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52 lines
1.8 KiB
52 lines
1.8 KiB
id: opt-ex-01a
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type: calculation
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difficulty: medium
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points: 15
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related_lesson: opt-01
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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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Calculate both R_opt_power and R_opt_phase.
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hints:
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- "R_opt_power = 1/[ω(C_mut + C_sh)]"
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- "R_opt_phase = 1/[ω√(C_mut(C_mut + C_sh))]"
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- "Calculate ω = 2πf first"
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- "R_opt_power is always smaller than R_opt_phase"
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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 R_opt_power:"
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- "C_total = C_mut + C_sh = 10 + 8 = 18 pF"
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- "R_opt_power = 1/(ω × C_total)"
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- "= 1/(9.425×10⁵ × 18×10⁻¹²)"
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- "= 1/(16.965×10⁻⁶)"
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- "= 58.9 kΩ"
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- "Calculate R_opt_phase:"
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- "Product: C_mut × (C_mut + C_sh) = 10 × 18 = 180 pF²"
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- "Square root: √180 = 13.42 pF"
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- "R_opt_phase = 1/(ω × √180×10⁻¹²)"
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- "= 1/(9.425×10⁵ × 13.42×10⁻¹²)"
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- "= 1/(12.65×10⁻⁶)"
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- "= 79.1 kΩ"
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- "Compare: R_opt_power/R_opt_phase = 58.9/79.1 = 0.745"
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answer_power: "58.9"
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answer_phase: "79.1"
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unit: "kΩ"
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ratio: "0.745"
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tolerance: 3.0
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explanation: |
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This problem demonstrates the two critical resistances for spark optimization.
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R_opt_power (58.9 kΩ) maximizes real power transfer to the spark, while
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R_opt_phase (79.1 kΩ) minimizes the impedance phase angle magnitude. The ratio
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of 0.745 is typical - R_opt_power is usually 50-75% of R_opt_phase. These
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different values show that maximum power transfer and minimum phase angle are
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different optimization goals that cannot be achieved simultaneously.
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related_concepts: ["R_opt_power", "R_opt_phase", "power-optimization", "phase-optimization"]
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