id: fund-ex-03a
type: calculation
difficulty: medium
points: 15
related_lesson: fund-03
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
  - Plasma resistance: R = 80 kΩ

  Calculate Y_total in rectangular form (real and imaginary parts).

hints:
  - "First calculate ω = 2πf"
  - "Then calculate G = 1/R, B₁ = ωC_mut, B₂ = ωC_sh"
  - "Use the formulas: Re{Y} = GB₂²/[G² + (B₁+B₂)²]"
  - "And: Im{Y} = B₂[G² + B₁(B₁+B₂)]/[G² + (B₁+B₂)²]"

solution:
  steps:
    - "Calculate angular frequency: ω = 2π × 150×10³ = 9.425×10⁵ rad/s"
    - "Calculate conductance: G = 1/R = 1/(80×10³) = 12.5 μS"
    - "Calculate susceptances: B₁ = ω×C_mut = 9.425×10⁵ × 10×10⁻¹² = 9.425 μS"
    - "B₂ = ω×C_sh = 9.425×10⁵ × 8×10⁻¹² = 7.54 μS"
    - "Calculate denominator: G² + (B₁+B₂)² = 156.25 + (16.965)² = 156.25 + 287.8 = 444.05 μS²"
    - "Calculate Re{Y}: Re{Y} = 12.5 × (7.54)² / 444.05 = 12.5 × 56.85 / 444.05 = 710.6 / 444.05 = 1.60 μS"
    - "Calculate Im{Y} numerator: G² + B₁(B₁+B₂) = 156.25 + 9.425×16.965 = 156.25 + 159.9 = 316.15 μS²"
    - "Calculate Im{Y}: Im{Y} = 7.54 × 316.15 / 444.05 = 2383.8 / 444.05 = 5.37 μS"

  answer: "Y = 1.60 + j5.37 μS"
  real_part: "1.60"
  imaginary_part: "5.37"
  unit: "μS"
  tolerance: 3.0

explanation: |
  This calculation demonstrates the admittance analysis method for the spark circuit.
  The real part (1.60 μS) represents conductance - the component that dissipates
  power in the plasma resistance. The imaginary part (5.37 μS) is the susceptance,
  representing energy storage in the capacitances. The susceptance is 3.4× larger
  than the conductance, indicating a strongly capacitive circuit - typical for
  Tesla coil sparks.

related_concepts: ["admittance-calculation", "complex-numbers", "conductance", "susceptance"]