id: fund-ex-08-comprehensive
type: multi-part
difficulty: hard
points: 50
related_lesson: fund-08
question: |
  COMPREHENSIVE INTEGRATION EXERCISE

  A Tesla coil operates at 220 kHz with a 3.5-foot spark. FEMM analysis gives
  C_mut = 9 pF. Assume R = 60 kΩ.

  (a) Calculate C_sh, ω, G, B₁, B₂
  (b) Calculate Y_total and Z_total
  (c) Find φ_Z and compare to -45°
  (d) Calculate r and φ_Z,min
  (e) If V_top = 350 kV, find power dissipated

hints:
  - "Use C_sh ≈ 2 pF/foot for estimation"
  - "Calculate all intermediate values carefully"
  - "Use admittance formulas from fund-03"
  - "Compare actual φ_Z to φ_Z,min from topological constraint"
  - "Power = 0.5 × |I|² × R or 0.5 × |V|² × Re{Y}"

solution:
  steps:
    - "Part (a): Calculate components"
    - "C_sh = 2 pF/foot × 3.5 feet = 7 pF"
    - "ω = 2πf = 2π × 220×10³ = 1.382×10⁶ rad/s"
    - "G = 1/R = 1/(60×10³) = 16.67 μS"
    - "B₁ = ωC_mut = 1.382×10⁶ × 9×10⁻¹² = 12.44 μS"
    - "B₂ = ωC_sh = 1.382×10⁶ × 7×10⁻¹² = 9.67 μS"
    - "Part (b): Calculate Y_total"
    - "Denominator: G² + (B₁+B₂)² = 277.9 + (22.11)² = 277.9 + 488.9 = 766.8 μS²"
    - "Re{Y} = GB₂²/[G²+(B₁+B₂)²] = 16.67×93.5/766.8 = 1559/766.8 = 2.03 μS"
    - "Im{Y} = B₂[G²+B₁(B₁+B₂)]/[G²+(B₁+B₂)²]"
    - "= 9.67×[277.9+12.44×22.11]/766.8"
    - "= 9.67×[277.9+275.0]/766.8"
    - "= 9.67×552.9/766.8 = 6.98 μS"
    - "Y_total = 2.03 + j6.98 μS"
    - "Convert to impedance:"
    - "|Y| = √(2.03² + 6.98²) = √(4.12 + 48.72) = 7.27 μS"
    - "|Z| = 1/|Y| = 137.5 kΩ"
    - "φ_Y = atan(6.98/2.03) = 73.8°"
    - "φ_Z = -φ_Y = -73.8°"
    - "R_eq = 137.5 × cos(-73.8°) = 38.4 kΩ"
    - "X_eq = 137.5 × sin(-73.8°) = -132 kΩ"
    - "Z_total = 38.4 - j132 kΩ = 137.5 kΩ ∠-73.8°"
    - "Part (c): Compare to -45°"
    - "φ_Z = -73.8° is more capacitive than -45° (larger magnitude)"
    - "|X|/R = 132/38.4 = 3.44"
    - "Capacitive reactance is 3.44× the resistance"
    - "Part (d): Calculate topological constraint"
    - "r = C_mut/C_sh = 9/7 = 1.286"
    - "φ_Z,min = -atan(2√[1.286×2.286]) = -atan(2×1.716) = -atan(3.43) = -73.7°"
    - "Actual φ_Z = -73.8° ≈ φ_Z,min (operating near optimal phase!)"
    - "Part (e): Calculate power"
    - "Current: I = V/|Z| = 350×10³/137.5×10³ = 2.55 A peak"
    - "Power: P = 0.5 × I² × R_eq = 0.5 × 2.55² × 38.4×10³"
    - "= 0.5 × 6.50 × 38.4×10³ = 125 kW"
    - "Alternative: P = 0.5 × V² × Re{Y}"
    - "= 0.5 × (350×10³)² × 2.03×10⁻⁶ = 124 kW ✓"

  answer_a: "C_sh=7pF, ω=1.382e6 rad/s, G=16.67μS, B₁=12.44μS, B₂=9.67μS"
  answer_b: "Y=2.03+j6.98 μS, Z=137.5kΩ∠-73.8° or 38.4-j132 kΩ"
  answer_c: "φ_Z=-73.8°, more capacitive than -45°, ratio=3.44"
  answer_d: "r=1.286, φ_Z,min=-73.7°"
  answer_e: "125"
  unit_e: "kW"
  tolerance: 5.0

explanation: |
  This comprehensive problem integrates all fundamental concepts from Part 1. The
  solution demonstrates: (1) using empirical rules for estimation, (2) systematic
  admittance calculation, (3) conversion between Y and Z, (4) understanding phase
  constraints, and (5) power calculation methods.

  Key insights: The actual phase angle (-73.8°) is essentially at the minimum
  possible value (-73.7°), suggesting this R value is close to R_opt_phase. The
  power dissipated (125 kW) is substantial for a 3.5-foot spark. The capacitive
  reactance dominates (3.44× the resistance), which is typical for Tesla coil
  sparks with r > 1.

related_concepts: ["integration", "complete-analysis", "power-calculation", "phase-optimization"]