---
title: "Restriking voltage numerical (6.6 kV, 8 mH)"
subject: "Switchgear and Protection"
chapter: 6
key: "semester_6/switchgear_and_protection:restriking-6-6kv-8mh"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/restriking-6-6kv-8mh
site: BCT Engineering
---

# Restriking voltage numerical (6.6 kV, 8 mH)

Switchgear and Protection · Chapter 6: Circuit Breakers

- Asked in: 2075 Bhadra, 4 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/restriking-6-6kv-8mh
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6

## Question

The system inductance per phase of a system is 8 mH and the frequency of transient oscillation is 10.273 kHz. If a 50 Hz, 6.6 kV generator is connected to that system, calculate: i) Max voltage across the contacts of the CB at an instant when it passes through zero. ii) System capacitance per phase. iii) Average rate of rise of voltage up to the first peak of oscillation, neglect resistance.

## Short answer

Peak phase voltage is 5389 V, so maximum voltage across contacts is 10.78 kV. Capacitance is found as C=1/(4π²f_n²L)=0.03 μF. Time to first peak is 48.67 μs, giving average RRRV of 0.221 kV/μs.

## Answer

When the CB current passes through zero, the voltage across its contacts oscillates at the natural frequency of the system and can reach twice the peak system voltage.

#### Method (resistance neglected)
- Peak phase voltage *V_m* = √2 × *V_L*/√3.
- Restriking voltage *v* = *V_m*(1 − cos ω*ₙt*); maximum across contacts = 2*V_m*.
- Time to first peak *t* = π√(*LC*) = 1/(2*f_n*).
- Average rate of rise up to first peak = 2*V_m* / *t*.

#### Data
*L* = 8 mH, *f_n* = 10.273 kHz, *V_L* = 6.6 kV, 50 Hz.

#### i) Maximum voltage across contacts
```text
V_m = √2 × 6600/√3 = 5389 V
v_max = 2 V_m = 10 778 V ≈ 10.78 kV
```
#### ii) System capacitance per phase
```text
f_n = 1/(2π√(LC))  →  C = 1/(4π² f_n² L)
C = 1/(4π² × 10273² × 8×10⁻³)
  = 3.0 × 10⁻⁸ F = 0.03 μF
```
#### iii) Average rate of rise up to first peak
```text
t = 1/(2 f_n) = 1/(2 × 10273) = 48.67 μs
Average RRRV = 10 778 V / 48.67 μs
             = 221.4 V/μs ≈ 0.221 kV/μs
```
**Answer: i) 10.78 kV; ii) C = 0.03 μF per phase; iii) 0.221 kV/μs (2.214 × 10⁸ V/s).**

## Check yourself

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

1. For a 6.6 kV generator with no resistance, the maximum voltage across the contacts is
   A. 5.39 kV
   B. 6.6 kV
   C. 10.78 kV
   D. 13.2 kV
   Answer: C. V_m = sqrt(2) x 6600/sqrt(3) = 5389 V, and the maximum is 2 V_m = 10.78 kV. (Key point: Maximum contact voltage)

2. C is obtained from f_n by
   A. C = 1/(2 pi f_n L)
   B. C = 1/(4 pi^2 f_n^2 L)
   C. C = f_n^2 L
   D. C = L/(4 pi f_n)
   Answer: B. f_n = 1/(2 pi sqrt(LC)) gives C = 1/(4 pi^2 f_n^2 L) = 0.03 uF. (Key point: Capacitance from frequency)

3. Time taken to reach the first peak at f_n = 10.273 kHz is
   A. 97.3 us
   B. 24.3 us
   C. 48.67 us
   D. 4.87 us
   Answer: C. t = 1/(2 f_n) = pi sqrt(LC) = 48.67 us. (Key point: Time to first peak)

4. The average rate of rise up to the first peak is
   A. 0.221 kV/us
   B. 0.11 kV/us
   C. 2.21 kV/us
   D. 0.44 kV/us
   Answer: A. 2 V_m / t = 10 778 V / 48.67 us = 221 V/us. (Key point: Average RRRV)

## Close variants set in other exams

- [Restriking voltage numerical (11 kV, 10 mH)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-11kv-10mh) — Switchgear and Protection, chapter 6, asked 2×

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