--- title: "3-unit string with guard ring: redistribution of voltage" subject: "Power System Analysis I" chapter: 2 key: "semester_4/power_system_analysis_i:guard-ring-3-units-redistribution" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/guard-ring-3-units-redistribution site: BCT Engineering --- # 3-unit string with guard ring: redistribution of voltage Power System Analysis I · Chapter 2: Overhead and Underground Transmission - Asked in: 2080 Chaitra, 10 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-3-units-redistribution - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2 ## Question Determine the voltage across each disc of a suspension insulator comprising of 3 units as a percentage of the line voltage to earth. The self and capacitance to ground of each disc is C and 0.2C respectively. The capacitance between the link pin and the guard ring is 0.1C. If the capacitance to the line of the lower link pin were increased to 0.3C by means of a guard ring, determine the redistribution of voltage. Also determine the string efficiency in each case. ## Short answer With earth capacitance 0.2C and line 0.1C, KCL gives voltages 31.0%, 30.3%, 38.7% of phase voltage, efficiency 86.1%. Increasing lower pin line capacitance to 0.3C redistributes to 32.6%, 32.4%, 35.0%, raising efficiency to 95.4%. ## Answer A guard ring adds capacitance between each link pin and the line conductor. This line capacitance partly cancels the charging current lost to the tower, so the voltage spreads more evenly across the discs. #### Circuit and junction equations Number the discs 1 (tower end) to 3 (line end), with voltages $V_1, V_2, V_3$. Let $V$ be the voltage of the line to earth. Each link pin has $0.2C$ to earth and $g\,C$ to the line. ```text Tower ─[C]─ A ─[C]─ B ─[C]─ Line (V) | | earth ─0.2C 0.2C line ─gA·C gB·C ``` Kirchhoff's current law at a junction (current entering from the line side = current leaving towards the tower + current to earth): $$ V_{k+1} = V_k + m\,(V_1+\dots+V_k) - g\,[\,V-(V_1+\dots+V_k)\,] $$ #### Case 1: line capacitance 0.1C at both pins Junction A ($m=0.2,\ g=0.1$): $$ V_2 = V_1 + 0.2V_1 - 0.1(V - V_1) = 1.3V_1 - 0.1V $$ Junction B: $$ \begin{aligned} V_3 &= V_2 + 0.2(V_1+V_2) - 0.1(V - V_1 - V_2)\\ &= 1.99V_1 - 0.23V \end{aligned} $$ Since $V_1+V_2+V_3 = V$: $4.29V_1 - 0.33V = V$, so $V_1 = 0.3100V$. | Disc | Voltage (% of $V$) | |---|---| | 1 (top) | 31.00 % | | 2 | 30.30 % | | 3 (line end) | 38.69 % | $$ \eta = \frac{V}{3V_3} = \frac{1}{3\times 0.38695} = 86.14\ \% $$ #### Case 2: guard ring raises the lower pin's line capacitance to 0.3C Junction A is unchanged: $V_2 = 1.3V_1 - 0.1V$. Junction B ($g = 0.3$): $$ \begin{aligned} V_3 &= V_2 + 0.2(V_1+V_2) - 0.3(V - V_1 - V_2)\\ &= 2.45V_1 - 0.45V \end{aligned} $$ Sum: $4.75V_1 - 0.55V = V$, so $V_1 = 0.3263V$. | Disc | Case 1 | Case 2 (guard ring) | |---|---|---| | 1 (top) | 31.00 % | 32.63 % | | 2 | 30.30 % | 32.42 % | | 3 (line end) | 38.69 % | 34.95 % | | String efficiency | 86.14 % | 95.38 % | $$ \eta_2 = \frac{1}{3\times 0.34947} = 95.38\ \% $$ **Answer:** Without the larger guard ring capacitance the discs carry 31.00 %, 30.30 % and 38.69 % of $V$, and the string efficiency is 86.14 %. With 0.3C at the lower pin the voltages become 32.63 %, 32.42 % and 34.95 % of $V$, and the efficiency rises to 95.38 %. The guard ring takes stress off the line-end disc. (For comparison, with no line capacitance at all the same string would carry 26.0 %, 31.3 % and 42.7 %, giving an efficiency of 78.0 %.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does a guard ring improve voltage distribution across discs? A. It adds capacitance to the line that cancels part of the charging current lost to earth B. It increases the capacitance to earth C. It adds resistance in series with each disc D. It increases the self capacitance of the top disc Answer: A. Line-side capacitance supplies current to the pins, offsetting the shunt current to the tower. (Key point: Purpose of guard ring) 2. Which junction relation is used with earth ratio m and line ratio g? A. V(k+1) = V(k) + m(V1+...+Vk) + g[V - (V1+...+Vk)] B. V(k+1) = V(k) + m(V1+...+Vk) - g[V - (V1+...+Vk)] C. V(k+1) = V(k) - m(V1+...+Vk) - g V D. V(k+1) = V(k) x (1+m) x (1+g) Answer: B. Line capacitance current reduces the extra current the lower disc must carry, hence the minus sign. (Key point: Junction equation) 3. With 0.1C line capacitance at both pins, what are the disc voltages (top to line end) as % of V? A. 33.3 %, 33.3 %, 33.3 % B. 26.0 %, 31.3 %, 42.7 % C. 31.00 %, 30.30 %, 38.69 % D. 32.63 %, 32.42 %, 34.95 % Answer: C. Solving gives V1 = 0.31V, V2 = 0.303V, V3 = 0.3869V. (Key point: Case 1 distribution) 4. When the guard ring raises the lower pin's line capacitance to 0.3C, what happens to the line-end disc voltage and string efficiency? A. Line-end disc rises to 42.7 % and efficiency falls to 78.0 % B. Line-end disc falls to 34.95 % and efficiency rises to 95.38 % C. Line-end disc falls to 34.95 % and efficiency stays 86.15 % D. Line-end disc stays 38.69 % and efficiency rises to 95.38 % Answer: B. A larger g takes stress off the line-end disc: V3 = 0.3495V, so efficiency = 1/(3 x 0.3495) = 95.38 %. 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