--- title: "Capacitor size, feeder with 500 kVA end load" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:capacitor-size-500kva-load-c" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/capacitor-size-500kva-load-c site: BCT Engineering --- # Capacitor size, feeder with 500 kVA end load Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2069 Bhadra (old course), 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/capacitor-size-500kva-load-c - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question For an 11 kV radial primary distribution feeder shown in figure below, determine the size of the capacitor at the stated location in the feeder to achieve the maximum loss reduction. For simplifying the analysis assume voltage at each node is 1 p.u. [Figure: radial feeder from a 1000 kVA, 33/11 kV source substation S to node A through 3+j2, A to B through 2+j4, B to C through 1+j2. Loads through distribution transformers: 200 kVA at 0.8 p.f. lag at A, 100 kVA at 0.7 p.f. lag at B, 500 kVA at 0.6 p.f. lag at C. The capacitor is placed at node C.] ## Short answer Optimum capacitor size is the resistance-weighted average of section reactive flows: Q_c = (∑R_i Q_i)/(∑R_i). For the given feeder, Q_c = 519.5 kVAR, approximately 520 kVAR. ## Answer #### Method (optimum single capacitor) With all node voltages taken as 1 p.u., the reactive current in a section equals its reactive power flow $Q_i$ (p.u.). A shunt capacitor $Q_c$ at the end node reduces the reactive flow in **every section between the source and that node** from $Q_i$ to $(Q_i - Q_c)$. The loss reduction is $$ \begin{aligned} \Delta P_L &= \sum_i R_i\left[Q_i^2-(Q_i-Q_c)^2\right] = \sum_i R_i\left(2Q_iQ_c-Q_c^2\right) \end{aligned} $$ For maximum loss reduction, $\dfrac{d(\Delta P_L)}{dQ_c} = \sum_i R_i(2Q_i-2Q_c)=0$, so $$ Q_{c,opt} = \frac{\sum_i R_iQ_i}{\sum_i R_i} $$ i.e. the optimum capacitor equals the **resistance-weighted average** of the reactive flows in the sections it relieves. The ratio does not depend on whether $R$ is in ohms or p.u., so kVAR can be used directly. ```text S --(3+j2)-- A --(2+j4)-- B --(1+j2)-- C | | | 200 kVA 100 kVA 500 kVA 0.8 lag 0.7 lag 0.6 lag [Qc] ``` #### Reactive loads | Node | S (kVA) | p.f. | $\sin\phi$ | Q (kVAR) | |---|---|---|---|---| | A | 200 | 0.8 | 0.6 | 120.00 | | B | 100 | 0.7 | 0.7141 | 71.41 | | C | 500 | 0.6 | 0.8 | 400.00 | #### Section flows | Section | R | Q flow (kVAR) | R × Q | |---|---|---|---| | S–A | 3 | 591.41 | 1774.24 | | A–B | 2 | 471.41 | 942.83 | | B–C | 1 | 400.00 | 400.00 | | Total | 6 | | 3117.07 | #### Optimum capacitor $$ Q_c = \frac{1774.24+942.83+400}{3+2+1} = \frac{3117.07}{6} = 519.5\ \text{kVAR} $$ Because the largest resistance (S–A) carries the largest reactive flow, the optimum capacitor exceeds the 400 kVAR load at C; it slightly over-compensates section B–C but gives the largest total loss reduction. **Answer: Capacitor at node C ≈ 519.5 kVAR (≈ 520 kVAR).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a single capacitor at the end node, the optimum size is: A. Reactive load at that node B. Sum of R_i Q_i / sum of R_i over sections it relieves C. Sum of Q_i of all nodes D. Half the reactive load Answer: B. Minimising loss gives the resistance-weighted average of section reactive flows. (Key point: Optimum formula) 2. Reactive load of node B (100 kVA, p.f. 0.7 lag) is about: A. 70 kVAR B. 71.4 kVAR C. 49 kVAR D. 100 kVAR Answer: B. Q = 100 x sinφ = 100 x 0.7141 = 71.41 kVAR. (Key point: Reactive loads) 3. The reactive flow in S-A is: A. 591.41 kVAR B. 120 kVAR C. 471.41 kVAR D. 400 kVAR Answer: A. S-A carries all loads: 120 + 71.41 + 400 = 591.41 kVAR. (Key point: Section flows) 4. The optimum capacitor of 519.5 kVAR exceeds the 400 kVAR load at C because: A. A mistake in p.f. B. The largest resistance (S-A) carries the largest reactive flow, so it dominates the weighting C. Capacitors must exceed load D. Voltage is above 1 p.u. Answer: B. Heavily weighted upstream flows pull the optimum above 400 kVAR, over-compensating B-C slightly for greater total reduction. (Key point: Result and meaning) ## Close variants set in other exams - [Capacitor size, four-node 11 kV feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-four-node-feeder) — Transmission and Distribution System Design, chapter 7, asked 5× - [Capacitor size for maximum loss reduction (variant)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-11kv-network-b) — Transmission and Distribution System Design, chapter 7, asked 5× - [Capacitor size for maximum loss reduction](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-11kv-network) — Transmission and Distribution System Design, chapter 7, asked 5× ## Related in Transmission and Distribution System Design - [Capacitor size, feeder with three equal loads](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-equal-loads-ohms) — Transmission and Distribution System Design, chapter 7 - [Loss and drop, 15 km 11 kV feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Audl-feeder-15km-11kv) — Transmission and Distribution System Design, chapter 7, asked 2× - [Power loss, 10 km 11 kV UDL feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Audl-feeder-10km-11kv) — Transmission and Distribution System Design, chapter 7, asked 2× ## Related in other subjects - [Design: 500 kVA 66/11 kV power transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-66-11kv) — Electric Machine Design, chapter 3, asked 6× - [3-phase 500 kVA transformer primary current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-11kv-400v-primary-current) — Electrical Machines, chapter 2 - [Design: 100 kVA 11/132 kV transformer with winding data](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-11-132kv) — Electric Machine Design, chapter 3, asked 3× - [Primary current and efficiency of 11 kV/380 V Δ/Y transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Adelta-star-11kv-380v-40kw-efficiency) — Electrical Machines I, chapter 2, asked 3×