--- title: "Recovery, restriking voltage, RRRV and derivation" subject: "Switchgear and Protection" chapter: 6 key: "semester_6/switchgear_and_protection:restriking-voltage-derivation" source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/restriking-voltage-derivation site: BCT Engineering --- # Recovery, restriking voltage, RRRV and derivation Switchgear and Protection · Chapter 6: Circuit Breakers - Asked in: 2077 Chaitra, 6 marks - Video: [SGP205 Derivation for Restriking Voltage and RRRV](https://www.youtube.com/watch?v=R5HaOtZ21lc) — Lectures in Electrical Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/restriking-voltage-derivation - Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6 ## Question Define following terms: (i) Recovery voltage (ii) Re-striking voltage (iii) RRRV. Also, derive an expression for the restriking voltage in terms of system inductance and capacitance. ## Short answer Recovery voltage: power-frequency voltage after transients. Restriking voltage: v(t) = V_m(1−cos ω_n t) across contacts, ω_n=1/√(LC), maximum 2V_m. RRRV: dV/dt, peak V_m/√(LC), determines whether arc restrikes. ## Answer #### (i) Recovery voltage The normal-frequency (50 Hz) rms voltage that appears across the CB contacts after the final arc extinction and after the high-frequency transients have died out. Its peak is about *V_m* sin φ. #### (ii) Restriking voltage The transient, high-frequency voltage that appears across the contacts at or near current zero during arc interruption. It oscillates about the recovery voltage and can reach 2*V_m*. #### (iii) RRRV The **rate of rise of restriking voltage**, d*v*/d*t*, in kV/μs. If the voltage across the gap rises faster than the gap regains dielectric strength, the arc restrikes. It is the main factor in arc extinction. #### Expression for restriking voltage Consider a short circuit at the CB terminals of a generator with inductance *L* and stray capacitance *C* across the contacts (resistance neglected). ```text e = V_m cos ωt (~)----^^^L^^^----+----o/ o----X fault | | earth === C | earth ``` 1. Before interruption, *C* is shorted by the fault. The fault current lags *e* by 90°, so at current zero *e* = *V_m* (taken constant during the short transient). 2. After the arc goes out at current zero, *L* and *C* form a series circuit driven by *V_m*. Let *v* be the voltage across *C* (= across the contacts) and *i* the current: ```text V_m = L di/dt + v, i = C dv/dt V_m = LC d²v/dt² + v ``` 3. Taking the Laplace transform with *v*(0) = 0, *i*(0) = 0: ```text V_m/s = LC s² V(s) + V(s) V(s) = V_m / [ s (1 + s² LC) ] = V_m ω_n² / [ s (s² + ω_n²) ], ω_n = 1/√(LC) = V_m [ 1/s − s/(s² + ω_n²) ] ``` 4. Inverse transform: ```text v(t) = V_m (1 − cos ω_n t), ω_n = 1/√(LC) ``` #### Results ```text f_n = 1 / (2π √(LC)) Peak value = 2 V_m at t = π √(LC) RRRV = dv/dt = V_m ω_n sin ω_n t RRRV (max) = V_m / √(LC) at t = (π/2) √(LC) ``` So a smaller capacitance or inductance raises the natural frequency and the RRRV, making interruption harder. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the derivation, why is e = V_m at current zero? A. Because the fault current lags the voltage by 90 degrees B. Because the circuit is purely resistive C. Because the capacitor is charged to V_m D. Because the frequency is zero Answer: A. The fault current is inductive, lagging by 90 degrees, so voltage is at its peak when current is zero. (Key point: Assumption at current zero) 2. After arc extinction the LC circuit is described by A. V_m = L di/dt + v with i = C dv/dt, i.e. V_m = LC d^2v/dt^2 + v B. V_m = Ri + v C. V_m = LC dv/dt D. V_m = L/C d^2v/dt^2 Answer: A. Substituting i = C dv/dt into the series loop equation gives the second-order equation. (Key point: Differential equation) 3. The restriking voltage obtained is A. v = V_m sin(w_n t) B. v = V_m (1 - cos w_n t), w_n = 1/sqrt(LC) C. v = V_m e^(-t/LC) D. v = 2 V_m cos w_n t Answer: B. The inverse Laplace transform of V_m w_n^2 / [s (s^2 + w_n^2)] gives V_m (1 - cos w_n t). (Key point: Solution) 4. How does smaller L or C affect interruption? A. Lowers RRRV and eases interruption B. Raises natural frequency and RRRV, making interruption harder C. Has no effect D. Lowers the peak voltage to V_m Answer: B. Max RRRV = V_m/sqrt(LC) increases as L or C decreases. 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