id: opt-ex-01b
type: multi-part
difficulty: medium
points: 15
related_lesson: opt-01
question: |
  At 200 kHz, a spark has total capacitance C_total = 12 pF.

  (a) What is R_opt_power?
  (b) If V_top = 400 kV, estimate the maximum deliverable power (assume R is at
      optimal value and φ_Z ≈ -70°)

hints:
  - "R_opt_power = 1/(ω × C_total)"
  - "Power = 0.5 × |V|² × Re{Y}"
  - "Or use: P = 0.5 × |V|²/|Z| × cos(φ_Z)"
  - "At R_opt_power, typical phase is around -65° to -75°"

solution:
  steps:
    - "Part (a): Calculate R_opt_power"
    - "ω = 2π × 200×10³ = 1.257×10⁶ rad/s"
    - "R_opt_power = 1/(ω × C_total)"
    - "= 1/(1.257×10⁶ × 12×10⁻¹²)"
    - "= 1/(15.08×10⁻⁶)"
    - "= 66.3 kΩ"
    - "Part (b): Estimate maximum power"
    - "At R_opt_power with given capacitances, φ_Z ≈ -70° (typical)"
    - "Approximate |Z| ≈ R_opt_power / cos(-70°) = 66.3/0.342 ≈ 194 kΩ"
    - "Current: I = V/|Z| = 400×10³/194×10³ = 2.06 A peak"
    - "Power: P = 0.5 × V × I × cos(φ_Z)"
    - "= 0.5 × 400×10³ × 2.06 × cos(-70°)"
    - "= 0.5 × 400×10³ × 2.06 × 0.342"
    - "= 141 kW"
    - "Alternative: P ≈ 0.5 × I² × R = 0.5 × 2.06² × 66.3×10³ ≈ 141 kW"

  answer_a: "66.3"
  answer_b: "141"
  unit_a: "kΩ"
  unit_b: "kW"
  tolerance: 5.0

explanation: |
  R_opt_power is determined solely by frequency and total capacitance. At this
  resistance, power transfer is maximized. The estimated power (141 kW) is
  substantial, but achievable for medium-to-large DRSSTCs. This calculation shows
  why R_opt_power is critical for performance - operating far from this value
  significantly reduces delivered power. The estimate uses typical phase angle
  for operation at R_opt_power; exact value would require full admittance calculation.

related_concepts: ["R_opt_power", "maximum-power-transfer", "power-estimation"]