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46 lines
1.9 KiB
46 lines
1.9 KiB
id: fund-ex-03a
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type: calculation
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difficulty: medium
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points: 15
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related_lesson: fund-03
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question: |
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For a spark circuit with the following parameters:
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- Frequency: f = 150 kHz
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- Mutual capacitance: C_mut = 10 pF
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- Shunt capacitance: C_sh = 8 pF
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- Plasma resistance: R = 80 kΩ
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Calculate Y_total in rectangular form (real and imaginary parts).
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hints:
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- "First calculate ω = 2πf"
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- "Then calculate G = 1/R, B₁ = ωC_mut, B₂ = ωC_sh"
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- "Use the formulas: Re{Y} = GB₂²/[G² + (B₁+B₂)²]"
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- "And: Im{Y} = B₂[G² + B₁(B₁+B₂)]/[G² + (B₁+B₂)²]"
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solution:
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steps:
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- "Calculate angular frequency: ω = 2π × 150×10³ = 9.425×10⁵ rad/s"
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- "Calculate conductance: G = 1/R = 1/(80×10³) = 12.5 μS"
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- "Calculate susceptances: B₁ = ω×C_mut = 9.425×10⁵ × 10×10⁻¹² = 9.425 μS"
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- "B₂ = ω×C_sh = 9.425×10⁵ × 8×10⁻¹² = 7.54 μS"
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- "Calculate denominator: G² + (B₁+B₂)² = 156.25 + (16.965)² = 156.25 + 287.8 = 444.05 μS²"
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- "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"
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- "Calculate Im{Y} numerator: G² + B₁(B₁+B₂) = 156.25 + 9.425×16.965 = 156.25 + 159.9 = 316.15 μS²"
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- "Calculate Im{Y}: Im{Y} = 7.54 × 316.15 / 444.05 = 2383.8 / 444.05 = 5.37 μS"
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answer: "Y = 1.60 + j5.37 μS"
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real_part: "1.60"
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imaginary_part: "5.37"
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unit: "μS"
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tolerance: 3.0
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explanation: |
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This calculation demonstrates the admittance analysis method for the spark circuit.
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The real part (1.60 μS) represents conductance - the component that dissipates
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power in the plasma resistance. The imaginary part (5.37 μS) is the susceptance,
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representing energy storage in the capacitances. The susceptance is 3.4× larger
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than the conductance, indicating a strongly capacitive circuit - typical for
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Tesla coil sparks.
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related_concepts: ["admittance-calculation", "complex-numbers", "conductance", "susceptance"]
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