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Arc suppression coil for 0.0125 μF line

Short answer

Total capacitance per phase is 0.625 μF (0.0125 μF/km × 50 km). Using L = 1/(3ω²C) at 50 Hz gives inductance 5.404 H. Phase voltage is 19.05 kV; inductive current I_L = 3V_phωC = 11.22 A, so kVA rating = V_ph × I_L ≈ 213.8 kVA.

Calculate the inductance and kVA rating of the arc suppression coil of a 50 km long 33 kV, 50 Hz, 3 phase overhead transmission system having line to earth capacitance of each phase equal to 0.0125 μF.

An arc suppression coil (Peterson coil) is a reactor between neutral and earth whose inductive current cancels the capacitive earth-fault current, so the fault arc is self-extinguished.

Principle: during an earth fault on one phase, the healthy phases send a capacitive current I_C = 3 V_ph ωC into the fault. The Peterson coil in the neutral sends an equal inductive current I_L = V_ph / (ωL) through the fault, which cancels I_C, so the arc goes out. For resonance:

I_L = I_C
V_ph/(ωL) = 3 V_ph ωC
        L = 1 / (3 ω² C)

Rating of the coil = V_ph × I_L (the coil sees phase voltage during the fault).

Data and assumption

The length (50 km) is given, so the capacitance 0.0125 μF is taken as per km of each phase:

  • C = 0.0125 μF/km × 50 km = 0.625 μF per phase
  • V_L = 33 kV, f = 50 Hz, ω = 314.16 rad/s

Inductance

L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 0.625 × 10⁻⁶)
  = 5.404 H

kVA rating

V_ph = 33000/√3              = 19052.6 V
I_L  = 3 V_ph ω C
     = 3 × 19052.6 × 314.16 × 0.625×10⁻⁶
     = 11.22 A
Rating = V_ph × I_L = 19052.6 × 11.22
       = 213.8 kVA

Answer: L = 5.404 H; kVA rating ≈ 213.8 kVA.

(If 0.0125 μF were the total capacitance of the line, the same formula gives L ≈ 270 H and about 4.3 kVA, which is unrealistically small; the per-km reading is the intended one.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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