---
title: "Restriking voltage numerical (11 kV, 10 mH)"
subject: "Switchgear and Protection"
chapter: 6
key: "semester_6/switchgear_and_protection:restriking-11kv-10mh"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/restriking-11kv-10mh
site: BCT Engineering
---

# Restriking voltage numerical (11 kV, 10 mH)

Switchgear and Protection · Chapter 6: Circuit Breakers

- Asked in: 2079 Chaitra, 6 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-11kv-10mh
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6

## Question

A 50 Hz, 11 kV generator is connected to a power system. The system inductance and capacitance per phase are 10 mH and 0.02 μF respectively. Calculate: i) Maximum voltage across the contacts of the CB at an instant when it passes through zero. ii) Frequency of transient oscillation. iii) Average rate of rise of voltage up to the first peak of oscillation, neglect the resistance.

## Short answer

i) Vmax = 2×√2×11000/√3 = 17.96 kV. ii) f_n = 1/(2π√(10mH×0.02μF)) = 11.25 kHz. iii) Average RRRV = Vmax/(π√(LC)) = 0.404 kV/μs.

## Answer

The restriking voltage across the CB contacts after current zero is an *L*–*C* oscillation that rises to twice the peak phase voltage when resistance is neglected.

#### 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
*V_L* = 11 kV, 50 Hz, *L* = 10 mH, *C* = 0.02 μF per phase.

#### i) Maximum voltage across the contacts
```text
V_m = √2 × 11000/√3 = 8981 V
v_max = 2 V_m = 17 963 V ≈ 17.96 kV
```
#### ii) Frequency of transient oscillation
```text
√(LC) = √(10×10⁻³ × 0.02×10⁻⁶) = 1.4142×10⁻⁵ s
f_n = 1/(2π × 1.4142×10⁻⁵)
    = 11 254 Hz ≈ 11.25 kHz
```
#### iii) Average rate of rise up to first peak
```text
t = 1/(2 f_n) = 44.43 μs
Average RRRV = 17 963 V / 44.43 μs
             = 404.3 V/μs ≈ 0.404 kV/μs
```
**Answer: i) 17.96 kV; ii) 11.25 kHz; iii) 0.404 kV/μs (4.04 × 10⁸ V/s).**

## Check yourself

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

1. For 11 kV, the maximum voltage across the contacts is
   A. 8.98 kV
   B. 11 kV
   C. 17.96 kV
   D. 22 kV
   Answer: C. V_m = sqrt(2) x 11 000/sqrt(3) = 8981 V; maximum 2 V_m = 17.96 kV. (Key point: Maximum voltage across contacts)

2. Why can the contact voltage reach 2 V_m?
   A. The LC oscillation v = V_m (1 - cos w_n t) overshoots to twice the driving value when resistance is neglected
   B. The generator voltage doubles at current zero
   C. Because of the 50 Hz frequency
   D. Because of the three phases
   Answer: A. An undamped LC response to a step V_m oscillates between 0 and 2 V_m. (Key point: Why 2 V_m)

3. With L = 10 mH and C = 0.02 uF the transient frequency is
   A. 1.125 kHz
   B. 112.5 kHz
   C. 5.6 kHz
   D. 11.25 kHz
   Answer: D. f_n = 1/(2 pi x 1.4142e-5) = 11.25 kHz. (Key point: Natural frequency)

4. The average rate of rise to first peak is about
   A. 0.404 kV/us
   B. 0.81 kV/us
   C. 0.2 kV/us
   D. 4.04 kV/us
   Answer: A. t = 44.43 us, so 17 963 V / 44.43 us = 404 V/us. (Key point: Average RRRV)

## Close variants set in other exams

- [Restriking voltage numerical (6.6 kV, 8 mH)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Arestriking-6-6kv-8mh) — Switchgear and Protection, chapter 6, asked 2×

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