- Asked 2/16
- 2073 Magh · 8 marks
Resistance switching numerical (132 kV, 350 Ω)
Short answer
With L=15.915 mH, natural frequency is 6.74 kHz. With 350 Ω damping, damped frequency is 1.81 kHz. Critical resistance is 337.2 Ω, and the resistance giving f_d=f_n/3 is 357.6 Ω.
For a 132 kV system, the reactance and capacitance up to the location of circuit breaker is 5 Ω and 0.035 μF, respectively. A resistance of 350 Ω is connected across the contacts of the circuit breaker. Determine the (a) Natural frequency of oscillation (b) Damped frequency of oscillation (c) Critical value of resistance which will give no transient oscillation and (d) The value of resistance which will give damped frequency of oscillation one-third of the natural frequency of oscillation.
A resistance across the CB contacts damps the restriking voltage oscillation; if it is at or below the critical value, the voltage rises without oscillation.
Formulas (resistance R across the contacts, parallel R–L–C circuit)
Natural frequency f_n = 1 / (2π √(LC))
Damped frequency f_d = (1/2π) √(1/LC − 1/(4R²C²))
Critical resistance R_c = ½ √(L/C)
(R ≤ R_c gives no oscillation)
For f_d = f_n / k:
1/(4R²C²) = (1/LC)(1 − 1/k²)
R = R_c / √(1 − 1/k²)
Inductance from reactance: L = X/(2π × 50).
Data
X = 5 Ω, C = 0.035 μF, R = 350 Ω, f = 50 Hz.
L = 5 / 314.16 = 15.915 mH
(a) Natural frequency
1/LC = 1/(15.915×10⁻³ × 0.035×10⁻⁶) = 1.7952×10⁹
f_n = √(1.7952×10⁹) / 2π
= 6743 Hz ≈ 6.74 kHz
(b) Damped frequency
1/(4R²C²) = 1/(4 × 350² × (0.035×10⁻⁶)²) = 1.6660×10⁹
f_d = √(1.7952×10⁹ − 1.6660×10⁹) / 2π
= 1809 Hz ≈ 1.81 kHz
(c) Critical resistance
R_c = ½ √(15.915×10⁻³ / 0.035×10⁻⁶)
= ½ × 674.34 = 337.2 Ω
Since 350 Ω > 337.2 Ω, the given resistor still allows a (heavily damped) oscillation, as found in (b).
(d) Resistance for f_d = f_n/3
R = R_c / √(1 − 1/9) = 337.2 / 0.9428
= 357.6 Ω
Check: f_d = 2248 Hz = 6743/3 ✓
Answer: (a) f_n = 6.74 kHz; (b) f_d = 1.81 kHz; (c) R_c = 337.2 Ω; (d) R = 357.6 Ω.
✓ Answer checked by Claude Opus 5.5 (high effort)
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