---
title: "Arrester with 1600 kV BIL on 300 Ω line"
subject: "High Voltage Engineering"
chapter: 4
key: "semester_8/high_voltage_engineering:arrester-bil-1600kv"
source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/arrester-bil-1600kv
site: BCT Engineering
---

# Arrester with 1600 kV BIL on 300 Ω line

High Voltage Engineering · Chapter 4: Insulation Coordination

- Asked in: 2067 Mangsir (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/arrester-bil-1600kv
- Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-4

## Question

A lightning arrestor with BIL rating of 1600 kV located at the end of the line having surge impedance of 300 Ω receives a traveling wave of 4200 kV, determine the following:
i) Reflected voltage and current at the arrestor location
ii) Arrestor discharge current and resistance offered by the arrestor

## Short answer

Reflected voltage is -2600 kV, reflected current 8.67 kA. Arrester discharge current is 22.67 kA and arrester resistance is 70.59 Ω.

## Answer

When the travelling wave reaches the arrester at the line end, the arrester conducts and holds the voltage at its rated level of 1600 kV. It behaves as a resistance $R_a$ terminating the line.

#### Data
- Incident voltage, $V_i = 4200$ kV
- Arrester voltage (BIL rating / protective level), $V_a = 1600$ kV
- Surge impedance, $Z = 300\ \Omega$

#### Incident current
$$
I_i = \frac{V_i}{Z} = \frac{4200}{300} = 14\ \text{kA}
$$

#### i) Reflected voltage and current
$$
\begin{aligned}
V_r &= V_a - V_i = 1600 - 4200 = -2600\ \text{kV} \\
I_r &= -\frac{V_r}{Z} = \frac{2600}{300} = 8.67\ \text{kA}
\end{aligned}
$$

#### ii) Arrester discharge current and resistance
$$
\begin{aligned}
I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\
&= \frac{8400 - 1600}{300} = 22.67\ \text{kA}
\end{aligned}
$$
$$
R_a = \frac{V_a}{I_a} = \frac{1600}{22.667} = 70.59\ \Omega
$$

#### Check
$$
\beta = \frac{R_a - Z}{R_a + Z} = \frac{70.59 - 300}{370.59} = -0.619, \quad \beta V_i = -0.619 \times 4200 = -2600\ \text{kV}
$$

| Quantity | Value |
|---|---|
| Incident current | 14 kA |
| Reflected voltage | −2600 kV |
| Reflected current | +8.67 kA |
| Arrester current | 22.67 kA |
| Arrester resistance | 70.59 Ω |

**Answer: $V_r = -2600$ kV, $I_r = 8.67$ kA, $I_a = 22.67$ kA, $R_a = 70.59\ \Omega$.**

## Check yourself

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

1. What is the incident current?
   A. 14 kA
   B. 5.33 kA
   C. 22.67 kA
   D. 8.67 kA
   Answer: A. Ii = 4200 / 300 = 14 kA. (Key point: Incident current)

2. What are the reflected voltage and current?
   A. -2600 kV and -8.67 kA
   B. +2600 kV and +8.67 kA
   C. -2600 kV and +8.67 kA
   D. -1600 kV and +5.33 kA
   Answer: C. Vr = 1600 - 4200 = -2600 kV, Ir = 2600/300 = +8.67 kA. (Key point: Reflected voltage and current)

3. What is the arrester discharge current?
   A. 14 kA
   B. 22.67 kA
   C. 8.67 kA
   D. 28 kA
   Answer: B. Ia = (2 x 4200 - 1600)/300 = 22.67 kA. (Key point: Discharge current)

4. What resistance results, and how is it checked?
   A. 70.59 ohm; beta = -0.619 and beta x Vi = -2600 kV
   B. 70.59 ohm; beta = +0.619
   C. 300 ohm; beta = 0
   D. 142 ohm; beta = -1
   Answer: A. Ra = 1600/22.667 = 70.59 ohm; beta = (70.59 - 300)/370.59 = -0.619 and -0.619 x 4200 = -2600 kV. (Key point: Resistance and check)

## Close variants set in other exams

- [Arrester with 1000 kV BIL on 300 Ω line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-bil-1000kv-300-ohm) — High Voltage Engineering, chapter 4, asked 3×
- [Arrester with 1000 kV BIL on 350 Ω line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-bil-1000kv-350-ohm) — High Voltage Engineering, chapter 4, asked 3×

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