--- title: "TRV numerical: 1.7 mH, 0.0025 μF, 7500 A" subject: "Switchgear and Protection" chapter: 6 key: "semester_6/switchgear_and_protection:trv-1-7mh-7500a" source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/trv-1-7mh-7500a site: BCT Engineering --- # TRV numerical: 1.7 mH, 0.0025 μF, 7500 A Switchgear and Protection · Chapter 6: Circuit Breakers - Asked in: 2074 Bhadra, 8 marks - Video: [SGP206 Numerical on Restriking Voltage and RRRV](https://www.youtube.com/watch?v=dVQWEEci8XY) — 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/trv-1-7mh-7500a - Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6 ## Question A 50 Hz, three-phase synchronous generator has an inductance per phase of 1.7 mH and its neutral is grounded. It feeds a line through a circuit breaker. The total stray capacitance to ground of the generator and circuit breaker is 0.0025 μF. A fault occurs just beyond the circuit breaker, which opens when the symmetrical short circuit current is 7500 A (rms). Ignoring the first pole to clear factor, determine the following: (i) Natural frequency of oscillations (ii) Peak value of TRV (iii) Time at which peak value of TRV occurs (iv) Maximum rate of rise of TRV (v) Time at which the maximum in part (iv) occurs. ## Short answer Using V_m=√2×7500×0.5341=5665 V and the LC transient v=V_m(1−cosωt), the natural frequency is 77.2 kHz, peak TRV is 11.33 kV at 6.48 μs, and maximum RRRV is 2.75 kV/μs at 3.24 μs. ## Answer The **transient recovery voltage (TRV)** is the oscillatory voltage that appears across the CB contacts just after the arc is extinguished at current zero. #### Method At fault current zero the system voltage is at its peak *V_m* (current lags by about 90°). The voltage across the opening contacts then follows the *L*–*C* transient (resistance neglected): ```text v(t) = V_m (1 − cos ω_n t), ω_n = 1/√(LC) peak = 2 V_m at ω_n t = π (t = 1/(2 f_n)) dv/dt = V_m ω_n sin ω_n t max = V_m ω_n at ω_n t = π/2 (t = 1/(4 f_n)) ``` The source voltage is not given, so it is obtained from the fault current limited by the generator reactance: *V_m* = √2 × *I* × ω*L* (phase peak). #### Data *L* = 1.7 mH, *C* = 0.0025 μF, *I* = 7500 A rms, *f* = 50 Hz, neutral grounded (first-pole-to-clear factor ignored). ```text X_L = 2π × 50 × 1.7×10⁻³ = 0.5341 Ω V_m = √2 × 7500 × 0.5341 = 5665 V (peak) √(LC) = √(1.7×10⁻³ × 0.0025×10⁻⁶) = 2.0616×10⁻⁶ s ``` #### (i) Natural frequency ```text f_n = 1 / (2π × 2.0616×10⁻⁶) = 77 201 Hz ≈ 77.2 kHz ω_n = 4.8507 × 10⁵ rad/s ``` #### (ii) Peak value of TRV ```text V_peak = 2 V_m = 2 × 5665 = 11 329 V ≈ 11.33 kV ``` #### (iii) Time of peak TRV ```text t = π / ω_n = 1 / (2 f_n) = 1 / (2 × 77 201) = 6.477 × 10⁻⁶ s = 6.48 μs ``` #### (iv) Maximum rate of rise of TRV ```text RRRV_max = V_m ω_n = 5665 × 4.8507×10⁵ = 2.748 × 10⁹ V/s ≈ 2.75 kV/μs ``` #### (v) Time of maximum RRRV ```text t = π / (2 ω_n) = 1 / (4 f_n) = 3.238 × 10⁻⁶ s = 3.24 μs ``` **Answer: (i) 77.2 kHz; (ii) 11.33 kV; (iii) 6.48 μs; (iv) 2.75 kV/μs; (v) 3.24 μs.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With resistance neglected, the voltage across the opening contacts after current zero is A. V_m sin(w_n t) B. V_m (1 - cos w_n t) C. V_m e^(-t/RC) D. 2 V_m cos(w_n t) Answer: B. The LC transient gives v(t) = V_m (1 - cos w_n t), which peaks at 2 V_m. (Key point: TRV waveform) 2. Since V_m is not given, it is obtained as A. V_m = I x wL (peak: sqrt(2) x I x wL) B. V_m = I / wL C. V_m = I x C D. V_m = I x R Answer: A. The fault current is limited by the generator reactance, so V_m = sqrt(2) x 7500 x 0.5341 = 5665 V. (Key point: Source voltage from fault current) 3. With L = 1.7 mH and C = 0.0025 uF, the natural frequency is about A. 7.7 kHz B. 77.2 kHz C. 772 kHz D. 0.77 kHz Answer: B. f_n = 1/(2 pi sqrt(LC)) = 1/(2 pi x 2.0616e-6) = 77.2 kHz. (Key point: Natural frequency) 4. The peak TRV and time at which it occurs are A. 5.67 kV at 3.24 us B. 11.33 kV at 3.24 us C. 11.33 kV at 6.48 us D. 5.67 kV at 6.48 us Answer: C. Peak = 2 V_m = 11.33 kV, at w_n t = pi, i.e. t = 1/(2 f_n) = 6.48 us. 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