(a) (i) Draw the neat and properly labelled output voltage waveform of a three-phase, phase-controlled rectifier having firing…

(a)(i) Output waveform of a three-phase fully controlled bridge and derivation of V₀

Circuit and assumptions. A six-pulse bridge of thyristors is fed from lines a, b and c. The load current is continuous (a large inductance), and the devices and supply are ideal. The output at any instant is one of the six line voltages v_ab, v_ac, v_bc, v_ba, v_ca, v_cb, which are 60° apart. With α = 0 each line voltage is connected to the load for the 60° around its peak. The firing angle α is measured from the natural commutation point, where two adjacent line voltages cross.

The upper plot (α = 30°) shows the output staying positive; the lower plot (α = 75°) shows negative portions of every pulse. The red dashed line is the average V₀. Each thyristor conducts for 120°, and a new pair is fired every 60° in the sequence T1 T2, T2 T3, T3 T4, …, so the output has six pulses per cycle (300 Hz ripple on a 50 Hz supply). For α greater than 60°, parts of each pulse go negative. For α greater than 90°, the average becomes negative and the bridge inverts.

Derivation. Take one line voltage as v = √2 V_LL sin θ, with θ measured from its zero crossing. With α = 0 it is applied from θ = 60° to 120°. With firing angle α it is applied from π/3 + α to 2π/3 + α. The output repeats every π/3, so

V₀ = (3/π) ∫ from (π/3 + α) to (2π/3 + α) of √2 V_LL sin θ dθ = (3√2 V_LL/π)[cos(π/3 + α) − cos(2π/3 + α)]

Using cos A − cos B = 2 sin[(A + B)/2] sin[(B − A)/2] = 2 sin(π/2 + α) sin(π/6) = cos α:

V₀ = (3√2/π) V_LL cos α ≈ 1.35 V_LL cos α

Check: α = 0 gives the uncontrolled bridge value 1.35 V_LL, and α = 90° gives zero.

(a)(ii) Firing angle, line current and input power factor

Maximum output voltage (α = 0) with V_LL = 415 V: V₀,max = 1.35 × 415 = 3√2 × 415/π = 560.4 V

(1) Firing angle. V₀ = 0.8V₀,max ⇒ cos α = 0.8 ⇒ α = 36.87°; V₀ = 448.4 V.

(2) Load current. Constant current: I_d = P/V₀ = 15 000/448.4 = 33.46 A.

(3) RMS line current. Each line carries +I_d for 120° and −I_d for 120° per cycle, i.e. a quasi-square wave: I_L = I_d√(2/3) = 0.8165 × 33.46 = 27.32 A

(4) Input power factor. Taking the devices as lossless, input power = output power: PF = P/(√3 V_LL I_L) = 15 000/(√3 × 415 × 27.32) = 0.764 lagging

Equivalently, PF = (3/π)cos α = 0.955 × 0.8 = 0.764. Here 0.955 is the distortion factor (fundamental 26.09 A ÷ 27.32 A), and cos α = 0.8 is the displacement factor, since the fundamental lags the phase voltage by α.

(b)(i) Maximum power of a synchronous generator increases with excitation

Claim: at fixed busbar voltage V and frequency, the steady-state power limit P_max is proportional to the excitation emf E.

1. Power-angle equation. Neglect armature resistance and let E lead V by the load angle δ. The phasor diagram gives E = V + jX_sI, so X_sI cos φ = E sin δ, and per phase P = VI cos φ = (EV/X_s) sin δ

2. Maximum. V and X_s are fixed by the infinite bus and the machine. P varies only as sin δ and is greatest at δ = 90°: P_max = EV/X_s (per phase), or 3EV/X_s in total

3. Role of excitation. E is set by the field current through the open-circuit characteristic, E = 4.44fNΦ. Raising the field current raises E and therefore raises P_max in direct proportion (see curves). If R is included, P_max = EV/Z_s − V²R/Z_s², which also increases linearly with E.

