id: opt-ex-01a
type: calculation
difficulty: medium
points: 15
related_lesson: opt-01
question: |
  For a spark circuit with the following parameters:
  - Frequency: f = 150 kHz
  - Mutual capacitance: C_mut = 10 pF
  - Shunt capacitance: C_sh = 8 pF

  Calculate both R_opt_power and R_opt_phase.

hints:
  - "R_opt_power = 1/[ω(C_mut + C_sh)]"
  - "R_opt_phase = 1/[ω√(C_mut(C_mut + C_sh))]"
  - "Calculate ω = 2πf first"
  - "R_opt_power is always smaller than R_opt_phase"

solution:
  steps:
    - "Calculate angular frequency: ω = 2π × 150×10³ = 9.425×10⁵ rad/s"
    - "Calculate R_opt_power:"
    - "C_total = C_mut + C_sh = 10 + 8 = 18 pF"
    - "R_opt_power = 1/(ω × C_total)"
    - "= 1/(9.425×10⁵ × 18×10⁻¹²)"
    - "= 1/(16.965×10⁻⁶)"
    - "= 58.9 kΩ"
    - "Calculate R_opt_phase:"
    - "Product: C_mut × (C_mut + C_sh) = 10 × 18 = 180 pF²"
    - "Square root: √180 = 13.42 pF"
    - "R_opt_phase = 1/(ω × √180×10⁻¹²)"
    - "= 1/(9.425×10⁵ × 13.42×10⁻¹²)"
    - "= 1/(12.65×10⁻⁶)"
    - "= 79.1 kΩ"
    - "Compare: R_opt_power/R_opt_phase = 58.9/79.1 = 0.745"

  answer_power: "58.9"
  answer_phase: "79.1"
  unit: "kΩ"
  ratio: "0.745"
  tolerance: 3.0

explanation: |
  This problem demonstrates the two critical resistances for spark optimization.
  R_opt_power (58.9 kΩ) maximizes real power transfer to the spark, while
  R_opt_phase (79.1 kΩ) minimizes the impedance phase angle magnitude. The ratio
  of 0.745 is typical - R_opt_power is usually 50-75% of R_opt_phase. These
  different values show that maximum power transfer and minimum phase angle are
  different optimization goals that cannot be achieved simultaneously.

related_concepts: ["R_opt_power", "R_opt_phase", "power-optimization", "phase-optimization"]