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

# String efficiency: 66 kV, 4 discs, 10%

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

- Asked in: 2082 Kartik (new course), 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-efficiency-66kv-4-discs
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

An insulator string for 66 kV line has 4 discs. The shunt capacitance between each joint and metal work is 10% of the capacitance of each disc. Find the voltage across the different discs and string efficiency.

## Short answer

For phase voltage 38.105 kV and m=0.1, set V2=V1(1+m), V3=V2+m(V1+V2), V4=V3+m(V1+V2+V3). Solving: V1=7.53 kV, V2=8.28 kV, V3=9.86 kV, V4=12.43 kV. String efficiency = V/(4×V4) = 76.6%.

## Answer

#### Data
- Line voltage 66 kV, so voltage across the string (phase to earth):
$$
V = \frac{66}{\sqrt3} = 38.105\ \text{kV}
$$
- Number of discs $n = 4$; shunt-to-self capacitance ratio $m = 0.1$.

Let $V_1$ (top, at cross-arm) to $V_4$ (bottom, at line) be the disc voltages.

#### Equations at each joint
Current through lower disc = current through upper disc + shunt current to tower:
$$
\begin{aligned}
V_2 &= V_1(1+m) = 1.1V_1\\
V_3 &= V_2 + m(V_1+V_2) = [1.1 + 0.1(2.1)]V_1 = 1.31V_1\\
V_4 &= V_3 + m(V_1+V_2+V_3) = [1.31 + 0.1(3.41)]V_1 = 1.651V_1
\end{aligned}
$$

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

| Disc (from cross-arm) | Ratio | Voltage (kV) |
|---|---|---|
| 1 (top) | 1.000 | 7.529 |
| 2 | 1.100 | 8.282 |
| 3 | 1.310 | 9.863 |
| 4 (line end) | 1.651 | 12.431 |
| Total | 5.061 | 38.105 |

#### String efficiency
$$
\eta = \frac{V}{n\times V_4} = \frac{38.105}{4\times12.431}\times100 = 76.6\ \%
$$

The disc next to the line carries about 33 % of the string voltage, so it is the most stressed. Guard rings or longer cross-arms would improve the distribution.

**Answer: $V_1 = 7.53$, $V_2 = 8.28$, $V_3 = 9.86$, $V_4 = 12.43$ kV; string efficiency = 76.6 %.**

## 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 taken across the insulator string?
   A. 66/sqrt(3) = 38.1 kV (phase to earth)
   B. 66 kV
   C. 66 x sqrt(3) = 114.3 kV
   D. 66/3 = 22 kV
   Answer: A. The string is between the line conductor and the earthed tower, so the phase voltage is used. (Key point: Voltage across string)

2. With shunt-to-disc capacitance ratio m, how is V2 related to V1 (disc 1 at the cross-arm)?
   A. V2 = V1(1-m)
   B. V2 = V1(1+m)
   C. V2 = V1/m
   D. V2 = V1 + m
   Answer: B. Current through disc 2 equals current through disc 1 plus the shunt current mC V1, so V2 = V1(1+m). (Key point: Junction relation)

3. Which disc carries the highest voltage and about how much?
   A. The disc next to the cross-arm, 7.53 kV
   B. The second disc, 8.28 kV
   C. The disc next to the line, 12.43 kV
   D. All discs equally, 9.53 kV
   Answer: C. Shunt currents accumulate toward the line, so V4 = 1.651 V1 = 12.43 kV is greatest. (Key point: Disc with highest voltage)

4. How is string efficiency calculated here?
   A. V4 / V x 100
   B. V / (n x V1) x 100
   C. (V1 + V4) / V x 100
   D. V / (n x voltage across the disc nearest the line) x 100 = 76.6 %
   Answer: D. Efficiency = voltage across string / (number of discs x voltage across the worst-stressed disc) = 38.105/(4 x 12.431). (Key point: String efficiency formula)

## Close variants set in other exams

- [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×

## 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
- [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
- [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
- [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
- [Line voltage from disc voltages of 4-disc string](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aline-voltage-from-disc-voltages) — 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×

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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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