Skip to main content

Earth fault neutralizer reactance

Short answer

Reactance X_L = 1/(3ωC). For full 1.5 μF, X_L = 707.4 Ω. For 90% length (1.35 μF), X_L = 786.0 Ω. Reactance increases as capacitance decreases to maintain current balance during earth faults.

In a 50 Hz overhead line, the capacitance of one line to earth was 1.5 μF. It was decided to use an earth fault neutralizer. Calculate the reactance to neutralize the capacitance of (i) 100% of length of the line (ii) 90% of length of line.

An earth fault neutraliser (Peterson coil) is a reactor connected between the neutral and earth. During a line-to-earth fault its inductive current equals the capacitive current 3V_phωC of the healthy phases, so:

V_ph / X_L = 3 V_ph ω C   →   X_L = 1 / (3 ω C)

For a fraction k of the line, C is replaced by kC.

Data

  • C = 1.5 μF (line to earth, per phase), f = 50 Hz, ω = 314.16 rad/s

(i) 100% of the line

X_L = 1 / (3 × 314.16 × 1.5×10⁻⁶)
    = 1 / 1.4137×10⁻³
    = 707.4 Ω
(L = X_L/ω = 2.252 H)

(ii) 90% of the line

C' = 0.9 × 1.5 = 1.35 μF
X_L = 1 / (3 × 314.16 × 1.35×10⁻⁶)
    = 1 / 1.2723×10⁻³
    = 786.0 Ω
(L = 2.502 H)

Answer: Reactance of the neutraliser = 707.4 Ω for 100% of the line and 786.0 Ω for 90% of the line.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Switchgear and Protection chapter 5 questions →