Overvoltage from interrupting low inductive current

When a circuit breaker interrupts a small inductive current, such as the magnetising current of an unloaded transformer or a shunt reactor current, the arc is unstable. The breaker may force the current to zero before its natural zero. This is called current chopping, and it produces a high overvoltage.

How the overvoltage builds up

• Just before chopping, the current in the inductance is $i_0$. The energy stored in the magnetic field is $\tfrac{1}{2} L i_0^2$.
• After the breaker opens, the inductive current cannot stop at once. It has no path except the stray capacitance $C$ of the transformer winding and bushings. The magnetic energy is transferred to this capacitance and oscillates at the frequency

$$f = \frac{1}{2\pi\sqrt{LC}}$$

• At the peak of the oscillation all the energy is in the capacitance:

$$\tfrac{1}{2} C V_m^2 = \tfrac{1}{2} L i_0^2 \;\Rightarrow\; V_m = i_0\sqrt{\frac{L}{C}}$$

Because $L$ is large (henries) and $C$ is small (nanofarads), $\sqrt{L/C}$ is tens of kilo-ohms. So even a few amperes of chopped current gives hundreds of kV. For example, $i_0 = 10$ A, $L = 5$ H and $C = 0.01\ \mu$F give $V_m = 223.6$ kV.

    CB          i0 chopped
 ---/ ---+----------+
         |          |
         C (stray)  L (transformer)
         |          |
 --------+----------+--- ground

Consequences

• The high voltage across the breaker contacts can cause restrikes. Each restrike starts a new oscillation and the voltage may escalate.
• The transformer insulation is stressed.

Remedies: resistance switching (a resistor across the contacts damps the oscillation), surge arresters at the transformer, and breakers that do not chop current (for example SF$_6$ breakers with suitable design).