When a circuit breaker interrupts a fault, the voltage that appears across its contacts immediately after the arc goes out is called the restriking voltage — the transient recovery voltage. If this voltage rises faster than the contact gap can recover its dielectric strength, the arc restrikes and the breaker fails. This post derives the restriking voltage from first principles and proves that its rate of rise (RRRV) is directly proportional to the natural frequency of the circuit.

The circuit after contact separation

After the contacts separate, the faulted circuit reduces to a simple series L–C loop fed by the supply voltage: the system inductance L and the stray capacitance C across the breaker, with the resistance neglected. Two facts set the initial conditions:

• The fault current lags the supply voltage by 90°, so at the instant of arc extinction (t = 0, when the current is zero) the supply voltage is at its peak value Vm.
• At t = 0 the capacitor voltage is zero, because the arc had shorted the contacts: v_c(0) = 0, and the current is zero: i(0) = 0.

Figure 3.3: Transient restriking voltage after arc extinction. The voltage rises from zero to a peak of 2Vm in half a natural cycle; its steepest slope (maximum RRRV) occurs at ω_nt = π/2. (Schematic, not to scale — one natural oscillation cycle shown.)

Derivation

Applying KVL to the L–C loop after separation, with the supply at its peak Vm:

L(di/dt) + v_c = Vm

Since the capacitor current is i = C(dv_c/dt), this becomes a second-order equation in v_c:

LC(d²v_c/dt²) + v_c = Vm

Solving by Laplace transform with the initial conditions v_c(0) = 0 and i(0) = 0, and defining the natural angular frequency ω_n = 1/√(LC) and natural frequency f_n = 1/(2π√(LC)), the restriking voltage is:

v_r(t) = v_c(t) = Vm(1 − cos ω_nt) = Vm(1 − cos 2πf_nt)

Peak value and time to peak

The cosine term is −1 when ω_nt = π, so the restriking voltage reaches its maximum there:

Peak v_r = 2Vm at ω_nt_p = π, so t_p = π/ω_n = 1/(2f_n)

In words: half a natural cycle after the arc goes out, the voltage across the breaker is twice the supply peak — the capacitor has been charged to 2Vm. This is why the first half-cycle after interruption is the most dangerous moment for the breaker.

Rate of rise of restriking voltage (RRRV)

Differentiating the restriking voltage gives the rate at which it rises:

RRRV(t) = dv_r/dt = Vm ω_n sin(ω_nt) = 2π f_n Vm sin(2πf_nt)

The sine term is 1 when ω_nt = π/2, so the maximum rate of rise is:

RRRV_max = 2π f_n Vm   (at ω_nt = π/2)

The average rate of rise is the peak voltage divided by the time to reach it:

Average RRRV = v_r(max) / t_p = 2Vm / (1/(2f_n)) = 4 Vm f_n

Both the maximum and the average RRRV contain f_n as a direct multiplier — so RRRV ∝ f_n, which is what was to be proved. A circuit with a higher natural frequency (small L or small C) produces a faster-rising restriking voltage, and is therefore harder for the breaker to handle. This is exactly why resistance switching, which lowers the effective natural frequency and damps the oscillation, is used on breakers facing steep RRRV.

Key points to remember

• After arc extinction the circuit is an L–C loop with v_c(0) = 0 and i(0) = 0, while the supply sits at its peak Vm.
• Restriking voltage: v_r(t) = Vm(1 − cos ω_nt), with ω_n = 1/√(LC) and f_n = 1/(2π√(LC)).
• Peak v_r = 2Vm at t_p = 1/(2f_n) — half a natural cycle after extinction.
• Maximum RRRV = 2π f_n Vm at ω_nt = π/2; average RRRV = 4 Vm f_n.
• Therefore RRRV ∝ f_n: higher natural frequency means a steeper, more dangerous recovery voltage.

Based on Power System Protection & Switchgear, Paper II, Q3(c), Exam 11 June 2026.