---
title: "TRV numerical: 1.6 mH, 0.0032 μF, 8000 A"
subject: "Switchgear and Protection"
chapter: 6
key: "semester_6/switchgear_and_protection:restriking-1-6mh-8000a"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/restriking-1-6mh-8000a
site: BCT Engineering
---

# TRV numerical: 1.6 mH, 0.0032 μF, 8000 A

Switchgear and Protection · Chapter 6: Circuit Breakers

- Asked in: 2071 Bhadra, 8 marks; 2076 Bhadra, 6 marks (asked 2 times)
- 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/restriking-1-6mh-8000a
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6

## Question

A 50 Hz 3-phase alternator with grounded neutral has an inductance of 1.6 mH per phase and is connected to the bus-bars through a circuit breaker. The capacitance to earth of the circuit between the alternator and the circuit breaker is 0.0032 μF per phase. Due to a short on the bus-bars the breaker opens when the rms value of the current is 8000 A. Determine the following (i) frequency of oscillation (ii) Active recovery voltage (iii) time for maximum RRRV and (iv) maximum RRRV.

## Short answer

Compute X_L = 2πfL = 0.5027 Ω, then V_m = √2 × I × X_L = 5.69 kV peak. Oscillation frequency f_n = 1/(2π√(LC)) = 70.34 kHz. Time to max RRRV = 1/(4f_n) = 3.55 μs. Max RRRV = V_m × 2πf_n = 2.51 kV/μs.

## Answer

When the breaker interrupts a bus fault fed by an alternator, the voltage across its contacts rises as a transient oscillation of the system *L* and *C*. The fault current lags the voltage by nearly 90°, so at current zero the system voltage is at its peak *V_m* (the **active recovery voltage**). The restriking voltage is *v* = *V_m*(1 − cos ω*ₙt*), which peaks at 2*V_m*.

The system voltage is not given, so it is found from the fault current. With only the alternator reactance limiting the current:
```text
V_ph(rms) = I × X_L,   V_m = √2 × I × X_L
```
#### Data
*L* = 1.6 mH, *C* = 0.0032 μF, *I* = 8000 A rms, *f* = 50 Hz.

#### (i) Frequency of oscillation
```text
√(LC) = √(1.6×10⁻³ × 0.0032×10⁻⁶)
      = 2.2627 × 10⁻⁶ s
f_n   = 1 / (2π √(LC))
      = 1 / (2π × 2.2627×10⁻⁶)
      = 70 337 Hz ≈ 70.34 kHz
ω_n   = 2π f_n = 4.4194 × 10⁵ rad/s
```
#### (ii) Active recovery voltage
```text
X_L = 2π × 50 × 1.6×10⁻³ = 0.5027 Ω
V_m = √2 × 8000 × 0.5027
    = 5687 V ≈ 5.69 kV (peak, phase value)
```
(The maximum restriking voltage would be 2*V_m* = 11.37 kV.)

#### (iii) Time for maximum RRRV
```text
v      = V_m (1 − cos ω_n t)
dv/dt  = V_m ω_n sin ω_n t
Maximum when ω_n t = π/2:
t = π / (2 ω_n) = 1 / (4 f_n)
  = 1 / (4 × 70 337)
  = 3.554 × 10⁻⁶ s = 3.55 μs
```
#### (iv) Maximum RRRV
```text
RRRV_max = V_m ω_n
         = 5687 × 4.4194×10⁵
         = 2.513 × 10⁹ V/s
         = 2513 V/μs ≈ 2.51 kV/μs
```
**Answer: (i) f_n = 70.34 kHz; (ii) active recovery voltage = 5.69 kV (peak); (iii) t = 3.55 μs; (iv) maximum RRRV = 2.51 kV/μs.**

## Check yourself

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

1. Which formula gives the natural frequency?
   A. f = 1/(2 pi sqrt(LC))
   B. f = 2 pi sqrt(LC)
   C. f = sqrt(LC)/2pi
   D. f = 1/(LC)
   Answer: A. With L = 1.6 mH and C = 0.0032 uF, f = 70.34 kHz. (Key point: Frequency of oscillation)

2. How is the active recovery voltage found here?
   A. V_m = sqrt(2) x I x X_L, since the voltage is at its peak at current zero
   B. V_m = I x X_L
   C. V_m = 2 x I x X_L
   D. V_m = I / X_L
   Answer: A. The current lags by about 90 degrees, so at zero current the voltage is at peak; V_m = 5.69 kV. (Key point: Active recovery voltage)

3. At what time is RRRV maximum?
   A. When w_n t = pi, t = 1/(2 f_n)
   B. When w_n t = pi/2, t = 1/(4 f_n), about 3.55 us
   C. At t = 0 us
   D. At t = 1/f_n
   Answer: B. dv/dt = V_m w_n sin(w_n t) is maximum at w_n t = pi/2. (Key point: Time for maximum RRRV)

4. What is the maximum RRRV?
   A. V_m w_n, about 2.51 kV/us
   B. 2 V_m w_n
   C. V_m / w_n
   D. V_m f_n / 2
   Answer: A. RRRV_max = 5687 x 4.4194e5 = 2.51e9 V/s. (Key point: Maximum RRRV)

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