.2) = 1.95 µF

The chosen L = 150 µH is 3.3 × L_c and C = 250 µF is far above C_c, so operation is safely continuous. The actual output ripple is only ΔV_c = I₀D/(fC) = 1.25 × 0.25/(20 × 10³ × 250 × 10⁻⁶) = 0.0625 V, which is 1.6 % of |V₀|.