---
title: "TRV numerical: 1.5 mH, 0.0025 μF, 7000 A"
subject: "Switchgear and Protection"
chapter: 6
key: "semester_6/switchgear_and_protection:trv-1-5mh-7000a"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/trv-1-5mh-7000a
site: BCT Engineering
---

# TRV numerical: 1.5 mH, 0.0025 μF, 7000 A

Switchgear and Protection · Chapter 6: Circuit Breakers

- Asked in: 2080 Chaitra, 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-1-5mh-7000a
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6

## Question

A 3 phase 50 Hz alternator with grounded neutral is connected to a bus bar through a circuit breaker. Due to short circuit on the busbar the circuit breaker trips when the current is 7000 A (rms) as shown in figure. Determine (i) Active recovery voltage (ii) Frequency of oscillation (iii) Time to reach the peak restriking voltage (iv) Maximum RRRV.
[Figure: Alternator A with grounded neutral, series inductance L = 1.5 mH/phase, shunt capacitance C = 0.0025 μF/phase to ground, then the circuit breaker and a fault on the busbar.]

## Short answer

From V_m=√2×7000×0.4712=4665 V, the active recovery voltage is 4.67 kV peak, oscillation frequency is 82.19 kHz, peak restriking voltage occurs at 6.08 μs, and maximum RRRV is 2.41 kV/μs.

## Answer

When the CB interrupts the busbar fault, the voltage across its contacts rises from zero towards the system voltage as an *L*–*C* oscillation (the **restriking voltage**). Its steady value after the transient is the **recovery voltage**.

```text
  A (alternator)    L=1.5 mH      CB
 (~)---------------^^^^^^----+---o/ o---+ bus
  |                          |          |  fault
 earth                    C=0.0025 μF   X
                             |
                            earth
```
#### 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* = 1.5 mH/phase, *C* = 0.0025 μF/phase, *I* = 7000 A rms, *f* = 50 Hz.

#### (i) Active recovery voltage
```text
X_L = 2π × 50 × 1.5×10⁻³ = 0.4712 Ω
V_m = √2 × 7000 × 0.4712
    = 4665 V ≈ 4.67 kV (peak, per phase)
```
#### (ii) Frequency of oscillation
```text
√(LC) = √(1.5×10⁻³ × 0.0025×10⁻⁶) = 1.9365×10⁻⁶ s
f_n   = 1 / (2π × 1.9365×10⁻⁶)
      = 82 187 Hz ≈ 82.19 kHz
ω_n   = 5.164 × 10⁵ rad/s
```
#### (iii) Time to reach peak restriking voltage
```text
t = 1 / (2 f_n) = 1 / (2 × 82 187)
  = 6.084 × 10⁻⁶ s = 6.08 μs
Peak restriking voltage = 2 V_m = 9.33 kV
```
#### (iv) Maximum RRRV
```text
RRRV_max = V_m ω_n
         = 4665 × 5.164×10⁵
         = 2.409 × 10⁹ V/s ≈ 2.41 kV/μs
(occurs at t = 1/(4 f_n) = 3.04 μs)
```
**Answer: (i) active recovery voltage = 4.67 kV (peak); (ii) f_n = 82.19 kHz; (iii) t = 6.08 μs; (iv) maximum RRRV = 2.41 kV/μs.**

## Check yourself

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

1. The active recovery voltage in this problem is found from
   A. sqrt(2) x I x wL = 4.67 kV peak
   B. I x C
   C. I / (wL)
   D. 2 x I x wL
   Answer: A. X_L = 0.4712 ohm, V_m = sqrt(2) x 7000 x 0.4712 = 4665 V. (Key point: Active recovery voltage)

2. L = 1.5 mH and C = 0.0025 uF give a natural frequency of about
   A. 8.2 kHz
   B. 82.2 kHz
   C. 822 kHz
   D. 41 kHz
   Answer: B. f_n = 1/(2 pi x 1.9365e-6) = 82.19 kHz. (Key point: Frequency of oscillation)

3. The peak restriking voltage occurs at
   A. t = 1/(4 f_n) = 3.04 us
   B. t = 1/f_n = 12.2 us
   C. t = 1/(2 f_n) = 6.08 us
   D. t = 2/f_n = 24.3 us
   Answer: C. Peak 2 V_m occurs when w_n t = pi, i.e. t = 1/(2 f_n) = 6.08 us. (Key point: Time to peak restriking voltage)

4. Maximum RRRV equals
   A. 2 V_m / w_n
   B. V_m x w_n = 2.41 kV/us
   C. V_m / w_n
   D. 2 V_m x w_n
   Answer: B. dv/dt = V_m w_n sin(w_n t) has maximum V_m w_n = 4665 x 5.164e5 = 2.41 kV/us. (Key point: Maximum RRRV)

## Related in Switchgear and Protection

- [TRV numerical: 1.7 mH, 0.0025 μF, 7500 A](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Atrv-1-7mh-7500a) — Switchgear and Protection, chapter 6
- [TRV numerical: 1.6 mH, 0.0032 μF, 8000 A](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-1-6mh-8000a) — Switchgear and Protection, chapter 6, asked 2×

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