4. Stability consequence. For a given turbine output, a larger E means a smaller δ and a larger margin between operating power and P_max. The steady-state stability limit therefore rises. An under-excited machine works near its limit and can pull out of step.

Conclusion: because P_max = EV/X_s with V fixed, the maximum power a generator can deliver to constant-voltage, constant-frequency busbars increases with excitation.

(b)(ii) Turbo-alternator: new power factor and increase in emf

Given. V_ph = 11 000/√3 = 6350.9 V; R = 0.5 Ω, X_s = 10 Ω; I₁ = 250 A at unity pf.

Constant power. Steam input is constant and efficiency unchanged, so the electrical output is constant: √3 × 11 000 × 250 × 1 = √3 × 11 000 × 300 × cos φ₂ ⇒ cos φ₂ = 250/300 = 0.833 lagging Raising the excitation makes the generator over-excited, so it now delivers lagging reactive power. φ₂ = 33.56° and I₂ = 300∠−33.56° = 250 − j165.8 A.

EMF before (generator: E = V + IZ): E₁ = 6350.9 + 250(0.5 + j10) = 6475.9 + j2500 = 6941.7 V per phase (δ₁ = 21.1°)

EMF after: E₂ = 6350.9 + (250 − j165.8)(0.5 + j10) = 6350.9 + (125 + 1658.3) + j(2500 − 82.9) = 8134.2 + j2417.1 = 8485.7 V per phase (δ₂ = 16.5°)

Percentage increase = (8485.7 − 6941.7)/6941.7 × 100 = 22.2%

The in-phase (power) component of the current is unchanged at 250 A. Only the reactive component of 165.8 A is added, and this needs about 22% more induced emf (14.70 kV against 12.02 kV line). The load angle falls from 21.1° to 16.5°, which makes the machine more stable.

(c) Surface charge and surface current density on the conductor

Set-up. For z > 0 (ε_r = 7, μ_r = 18, σ = 0): E = 10 cos(3×10⁸t − 15x) ẑ V/m. The region z ≤ 0 is a perfect conductor, where E = H = 0. The unit normal from the conductor into the dielectric is ẑ.

Surface charge density. The boundary condition is ρ_s = n̂ · D at z = 0⁺. ρ_s = ε_rε₀E_z = 7 × (10⁻⁹/36π) × 10 cos(3×10⁸t − 15x) = 6.19×10⁻¹⁰ cos(3×10⁸t − 15x) C/m²

At (3, 4, 0) and t = 0.8 ns: the phase is 3×10⁸ × 0.8×10⁻⁹ − 15 × 3 = 0.24 − 45 = −44.76 rad, and cos(−44.76) = 0.7125. ρ_s = 6.19×10⁻¹⁰ × 0.7125 = 4.41×10⁻¹⁰ C/m² ≈ 0.441 nC/m²

Magnetic field (Faraday's law). ∇ × E = −μ ∂H/∂t. With E = E_z(x, t) ẑ: ∇ × E = −(∂E_z/∂x) ŷ = −150 sin(ωt − 15x) ŷ ∂H/∂t = (150/μ) sin(ωt − 15x) ŷ ⇒ H = −[150/(μω)] cos(ωt − 15x) ŷ μω = 18 × 4π×10⁻⁷ × 3×10⁸ = 6786 ⇒ H = −0.0221 cos(3×10⁸t − 15x) ŷ A/m

Surface current density. The boundary condition is K = n̂ × H at z = 0⁺. K = ẑ × [−0.0221 cos(·) ŷ] = +0.0221 cos(3×10⁸t − 15x) x̂ A/m

At (3, 4, 0), t = 0.8 ns: K = 0.0221 × 0.7125 = 0.01575 x̂ A/m ≈ 15.8 mA/m along +x

The y-coordinate (4) does not enter because the field is independent of y. The given phase constant (15 rad/m) differs from ω√(με) = 11.2 rad/m for this medium. I have used the field exactly as stated, since the boundary conditions depend only on the fields at the surface.