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Current chopping overvoltage in 132 kV breaker

Short answer

The surge impedance of the transformer-bushing circuit is sqrt(L/C) = sqrt(5/10⁻⁸) = 22.36 kΩ. Chopping 10 A gives V = 10 × 22.36 kΩ = 223.6 kV across the breaker poles. The oscillation frequency is 1/(2π√(LC)) ≈ 712 Hz.

In a 132 kV circuit breaker the bushing to ground capacitance is 0.01 μF and the transformer inductance succeeding the circuit breaker is 5 H. Calculate the voltage appearing across the poles of the circuit breaker if it breaks magnetizing current of 10 A flowing in the transformer.

When the breaker chops the magnetising current ii, the energy in the transformer inductance moves to the bushing capacitance:

12CV2=12Li2  ⇒  V=iLC\frac12 CV^2 = \frac12 Li^2 \;\Rightarrow\; V = i\sqrt{\frac{L}{C}}

Data: L=5L = 5 H, C=0.01 μF=1×10−8C = 0.01\ \mu\text{F} = 1\times10^{-8} F. The current is chopped at an instantaneous value of i=10i = 10 A.

Surge impedance of the transformer–bushing circuit

LC=51×10−8=5×108=22 360.7 Ω\sqrt{\frac{L}{C}} = \sqrt{\frac{5}{1\times10^{-8}}} = \sqrt{5\times10^{8}} = 22\,360.7\ \Omega

Voltage across the breaker poles

V=10×22 360.7=223 607 V≈223.6 kV\begin{aligned} V &= 10 \times 22\,360.7 \\ &= 223\,607\ \text{V} \approx 223.6\ \text{kV} \end{aligned}

Frequency of oscillation

f=12πLC=12π5×10−8=711.8 Hzf = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{5 \times 10^{-8}}} = 711.8\ \text{Hz}

For comparison, the normal peak phase voltage of a 132 kV system is 1322/3=107.8132\sqrt2/\sqrt3 = 107.8 kV. The chopping overvoltage is therefore about 2.07 times the normal peak. This voltage appears across the open contacts and can cause restriking.

Answer: voltage across the breaker poles ≈ 223.6 kV (oscillating at about 712 Hz).

✓ Answer checked by Claude Opus 5.5 (high effort)

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