---
title: "String efficiency: 3 units, 25% and 10%, 25 kV"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:string-efficiency-3-unit-25kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/string-efficiency-3-unit-25kv
site: BCT Engineering
---

# String efficiency: 3 units, 25% and 10%, 25 kV

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

- Asked in: 2081 Chaitra (new course), 3+3 marks
- Video: [Problem Solving on String Efficiency 7 || Power System Analysis](https://www.youtube.com/watch?v=1bdPBAWpzp8) — Rojib EEE Academy
- 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-3-unit-25kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

Calculate the string efficiency and individual voltage for a 3-unit suspension insulator if the capacitance of each unit to earth and line be 25% and 10% of the self-capacitance of the unit respectively for a voltage level of 25 kV.

## Short answer

With 25 kV string voltage, earth capacitance 0.25C, line 0.1C, KCL at junctions yields V1=7.38 kV, V2=7.46 kV, V3=10.16 kV. String efficiency = V/(3×V3) = 82.0%.

## Answer

Assumption: 25 kV is the voltage across the whole string (phase to earth). Each unit has self-capacitance $C$; each joint has capacitance $mC = 0.25C$ to earth (tower) and $nC = 0.1C$ to the line conductor.

```text
   Tower (earth)
      |
    [C] V1
      +-- 0.25C to tower, 0.1C to line (joint A)
    [C] V2
      +-- 0.25C to tower, 0.1C to line (joint B)
    [C] V3
      |
   Line (25 kV)
```

#### Joint equations
Current flows from line towards the tower. At each joint, current entering (through lower disc + from line capacitance) = current leaving (through upper disc + to earth capacitance).

Joint A (potential $V_1$; line-to-joint voltage $V_2+V_3$):
$$
V_2 + n(V_2+V_3) = V_1(1+m)
\quad\Rightarrow\quad 1.1V_2 + 0.1V_3 = 1.25V_1
$$
Joint B (potential $V_1+V_2$; line-to-joint voltage $V_3$):
$$
V_3(1+n) = V_2 + m(V_1+V_2)
\quad\Rightarrow\quad 1.1V_3 = 1.25V_2 + 0.25V_1
$$

#### Solving (put $V_1 = 1$)
From joint B: $V_3 = (1.25V_2 + 0.25)/1.1$. Substituting in joint A:
$$
\begin{aligned}
1.1V_2 + \frac{0.1(1.25V_2+0.25)}{1.1} &= 1.25\\
1.21V_2 + 0.125V_2 + 0.025 &= 1.375\\
V_2 &= \frac{1.35}{1.335} = 1.0112,\qquad V_3 = \frac{1.25(1.0112)+0.25}{1.1} = 1.3764
\end{aligned}
$$
Sum $= 1 + 1.0112 + 1.3764 = 3.3876$, so $V_1 = 25/3.3876 = 7.380$ kV.

| Unit (from tower) | Ratio | Voltage (kV) |
|---|---|---|
| 1 (top) | 1.0000 | 7.380 |
| 2 | 1.0112 | 7.463 |
| 3 (line end) | 1.3764 | 10.157 |
| Total | 3.3876 | 25.000 |

#### String efficiency
$$
\eta = \frac{V}{n\,V_3} = \frac{25}{3\times10.157}\times100 = 82.0\ \%
$$
The line capacitance partly cancels the earth capacitance current, so the distribution is better than with earth capacitance alone (with the 0.25 earth capacitance alone, $\eta$ would be only 74.7 %). If 25 kV were the line voltage, every disc voltage would be divided by $\sqrt3$, but the efficiency stays 82.0 %.

**Answer: $V_1 = 7.38$ kV, $V_2 = 7.46$ kV, $V_3 = 10.16$ kV; string efficiency ≈ 82.0 %.**

## Check yourself

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

1. At joint A (below the top unit), which equation holds when joints have 0.25C to tower and 0.1C to line?
   A. V2 = 1.25 V1
   B. V2 + 0.25(V2+V3) = 1.1 V1
   C. V2 - 0.1(V2+V3) = 1.25 V1
   D. V2 + 0.1(V2+V3) = 1.25 V1
   Answer: D. Current through the upper disc plus the earth branch equals current in the lower disc plus the line branch: 1.1V2 + 0.1V3 = 1.25V1. (Key point: Joint equation with line capacitance)

2. How does the capacitance of the joints to the line affect the voltage distribution?
   A. It makes the distribution worse
   B. It has no effect
   C. It partly cancels the earth current and makes the distribution better
   D. It raises the voltage on the top unit only
   Answer: C. The line capacitance feeds current into the joint, offsetting the loss to earth, so efficiency improves (82.0 % vs 74.7 %). (Key point: Effect of line capacitance)

3. Which unit carries the largest voltage, and how much?
   A. The tower-end unit, 7.38 kV
   B. The line-end unit, 10.16 kV
   C. The middle unit, 7.46 kV
   D. The line-end unit, 12.5 kV
   Answer: B. V3 = 1.3764 V1 = 10.157 kV. (Key point: Most stressed unit and value)

4. What is the string efficiency?
   A. 74.7 %
   B. 90.5 %
   C. 68.4 %
   D. 82.0 %
   Answer: D. Efficiency = 25/(3 x 10.157) = 82.0 %. (Key point: Efficiency)

## Close variants set in other exams

- [String efficiency: 4 units, 20% and 5%](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astring-efficiency-4-unit-20-5) — Power System Analysis I, chapter 2, asked 2×

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