4.(a)(i) The configuration of a 400 KV 3 phase line is shown in figure. The radius of each sub-conductor is 2 cm. Calculate the…

(a)(i) Charging MVA of the bundled-conductor line Each phase has two sub-conductors of radius r = 0.02 m, 0.5 m apart. Equivalent radius for capacitance: D_s = √(r d) = √(0.02 × 0.5) = 0.1 m. Phase spacings from the figure: 10 m, 10 m and 20 m, so GMD = (10 × 10 × 20)^(1/3) = 12.6 m (assuming transposition). C = 2πε₀/ln(GMD/D_s) = 2π × 8.854 × 10⁻¹²/ln(126) = 5.563 × 10⁻¹¹/4.836 = 1.15 × 10⁻¹¹ F/m per phase, so for 300 km C = 3.45 μF. Q = V_LL²ωC = (400 × 10³)² × 100π × 3.45 × 10⁻⁶ = 1.73 × 10⁸ var. Charging MVA ≈ 173 MVAR.

(a)(ii) Most economical cable insulation diameter For a single-core cable with conductor radius a and sheath radius b, the maximum stress at the conductor is g = V/(a ln(b/a)). For a given g and V, b = a e^(V/(ga)). Writing y = V/(ga), b = (V/g)(e^y/y), which is least when d(e^y/y)/dy = 0, i.e. y = 1. So a = V/g and b/a = e = 2.718 (the second derivative is positive, so this is a minimum). With V = 400/√3 = 230.9 kV (taking the 100 kV/cm limit as an rms value) and g = 100 kV/cm: a = 2.31 cm, b = e × 2.31 = 6.28 cm. Overall diameter 2b = 2eV/g ≈ 12.5 cm. (If the limit is a peak value, V is replaced by √2 × 230.9 kV, giving about 17.8 cm.)

(b) Anderson's bridge Source across a–c; arm ab: unknown coil (R₁, L₁) with series resistor r₁; arm bc: R₃; arm da: R₂; arm dc: R₄; from d a resistor r leads to e, and a capacitor C joins e to c; the detector is between b and e. At balance V_b = V_e, so I₁ flows through ab and bc, and the same voltage I₁R₃ appears across C: I_C = jωCR₃I₁. Across dc: I₄R₄ = I_C(r + 1/jωC), and I₂ = I₄ + I_C. Also I₁(R₁ + r₁ + jωL₁) = I₂R₂ + I_Cr. Substituting, R₁ + r₁ + jωL₁ = R₂R₃/R₄ + jωCR₃[r + R₂ + rR₂/R₄]. Equating real and imaginary parts: R₁ = R₂R₃/R₄ − r₁ and L₁ = CR₃[r + R₂(1 + r/R₄)]. Phasor diagram at balance: take I₁ as reference. V_bc = I₁R₃ is in phase with I₁ and equals V_ec, so I_C leads I₁ by 90°. Adding I_C r to V_ec gives V_dc = I₄R₄ and fixes I₄; I₂ = I₄ + I_C gives V_ad = I₂R₂; V_ab (I₁R₁ plus the quadrature drop jωL₁I₁) equals V_ae = V_ad + I_C r, and V_ab + V_bc is the supply.

Convergence: the balance equations do not contain ω, r₁ appears only in the resistance equation and r only in the inductance equation, so the two adjustments are nearly independent and balance is reached in few steps, even for low-Q coils.

(c) Phase of the non-minimum-phase system at 3 rad/s The asymptotic plot is flat to the corner at 2 rad/s and then falls at −20 dB/dec, the dashed lines marking the break at 20 rad/s. Reading the labelled breaks as a pole at 2 rad/s and a right-half-plane zero at 20 rad/s (non-minimum phase), G(s) = K(1 − s/20)/(1 + s/2). A right-half-plane zero has the magnitude of a left-half-plane zero but adds phase lag: φ(ω) = −tan⁻¹(ω/2) − tan⁻¹(ω/20). At ω = 3: φ = −56.3° − 8.5° = −64.8°. (The magnitude alone does not fix the phase of a non-minimum-phase system; if the −20 dB/dec slope were to continue beyond 20 rad/s, only the pole at 2 rad/s would be visible.)