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51 lines
1.9 KiB
51 lines
1.9 KiB
id: opt-ex-01b
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type: multi-part
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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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At 200 kHz, a spark has total capacitance C_total = 12 pF.
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(a) What is R_opt_power?
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(b) If V_top = 400 kV, estimate the maximum deliverable power (assume R is at
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optimal value and φ_Z ≈ -70°)
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hints:
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- "R_opt_power = 1/(ω × C_total)"
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- "Power = 0.5 × |V|² × Re{Y}"
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- "Or use: P = 0.5 × |V|²/|Z| × cos(φ_Z)"
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- "At R_opt_power, typical phase is around -65° to -75°"
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solution:
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steps:
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- "Part (a): Calculate R_opt_power"
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- "ω = 2π × 200×10³ = 1.257×10⁶ rad/s"
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- "R_opt_power = 1/(ω × C_total)"
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- "= 1/(1.257×10⁶ × 12×10⁻¹²)"
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- "= 1/(15.08×10⁻⁶)"
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- "= 66.3 kΩ"
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- "Part (b): Estimate maximum power"
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- "At R_opt_power with given capacitances, φ_Z ≈ -70° (typical)"
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- "Approximate |Z| ≈ R_opt_power / cos(-70°) = 66.3/0.342 ≈ 194 kΩ"
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- "Current: I = V/|Z| = 400×10³/194×10³ = 2.06 A peak"
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- "Power: P = 0.5 × V × I × cos(φ_Z)"
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- "= 0.5 × 400×10³ × 2.06 × cos(-70°)"
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- "= 0.5 × 400×10³ × 2.06 × 0.342"
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- "= 141 kW"
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- "Alternative: P ≈ 0.5 × I² × R = 0.5 × 2.06² × 66.3×10³ ≈ 141 kW"
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answer_a: "66.3"
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answer_b: "141"
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unit_a: "kΩ"
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unit_b: "kW"
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tolerance: 5.0
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explanation: |
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R_opt_power is determined solely by frequency and total capacitance. At this
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resistance, power transfer is maximized. The estimated power (141 kW) is
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substantial, but achievable for medium-to-large DRSSTCs. This calculation shows
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why R_opt_power is critical for performance - operating far from this value
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significantly reduces delivered power. The estimate uses typical phase angle
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for operation at R_opt_power; exact value would require full admittance calculation.
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related_concepts: ["R_opt_power", "maximum-power-transfer", "power-estimation"]
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