---
title: "String of 4 insulators at 66 kV, ratio 10"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:string-4-insulators-66kv-ratio-10"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/string-4-insulators-66kv-ratio-10
site: BCT Engineering
---

# String of 4 insulators at 66 kV, ratio 10

Power System Analysis I · Chapter 2: Overhead and Underground Transmission

- Asked in: 2072 Asoj, 6 marks
- Video: [String efficiency problems with solutions | string efficiency numerical problems |Future inventors-1](https://www.youtube.com/watch?v=U_AEQn7GxBA) — Future inventors
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/string-4-insulators-66kv-ratio-10
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

A string of 4 insulators is connected across a 66 kV line. It has a self-capacitance equal to 10 times the pin to earth capacitance. Calculate (i) the distribution of voltage on the insulator discs and (ii) the string efficiency.

## Short answer

Phase voltage 66/√3 = 38.105 kV, m = 0.1. Using V2=1.1V1 etc., disc voltages (tower to line) are V1=7.53 kV, V2=8.28 kV, V3=9.86 kV, V4=12.43 kV. String efficiency = 38.105/(4×12.43) = 76.64%.

## Answer

#### Data
- Phase voltage $V = 66/\sqrt3 = 38.105$ kV
- Self-capacitance $C = 10C_1$, so $m = C_1/C = 0.1$
- Discs numbered 1 (tower end) to 4 (line end)

#### (i) Voltage distribution
At each junction, the current through the lower disc equals the current through the upper disc plus the current to earth:

$$
V_{k+1} = V_k + m(V_1 + \dots + V_k)
$$

$$
\begin{aligned}
V_2 &= 1.1V_1\\
V_3 &= V_2 + 0.1(2.1V_1) = 1.31V_1\\
V_4 &= V_3 + 0.1(3.41V_1) = 1.651V_1
\end{aligned}
$$

$$
V = (1 + 1.1 + 1.31 + 1.651)V_1 = 5.061V_1
\;\Rightarrow\; V_1 = \frac{38.105}{5.061} = 7.529\ \text{kV}
$$

| Disc | Voltage | % of $V$ |
|---|---|---|
| 1 (tower end) | 7.529 kV | 19.76 % |
| 2 | 8.282 kV | 21.73 % |
| 3 | 9.863 kV | 25.88 % |
| 4 (line end) | 12.431 kV | 32.62 % |

#### (ii) String efficiency
$$
\eta = \frac{V}{4V_4} = \frac{38.105}{4\times 12.431} = 76.64\ \%
$$

**Answer:** The disc voltages are 7.53, 8.28, 9.86 and 12.43 kV (tower to line end), and the string efficiency is 76.64 %.

## Check yourself

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

1. For a 66 kV three-phase line, what voltage is distributed across the insulator string?
   A. 66 kV
   B. 66/sqrt(3) = 38.105 kV
   C. 66 x sqrt(3) kV
   D. 66/3 = 22 kV
   Answer: B. A string sits between one phase conductor and earth, so it carries the phase voltage 66/sqrt(3) = 38.105 kV. (Key point: Voltage across the string)

2. If the self-capacitance is 10 times the pin-to-earth capacitance, what is m = C1/C?
   A. 10
   B. 1
   C. 0.1
   D. 0.01
   Answer: C. C = 10 C1, so m = C1/C = 0.1. (Key point: Capacitance ratio m)

3. Which relation gives the voltage of the next disc from the tower end?
   A. V(k+1) = V_k + m(V_1 + ... + V_k)
   B. V(k+1) = V_k - m(V_1 + ... + V_k)
   C. V(k+1) = V_k + m V_k only
   D. V(k+1) = m(V_1 + ... + V_k)
   Answer: A. Current through the lower disc equals the upper disc current plus the current to earth through the pin capacitance, giving V(k+1) = V_k + m(sum of V1..Vk). (Key point: Junction equation)

4. String efficiency of this 4-disc string is calculated as
   A. V / V_1
   B. V / (4 V_4), about 76.64 %
   C. V_4 / V
   D. (V_1 + V_4) / V
   Answer: B. Efficiency = voltage across the string divided by (number of discs x voltage across the disc nearest the line) = 38.105/(4 x 12.431) = 76.64 %. (Key point: String efficiency)

## Related in Power System Analysis I

- [String efficiency: 66 kV, 4 discs, 10%](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-efficiency-66kv-4-discs) — Power System Analysis I, chapter 2, asked 2×
- [String efficiency: 33 kV, 3 discs, 10%](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-efficiency-33kv-3-discs) — Power System Analysis I, chapter 2, asked 2×
- [String of 5 insulators on 100 kV, ratio 0.1](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-5-insulators-100kv) — Power System Analysis I, chapter 2
- [String efficiency: 3 units, 25% and 10%, 25 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-efficiency-3-unit-25kv) — Power System Analysis I, chapter 2, asked 2×
- [5-disc string at 66 kV, ratios 5 and 6](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-efficiency-5-discs-66kv) — Power System Analysis I, chapter 2
- [Voltage across string discs is not uniform](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-voltage-not-uniform) — Power System Analysis I, chapter 2, asked 3×

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- [Number of insulator discs for 220 kV line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ainsulator-discs-220kv-appendix) — Transmission and Distribution System Design, chapter 3
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