---
title: "Current chopping overvoltage at zero and peak"
subject: "High Voltage Engineering"
chapter: 3
key: "semester_8/high_voltage_engineering:chopping-overvoltage-7a"
source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/chopping-overvoltage-7a
site: BCT Engineering
---

# Current chopping overvoltage at zero and peak

High Voltage Engineering · Chapter 3: Overvoltages in Power System

- Asked in: 2074 Magh, 8 marks
- Video: [Current Chopping in CB](https://www.youtube.com/watch?v=CcP6umOxUd8) — ShreeDevi Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/chopping-overvoltage-7a
- Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-3

## Question

In a 132 kV circuit breaker, the bushing to ground capacitance is 0.01 μF and the transformer impedance succeeding the CB is 5 H. The circuit breaker interrupts the magnetizing current of 7 A (rms). Calculate the possible overvoltage due to current chopping
i) When magnetizing current is interrupted at zero value
ii) When magnetizing current is interrupted at peak value

## Short answer

Peak magnetising current is 7√2 = 9.90 A and normal peak phase voltage is 107.8 kV. Interrupted at zero current, Vm = 107.8 kV with no chopping overvoltage. Interrupted at peak current, Vm = 9.90 × sqrt(5/0.01 μF) ≈ 221.4 kV.

## Answer

Current chopping transfers the energy stored in the transformer inductance to the bushing-to-ground capacitance:

$$
\frac12 CV_m^2 = \frac12 Li^2 + \frac12 Cv^2
$$

Here $i$ is the current and $v$ is the capacitor voltage at the instant of interruption.

**Data:** $L = 5$ H, $C = 0.01\ \mu$F, $I_{rms} = 7$ A.

$$
\sqrt{\frac{L}{C}} = \sqrt{\frac{5}{0.01\times10^{-6}}} = 22\,360.7\ \Omega
$$

Peak magnetising current: $I_m = 7\sqrt2 = 9.899$ A.

Peak system phase voltage: $V_p = 132\sqrt2/\sqrt3 = 107.78$ kV.

The magnetising current lags the voltage by about $90^\circ$. So when the current is zero the voltage is at its peak, and when the current is at its peak the voltage is zero.

#### (i) Interrupted at current zero

$i = 0$, so the magnetic energy is $\tfrac12 L(0)^2 = 0$.

$$
V_m = \sqrt{v^2 + 0} = v = V_p = 107.78\ \text{kV}
$$

No energy is released into the capacitance. The only voltage left is the normal system peak (107.8 kV), which decays. **Overvoltage due to chopping = 0.** This is the ideal case.

#### (ii) Interrupted at peak current

$i = I_m = 9.899$ A and $v = 0$.

$$
\begin{aligned}
V_m &= I_m\sqrt{\frac{L}{C}} = 9.899 \times 22\,360.7 \\
&= 221\,359\ \text{V} \approx 221.4\ \text{kV}
\end{aligned}
$$

This is $221.4/107.78 = 2.05$ times the normal peak phase voltage.

| Case | $i$ at cut | Voltage (kV) |
|---|---|---|
| Current zero | 0 A | 107.8 (normal peak, no overvoltage) |
| Current peak | 9.90 A | 221.4 |

The oscillation frequency is $f = 1/(2\pi\sqrt{LC}) = 711.8$ Hz.

**Answer:** (i) no chopping overvoltage; the voltage stays at the normal peak of 107.8 kV. (ii) **221.4 kV**.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. An rms magnetising current of 7 A has a peak of
   A. 7 A
   B. 4.95 A
   C. 12.2 A
   D. 9.9 A
   Answer: D. Im = 7 sqrt2 = 9.899 A. (Key point: Peak current)

2. The magnetising current lags voltage by about 90°. So at current zero the voltage is
   A. at its peak
   B. zero
   C. half of peak
   D. equal to chopping level
   Answer: A. Quadrature means one is zero when the other is maximum. (Key point: Phase of current and voltage)

3. What overvoltage results from interruption at current zero?
   A. 221.4 kV
   B. No chopping overvoltage; only the normal 107.8 kV peak remains
   C. 107.8 kV added to 221.4 kV
   D. 223.6 kV
   Answer: B. No magnetic energy is stored at i = 0. (Key point: Chopping at current zero)

4. Interruption at peak current gives V =
   A. Vp sqrt(L/C)
   B. Im sqrt(C/L)
   C. Im sqrt(L/C) ≈ 221.4 kV
   D. Im^2 L/C
   Answer: C. All the 1/2 L Im^2 energy goes to C, as v = 0 then. (Key point: Chopping at peak current)

## Related in High Voltage Engineering

- [Current chopping overvoltage in 132 kV breaker](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Achopping-overvoltage-10a) — High Voltage Engineering, chapter 3
- [Overvoltage due to chopping of magnetising current](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Amagnetising-current-chopping) — High Voltage Engineering, chapter 3

## Related in other subjects

- [Specifying breaker on 132/66 kV transformer](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Abreaker-x-specification) — Switchgear and Protection, chapter 6