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Restriking voltage numerical (11 kV, 10 mH)

Short answer

i) Vmax = 2×√2×11000/√3 = 17.96 kV. ii) f_n = 1/(2π√(10mH×0.02μF)) = 11.25 kHz. iii) Average RRRV = Vmax/(π√(LC)) = 0.404 kV/μs.

A 50 Hz, 11 kV generator is connected to a power system. The system inductance and capacitance per phase are 10 mH and 0.02 μF respectively. Calculate: i) Maximum voltage across the contacts of the CB at an instant when it passes through zero. ii) Frequency of transient oscillation. iii) Average rate of rise of voltage up to the first peak of oscillation, neglect the resistance.

The restriking voltage across the CB contacts after current zero is an L–C oscillation that rises to twice the peak phase voltage when resistance is neglected.

Method (resistance neglected)

  • Peak phase voltage V_m = √2 × V_L/√3.
  • Restriking voltage v = V_m(1 − cos ωₙt); maximum across contacts = 2V_m.
  • Time to first peak t = π√(LC) = 1/(2f_n).
  • Average rate of rise up to first peak = 2V_m / t.

Data

V_L = 11 kV, 50 Hz, L = 10 mH, C = 0.02 μF per phase.

i) Maximum voltage across the contacts

V_m = √2 × 11000/√3 = 8981 V
v_max = 2 V_m = 17 963 V ≈ 17.96 kV

ii) Frequency of transient oscillation

√(LC) = √(10×10⁻³ × 0.02×10⁻⁶) = 1.4142×10⁻⁵ s
f_n = 1/(2π × 1.4142×10⁻⁵)
    = 11 254 Hz ≈ 11.25 kHz

iii) Average rate of rise up to first peak

t = 1/(2 f_n) = 44.43 μs
Average RRRV = 17 963 V / 44.43 μs
             = 404.3 V/μs ≈ 0.404 kV/μs

Answer: i) 17.96 kV; ii) 11.25 kHz; iii) 0.404 kV/μs (4.04 × 10⁸ V/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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