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Peterson coil for 33 kV 60 km line
Short answer
Tune the Peterson coil so inductive fault current cancels capacitive current: L = 1/(3ω²C). With total capacitance 0.9 μF, inductance is 3.753 H. The coil carries 16.16 A at 19.05 kV, giving a rating of about 307.9 kVA.
A 33 kV, 3 phase, 50 Hz overhead line 60 km long has a capacitance to ground of each line equal to 0.015 μF per km. Determine the inductance and kVA rating of the Peterson coil (arc suppression coil).
A Peterson coil (arc suppression coil) is an iron-cored reactor connected between the system neutral and earth, tuned to the line-to-earth capacitance so that the earth-fault current is neutralised.
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
- Line voltage V_L = 33 kV, f = 50 Hz, ω = 2π × 50 = 314.16 rad/s
- Capacitance per phase C = 0.015 μF/km × 60 km = 0.9 μF = 0.9 × 10⁻⁶ F
Inductance of the coil
L = 1 / (3 ω² C)
= 1 / (3 × 314.16² × 0.9 × 10⁻⁶)
= 1 / 0.26648
= 3.753 H
kVA rating
V_ph = 33000/√3 = 19052.6 V
I_L = I_C = 3 V_ph ω C
= 3 × 19052.6 × 314.16 × 0.9×10⁻⁶
= 16.16 A
Rating = V_ph × I_L
= 19052.6 × 16.16
= 307.9 kVA
Answer: Inductance of Peterson coil = 3.753 H; kVA rating ≈ 307.9 kVA (coil current 16.16 A at 19.05 kV).
✓ Answer checked by Claude Opus 5.5 (high effort)
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