---
title: "5-disc string at 66 kV, ratios 5 and 6"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:string-efficiency-5-discs-66kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/string-efficiency-5-discs-66kv
site: BCT Engineering
---

# 5-disc string at 66 kV, ratios 5 and 6

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

- Asked in: 2076 Bhadra, 8 marks
- Video: [string efficiency calculation problem#1](https://www.youtube.com/watch?v=puVVp0YitWw) — BE Technical
- 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-efficiency-5-discs-66kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

Each conductor at 66 kV is suspended by a string of 5 similar insulators, the capacitance of each disc being 5 times the capacitance to earth. Determine the % string efficiency. If the distance between the insulators and tower structure is increased so as to make the capacitance of the disc 6 times the earth capacitance, what will happen to the string efficiency?

## Short answer

For m=0.2 (C=5C1), string efficiency is 54.16%. When distance increases so C=6C1 (m=1/6), efficiency rises to 57.47%. Reducing shunt earth capacitance improves voltage distribution uniformity.

## Answer

With no guard ring, the voltage across the discs increases towards the line, because each link pin leaks charging current to the tower through its shunt capacitance $C_1 = mC$.

#### Data
Phase voltage:

$$
V = \frac{66}{\sqrt 3} = 38.105\ \text{kV}
$$

Number the discs 1 (tower end) to 5 (line end). At each junction, the current through the lower disc equals the current through the upper disc plus the current to earth:

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

#### Case 1: $C = 5C_1$, so $m = 0.2$
$$
\begin{aligned}
V_2 &= 1.2V_1\\
V_3 &= V_2 + 0.2(2.2V_1) = 1.64V_1\\
V_4 &= V_3 + 0.2(3.84V_1) = 2.408V_1\\
V_5 &= V_4 + 0.2(6.248V_1) = 3.6576V_1
\end{aligned}
$$

$V = 9.9056V_1$, so $V_1 = 38.105/9.9056 = 3.847$ kV.

| Disc | kV ($m = 1/5$) | kV ($m = 1/6$) |
|---|---|---|
| 1 | 3.847 | 4.255 |
| 2 | 4.616 | 4.965 |
| 3 | 6.309 | 6.501 |
| 4 | 9.263 | 9.122 |
| 5 (line end) | 14.070 | 13.262 |

$$
\eta_1 = \frac{V}{5V_5} = \frac{38.105}{5\times 14.070} = 54.16\ \%
$$

#### Case 2: $C = 6C_1$, so $m = 1/6$
$$
\begin{aligned}
V_2 &= 1.1667V_1,\quad V_3 = 1.5278V_1\\
V_4 &= 2.1435V_1,\quad V_5 = 3.1165V_1
\end{aligned}
$$

$V = 8.9545V_1$, so $V_1 = 4.255$ kV and $V_5 = 13.262$ kV.

$$
\eta_2 = \frac{38.105}{5\times 13.262} = 57.47\ \%
$$

**Answer:** The string efficiency is 54.16 % when $C = 5C_1$. Moving the insulators further from the tower lowers the shunt capacitance so that $C = 6C_1$. The efficiency then rises to 57.47 %, and the line-end disc voltage falls from 14.07 kV to 13.26 kV. A lower ratio $m = C_1/C$ always gives a more uniform distribution.

## Check yourself

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

1. What relation holds at each junction when there is no guard ring?
   A. V(k+1) = V(k) - m(V1+...+Vk)
   B. V(k+1) = V(k) + m(V1+...+Vk)
   C. V(k+1) = V(k) x m
   D. V(k+1) = V(k) + m V
   Answer: B. Lower disc current equals upper disc current plus shunt current to earth. (Key point: Junction relation)

2. If the disc capacitance is 5 times the earth capacitance, what is m?
   A. 5
   B. 0.5
   C. 0.2
   D. 0.05
   Answer: C. m = C1/C = 1/5 = 0.2. (Key point: Value of m)

3. What is the string efficiency for C = 5 C1?
   A. 57.47 %
   B. 76.6 %
   C. 45.2 %
   D. 54.16 %
   Answer: D. V = 38.105 kV, V5 = 14.07 kV, eta = 38.105/(5 x 14.07) = 54.16 %. (Key point: Efficiency case 1)

4. What happens when the insulators are moved farther from the tower so that C = 6 C1?
   A. Efficiency falls to 50 %
   B. Shunt capacitance falls, efficiency rises to 57.47 %
   C. Efficiency is unchanged
   D. Efficiency rises to 100 %
   Answer: B. A smaller m gives a more uniform distribution; the line-end disc voltage drops from 14.07 to 13.26 kV. (Key point: Effect of increasing distance)

## Related in Power System Analysis I

- [String of 4 insulators at 66 kV, ratio 10](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-4-insulators-66kv-ratio-10) — Power System Analysis I, chapter 2
- [5-disc string at 66 kV: nearest and farthest disc](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Afive-discs-66kv-distribution) — Power System Analysis I, chapter 2
- [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×
- [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
- [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
- [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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