---
title: "4-unit string with guard ring (10% and 6%)"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:guard-ring-4-unit-10-6"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/guard-ring-4-unit-10-6
site: BCT Engineering
---

# 4-unit string with guard ring (10% and 6%)

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

- Asked in: 2079 Chaitra, 6 marks
- Video: [L54: Grading Ring in Hindi | Improving String Efficiency | Power System Series](https://www.youtube.com/watch?v=qmI7LRuQ9p8) — TechInsight
- 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/guard-ring-4-unit-10-6
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

A 4 unit insulator string is fitted with a guard ring. The capacitance of the link pins to metal work and guard ring can be assumed to be 10% and 6% of the capacitance of each unit. Determine the voltage distribution and string efficiency.

## Short answer

Using junction equations with m=0.1 and g=0.06, the disc voltages from tower to line end are 24.48%, 22.40%, 23.90%, 29.22% of V. String efficiency = V/(4*Vmax) = 85.55%.

## Answer

With a guard ring, each link pin has capacitance to the tower (earth), $mC$, and to the guard ring (which is at line potential), $gC$. Here $m = 0.1$ and $g = 0.06$.

#### Junction equation
Number the discs 1 (tower end) to 4 (line end). Let $V$ be the line-to-earth voltage and $S_k = V_1+\dots+V_k$. Current balance at the junction below disc $k$:

$$
V_{k+1} = V_k + mS_k - g(V - S_k)
$$

#### Solving in terms of $V_1$
$$
\begin{aligned}
V_2 &= V_1 + 0.1V_1 - 0.06(V - V_1) = 1.16V_1 - 0.06V\\
V_3 &= V_2 + 0.16(V_1+V_2) - 0.06V = 1.5056V_1 - 0.1296V\\
V_4 &= V_3 + 0.16(V_1+V_2+V_3) - 0.06V = 2.0921V_1 - 0.2199V
\end{aligned}
$$

Sum $= V$:

$$
5.7577V_1 - 0.4095V = V \;\Rightarrow\; V_1 = 0.2448V
$$

#### Voltage distribution
| Disc | Voltage (fraction of $V$) | % of $V$ |
|---|---|---|
| 1 (tower end) | 0.2448 | 24.48 % |
| 2 | 0.2240 | 22.40 % |
| 3 | 0.2390 | 23.90 % |
| 4 (line end) | 0.2922 | 29.22 % |

The guard ring makes the distribution much flatter than a plain string. Disc 2 has the smallest share, because the guard ring current is largest near the top.

#### String efficiency
$$
\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.29223} = 85.55\ \%
$$

**Answer:** The discs carry 24.48 %, 22.40 %, 23.90 % and 29.22 % of the phase voltage (tower end to line end), and the string efficiency is 85.55 %.

## Check yourself

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

1. What is the guard ring's role in this problem (m = 0.1, g = 0.06)?
   A. It is at earth potential and adds to the earth capacitance
   B. It is at line potential, so each pin gets capacitance gC to the line that offsets the pin-to-tower current
   C. It increases the self capacitance of each disc
   D. It adds a series resistance
   Answer: B. The ring is connected to the line, so g feeds current to the pins. (Key point: Guard ring capacitance)

2. Which equation relates the disc voltages?
   A. V(k+1) = V(k) + m S_k + g(V - S_k)
   B. V(k+1) = V(k) - m S_k
   C. V(k+1) = V(k) + m S_k - g(V - S_k)
   D. V(k+1) = V(k) + (m - g) V
   Answer: C. S_k is the sum of the first k disc voltages; the guard ring current is proportional to V - S_k and reduces V(k+1). (Key point: Junction relation)

3. What are the disc voltages (tower end to line end) as % of V?
   A. 25 %, 25 %, 25 %, 25 %
   B. 19.8 %, 19.8 %, 24.7 %, 35.7 %
   C. 24.48 %, 22.40 %, 23.90 %, 29.22 %
   D. 29.22 %, 23.90 %, 22.40 %, 24.48 %
   Answer: C. Solving the sum equation gives V1 = 0.2448V, then V2, V3, V4 as listed. (Key point: Voltage distribution)

4. What is the string efficiency?
   A. 70 %
   B. 85.55 %
   C. 77.1 %
   D. 92.3 %
   Answer: B. Efficiency = 1/(4 x 0.29223) = 85.55 %. (Key point: String efficiency)

## Close variants set in other exams

- [4-unit string with guard ring (10% and 5%)](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aguard-ring-4-unit-10-5) — Power System Analysis I, chapter 2, asked 2×

## Related in Power System Analysis I

- [3-unit string with guard ring: redistribution of voltage](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aguard-ring-3-units-redistribution) — Power System Analysis I, chapter 2
- [Guard ring sizing for 3-unit string at 20 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aguard-ring-3-units-20kv) — 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×
- [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×
- [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
- [Greatest working voltage of 3-unit string, 20 kV limit](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Agreatest-working-voltage-3-units) — Power System Analysis I, chapter 2

## Related in other subjects

- [Voltage distribution of 3-unit string with guard ring](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Astring-guard-ring-3-unit) — High Voltage Engineering, chapter 5
- [Guard ring capacitances for 5-disc string](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aguard-ring-five-disc-capacitance) — High Voltage Engineering, chapter 5