---
title: "String of 5 insulators on 100 kV, ratio 0.1"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:string-5-insulators-100kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/string-5-insulators-100kv
site: BCT Engineering
---

# String of 5 insulators on 100 kV, ratio 0.1

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

- Asked in: 2071 Magh, 6 marks
- 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-5-insulators-100kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

A string of 5 insulators is connected across a 100 kV line. If the ratio of shunt to self-capacitance is 0.1, determine the voltage distribution along the string and string efficiency.

## Short answer

With m = 0.1 and phase voltage 100/√3 = 57.735 kV, disc voltages (tower toline) are 8.00 kV, 8.80 kV, 10.48 kV, 13.21 kV, 17.25 kV. String efficiency = 57.735/(5×17.254) = 66.92%.

## Answer

Assumption: 100 kV is the line voltage of a three-phase line, so the voltage across each string is the phase voltage.

#### Data
- Phase voltage $V = 100/\sqrt3 = 57.735$ kV
- Ratio of shunt to self-capacitance $m = 0.1$
- Discs numbered 1 (tower end) to 5 (line end)

#### Voltage distribution
Using $V_{k+1} = V_k + m(V_1+\dots+V_k)$:

$$
\begin{aligned}
V_2 &= 1.1V_1\\
V_3 &= 1.1V_1 + 0.1(2.1V_1) = 1.31V_1\\
V_4 &= 1.31V_1 + 0.1(3.41V_1) = 1.651V_1\\
V_5 &= 1.651V_1 + 0.1(5.061V_1) = 2.1571V_1
\end{aligned}
$$

$$
V = 7.2181V_1 \;\Rightarrow\; V_1 = \frac{57.735}{7.2181} = 7.999\ \text{kV}
$$

| Disc | Voltage (kV) | % of $V$ |
|---|---|---|
| 1 (tower end) | 7.999 | 13.85 % |
| 2 | 8.799 | 15.24 % |
| 3 | 10.478 | 18.15 % |
| 4 | 13.206 | 22.87 % |
| 5 (line end) | 17.254 | 29.88 % |

#### String efficiency
$$
\eta = \frac{V}{5V_5} = \frac{57.735}{5\times 17.254} = 66.92\ \%
$$

**Answer:** The disc voltages are 8.00, 8.80, 10.48, 13.21 and 17.25 kV, and the string efficiency is 66.92 %. If 100 kV is instead taken as the voltage across the string, every disc voltage scales by $\sqrt3$ (to 13.85, 15.24, 18.15, 22.87 and 29.88 kV), and the efficiency stays at 66.92 %.

## Check yourself

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

1. With the assumption that 100 kV is the three-phase line voltage, the string voltage used is
   A. 100 kV
   B. 57.735 kV
   C. 173.2 kV
   D. 33.3 kV
   Answer: B. The string carries phase voltage 100/sqrt(3) = 57.735 kV. (Key point: Voltage across string)

2. With m = 0.1, how is V5 related to V1?
   A. V5 = 1.651 V1
   B. V5 = 2.1571 V1
   C. V5 = 5.061 V1
   D. V5 = 1.1 V1
   Answer: B. V5 = V4 + 0.1(V1+V2+V3+V4) = 1.651V1 + 0.1(5.061V1) = 2.1571V1. (Key point: Disc voltage ratios)

3. Which disc carries the largest voltage in the string?
   A. The disc at the tower end
   B. The middle disc
   C. The disc nearest the line conductor
   D. All discs carry equal voltage
   Answer: C. Shunt capacitance currents make the voltage increase toward the line end; disc 5 carries 17.25 kV. (Key point: Which disc is most stressed)

4. The string efficiency for the 5-disc string is
   A. 86.77 %
   B. 76.64 %
   C. 66.92 %
   D. 57.7 %
   Answer: C. eta = 57.735/(5 x 17.254) = 66.92 %. (Key point: Efficiency)

## Related in Power System Analysis I

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- [Voltage distribution on 4-disc string at 33 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-4-discs-33kv) — Power System Analysis I, chapter 2
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- [Maximum voltage of 3-disc string, 17.5 kV per unit](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amax-voltage-3-disc-17-5kv) — Power System Analysis I, chapter 2

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