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Peterson coil for full and 90% length

Short answer

For full line capacitance 0.1 μF, inductance L = 1/(3ω²C) = 33.77 H at 50 Hz. For 90% length, effective capacitance is 0.09 μF, giving L = 1/(3ω²×0.09 μF) = 37.53 H. Smaller capacitance requires larger inductance to maintain resonance.

A transmission line has a capacitance of 0.1 microfarad per phase. Determine the inductance of Peterson coil to neutralize the effect of capacitance of (i) complete length of (ii) 90% length of the line. f = 50 Hz.

For a Peterson coil to neutralise the capacitive earth-fault current of a line of capacitance C per phase, ωL = 1/(3ωC), i.e. L = 1/(3ω²C). If only a fraction k of the line length is to be neutralised, the capacitance to be tuned is kC, so L = 1/(3ω²kC).

Data

  • C = 0.1 μF per phase (whole line), f = 50 Hz
  • ω = 2π × 50 = 314.16 rad/s, ω² = 98696

(i) Complete length (k = 1)

L = 1 / (3 × 98696 × 0.1×10⁻⁶)
  = 1 / 0.029609
  = 33.77 H

(ii) 90% of the length (k = 0.9)

C' = 0.9 × 0.1 μF = 0.09 μF
L  = 1 / (3 × 98696 × 0.09×10⁻⁶)
   = 1 / 0.026648
   = 37.53 H

Answer: (i) L = 33.77 H for the complete line; (ii) L = 37.53 H for 90% of the line.

A smaller part of the line gives less capacitive current, so a larger inductance (smaller inductive current) is needed. In practice the coil has tappings so it can be retuned when the network length changes.

✓ Answer checked by Claude Opus 5.5 (high effort)

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