---
title: "Overvoltage due to chopping of magnetising current"
subject: "High Voltage Engineering"
chapter: 3
key: "semester_8/high_voltage_engineering:magnetising-current-chopping"
source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/magnetising-current-chopping
site: BCT Engineering
---

# Overvoltage due to chopping of magnetising current

High Voltage Engineering · Chapter 3: Overvoltages in Power System

- Asked in: 2070 Bhadra, 6 marks
- Video: [Current Chopping | 5-Minute Concept | Switchgear and Protection](https://www.youtube.com/watch?v=o_Gvlm2RMv0) — TechInsight
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/magnetising-current-chopping
- Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-3

## Question

Describe how the switching overvoltage takes place due to chopping of the magnetising current. Also show that the effect is minimum when switching takes place at zero crossing of current.

## Short answer

Chopping forces the magnetising current to zero early, trapping energy ½Li² in the transformer inductance. This energy oscillates with stray capacitance C, giving peak voltage Vm = sqrt(v² + (L/C)i²). Interrupting at natural current zero leaves i=0, so Vm equals normal peak voltage; no overvoltage.

## Answer

When a breaker interrupts the small magnetising current of an unloaded transformer, the arc is weak. The current may be forced to zero (chopped) at an instantaneous value $i$ before its natural zero. The energy then trapped in the transformer inductance produces a switching overvoltage.

#### Mechanism

```text
  CB (opens, chops i)
 ---/ ---+-----------+
         |           |
         C          L  (magnetising
     (winding +      |   inductance)
      bushing)       |
 --------+-----------+--- ground
```

At the chopping instant:
- Current in the magnetising inductance $= i$. Magnetic energy $= \tfrac12 L i^2$.
- Voltage on the stray capacitance $= v$. Electric energy $= \tfrac12 C v^2$.

After the breaker opens, $L$ and $C$ form a closed oscillating circuit. The inductor current cannot change suddenly, so it charges $C$. Energy swings between $L$ and $C$ at $f = 1/(2\pi\sqrt{LC})$, often a few hundred Hz to a few kHz. The peak voltage $V_m$ is reached when all the energy is in $C$:

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

$$
V_m = \sqrt{v^2 + \frac{L}{C}i^2}
$$

#### Effect of the chopping instant

The magnetising current lags the voltage by about $90^\circ$. With $i = I_m\sin\theta$ at the chopping angle $\theta$, the capacitor voltage is $v = V_p\cos\theta$, where $V_p$ is the system peak voltage.

**Chopping at the current peak** ($\theta = 90^\circ$): $i = I_m$, $v = 0$.

$$
V_m = I_m\sqrt{\frac{L}{C}}
$$

This is the maximum overvoltage. $\sqrt{L/C}$ is of the order of tens of kΩ, so it can be several times the system voltage.

**Interruption at current zero** ($\theta = 0$): $i = 0$, so the magnetic energy $\tfrac12Li^2 = 0$. Then

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

No extra energy is released. The voltage across the transformer does not exceed the normal system peak, and it simply decays through losses. **The overvoltage is minimum (zero excess) when interruption takes place at the natural current zero.**

**Example:** $L = 5$ H, $C = 0.01\ \mu$F, so $\sqrt{L/C} = 22.36$ kΩ. Chopping at 10 A gives 223.6 kV. Interrupting at current zero gives no overvoltage.

#### Remedies

- Breakers with low chopping current. Vacuum and SF$_6$ breakers are designed to interrupt near the natural zero.
- Resistance switching across the contacts to damp the oscillation.
- Surge arresters or RC snubbers at the transformer terminals.

## Check yourself

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

1. The peak voltage after chopping at current i and capacitor voltage v is:
   A. Vm = v + i sqrt(L/C)
   B. Vm = i sqrt(LC)
   C. Vm = sqrt(v^2 + (L/C) i^2)
   D. Vm = i^2 L
   Answer: C. (1/2)C Vm^2 = (1/2)L i^2 + (1/2)C v^2 gives Vm = sqrt(v^2 + (L/C) i^2). (Key point: Energy balance)

2. The magnetising current lags voltage by 90 degrees. The maximum overvoltage occurs when chopping at:
   A. Voltage peak with zero current
   B. The current peak, where v = 0
   C. Current zero
   D. Any instant equally
   Answer: B. At the current peak all energy is magnetic: Vm = Im sqrt(L/C). (Key point: Worst chopping instant)

3. Why is the overvoltage minimum when interrupting at current zero?
   A. Capacitance becomes infinite
   B. The arc resistance is high
   C. The inductance is zero
   D. The magnetic energy (1/2)L i^2 is zero, so Vm = v = Vp
   Answer: D. No extra energy is trapped, so voltage does not exceed the normal system peak. (Key point: Zero crossing)

## Related in High Voltage Engineering

- [Current chopping overvoltage at zero and peak](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Achopping-overvoltage-7a) — High Voltage Engineering, chapter 3
- [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
- [Causes of switching overvoltage; inductive current interruption](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aswitching-overvoltage-inductive-current) — High Voltage Engineering, chapter 3
- [Maximum switching overvoltage derivation](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aswitching-overvoltage-max-derivation) — High Voltage Engineering, chapter 3, asked 2×
- [Overvoltage from interrupting low inductive current](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Alow-inductive-current-interruption) — High Voltage Engineering, chapter 3
- [Switching overvoltage with trapped charge and Rs](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aswitching-overvoltage-trapped-charge-rs) — High Voltage Engineering, chapter 3, asked 3×

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