--- title: "Reactor function and ring bus-bar scheme" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:ring-bus-reactor-scheme" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/ring-bus-reactor-scheme site: BCT Engineering --- # Reactor function and ring bus-bar scheme Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2079 Bhadra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/ring-bus-reactor-scheme - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question What is the function of reactor in Power system network? Explain the ring type bus bar reactor scheme. ## Short answer Reactors limit short-circuit current, localize faults, and reduce stress. The ring bus-bar scheme splits the bus into sections joined by reactors in a closed ring, each with a generator. Fault current from non-faulted sections is limited by the reactors, keeping other sections' voltage high. ## Answer #### Function of reactor A **current-limiting reactor** is a coil with high inductive reactance and negligible resistance. It limits the short-circuit current to a value that circuit breakers can interrupt, localises the fault so that the rest of the bus keeps its voltage, and reduces mechanical and thermal stress on equipment. #### Ring type bus-bar reactor scheme The bus-bar is split into sections, one generator per section, and **adjacent sections are joined through reactors so that the sections form a closed ring**. Feeders are taken from each section. ```text G1 G2 G3 | | | ==[S1]===[X]===[S2]===[X]===[S3]== | | | | | | | feeders feeders feeders | | +-------------[X]--------------+ (reactor closing the ring) ``` **Working:** - In normal operation each generator mainly feeds the load on its own section. Only small balancing currents flow through the reactors, so their voltage drop and power loss are small. - When a fault occurs on a feeder of, say, section S2, generator G2 feeds it directly, but G1 and G3 feed it only **through the reactors**. The fault current from the other sections is therefore limited. - The faulty section is fed from both of its neighbours, each through its own reactor, so the fault current coming from the rest of the system is split between two reactors. **Advantages:** - Good fault-current limitation with low losses in normal running. - A fault is confined to one section; the voltage on the other sections stays fairly high. - Each section is supplied from both directions, so supply is more reliable than in a plain series (in-line) scheme. **Disadvantages:** - More reactors and a more complex layout than the simple series bus-bar scheme. - Load sharing between sections depends on the loads being balanced. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the ring-type bus-bar reactor scheme arranged? A. Bus sections, one generator each, are joined by reactors forming a closed ring B. All generators on one bus with one reactor on each feeder only C. Generators in series through a single reactor D. A reactor in the neutral of each generator Answer: A. Adjacent sections are tied through reactors, and the last section is joined back to the first. (Key point: Ring scheme layout) 2. For a fault on section S2, how do G1 and G3 contribute? A. Directly, with no limit B. Only through reactors, so their contribution is limited C. They do not contribute at all D. Through the generator G2 neutral Answer: B. Only the generator of the faulted section feeds directly; the others see reactor impedance. (Key point: Fault on one section) 3. Why is the ring scheme more reliable than a plain series scheme? A. It needs fewer reactors B. It has higher losses in normal running C. Each section is fed from two sides and each reactor carries less current D. Each section is fed from one side only Answer: C. Two paths to each section improve supply reliability and lower reactor current. (Key point: Advantage over a simple tie-bar) ## Related in Power Plant Equipment - [Bus-bar reactor scheme: 30 MVA and 50 MVA](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-30-50-mva) — Power Plant Equipment, chapter 4, asked 2× - [Bus-bar reactor scheme with four generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-four-generators) — Power Plant Equipment, chapter 4, asked 2× - [Function of reactor in power system](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Areactor-function) — Power Plant Equipment, chapter 4, asked 3× - [Bus reactor between 100 MVA and 200 MVA generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-reactor-100-200-mva) — Power Plant Equipment, chapter 4 - [Fault level: four 25 MVA generators, bus reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-four-25mva-bus-reactor) — Power Plant Equipment, chapter 4 ## Related in other subjects - [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3 - [HV bus fault and limiting reactor](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ahv-bus-fault-reactor-limit) — Power System Analysis II, chapter 3 - [Typical power system network and components](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atypical-power-system-network-components) — Power System Analysis I, chapter 1 - [Node equations and Ybus, six-bus network](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-six-bus-node-equations) — Power System Analysis II, chapter 1