---
title: "Capacitor size for maximum loss reduction"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:capacitor-size-11kv-network"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/capacitor-size-11kv-network
site: BCT Engineering
---

# Capacitor size for maximum loss reduction

Transmission and Distribution System Design · Chapter 7: Distribution System Design

- Asked in: 2077 Chaitra, 6 marks; 2074 Magh, 8 marks (asked 2 times)
- 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-11kv-network
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

For the 11 kV primary distribution network shown below, determine the size (kVAR) of the capacitor to be placed at the location shown in diagram to achieve maximum loss reduction. For simplicity of analysis assume voltage at each node is 1 p.u. All impedances are in p.u. at 1000 kVA.

[Figure: radial feeder from a 1000 kVA, 33/11 kV source substation S to node A through impedance 1+j2, A to node B through 2+j3, B to node C through 3+j4. Loads through distribution transformers: 100 kVA at 0.8 p.f. lag at A, 150 kVA at 0.9 p.f. lag at B, 200 kVA at 0.6 p.f. lag at C. The capacitor is placed at node C.]

## Short answer

Optimum capacitor size is the weighted average of section reactive flows: Qc = ΣRiQi/ΣRi = 0.2027 p.u. = 202.7 kVAR. Place a standard 200 kVAR bank at node C to maximize loss reduction.

## Answer

A shunt capacitor supplies reactive power locally, reducing the reactive current in every section upstream of it. The optimum size minimises the total $I^2R$ loss.

**Assumptions:** all node voltages = 1 p.u. (given), so section current magnitude² = $P^2 + Q^2$ in p.u.; base 1000 kVA; transformer impedances neglected.

#### Step 1: Loads in p.u.

| Node | S (p.u.) | p.f. | P (p.u.) | Q (p.u.) |
|---|---|---|---|---|
| A | 0.10 | 0.8 | 0.0800 | 0.10 × 0.6 = 0.0600 |
| B | 0.15 | 0.9 | 0.1350 | 0.15 × 0.4359 = 0.0654 |
| C | 0.20 | 0.6 | 0.1200 | 0.20 × 0.8 = 0.1600 |

#### Step 2: Section flows

| Section | R (p.u.) | P (p.u.) | Q (p.u.) |
|---|---|---|---|
| S–A | 1 | 0.335 | 0.2854 |
| A–B | 2 | 0.255 | 0.2254 |
| B–C | 3 | 0.120 | 0.1600 |

#### Step 3: Loss as a function of capacitor size

With capacitor $Q_c$ at C, the reactive flow in every section reduces by $Q_c$:

$$
P_L = \sum R_i \left[P_i^2 + (Q_i - Q_c)^2\right]
$$

For maximum loss reduction:

$$
\begin{aligned}
\frac{dP_L}{dQ_c} &= -2\sum R_i (Q_i - Q_c) = 0 \\
Q_c &= \frac{\sum R_i Q_i}{\sum R_i} \\
&= \frac{1(0.2854) + 2(0.2254) + 3(0.16)}{1 + 2 + 3} \\
&= \frac{1.2162}{6} = 0.2027 \text{ p.u.}
\end{aligned}
$$

$$
Q_c = 0.2027 \times 1000 = 202.7 \text{ kVAR}
$$

#### Step 4: Check of loss reduction (p.u.)

| Section | Loss before | Loss after |
|---|---|---|
| S–A | 0.1937 | 0.1191 |
| A–B | 0.2316 | 0.1311 |
| B–C | 0.1200 | 0.0487 |
| Total | 0.5453 | 0.2988 |

Loss falls from 0.545 p.u. to 0.299 p.u., a reduction of about 45%. (The given impedances are very large, so the absolute p.u. losses are only for comparison.)

Note: the optimum $Q_c$ (0.203 p.u.) is larger than the reactive load at C (0.16 p.u.) because upstream sections also carry the reactive power of A and B; the result is a weighted average of section reactive flows.

**Answer: Capacitor size ≈ 203 kVAR at node C** (a standard 200 kVAR bank would be used).

## Check yourself

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

1. How does a shunt capacitor at node C reduce loss?
   A. It reduces active power in all sections
   B. It supplies reactive power locally, reducing the reactive current in every upstream section
   C. It raises resistance
   D. It increases the load P
   Answer: B. Q flow in all sections upstream of C falls by Qc, reducing I^2R. (Key point: Effect of capacitor)

2. With capacitor Qc at C, total loss is:
   A. Sum of R_i (P_i^2 + Q_i^2) - Qc
   B. Sum of R_i [P_i^2 + (Q_i - Qc)^2]
   C. Sum of R_i (P_i - Qc)^2
   D. Sum of X_i Qc^2
   Answer: B. Only the reactive flow in each section is reduced by Qc. (Key point: Loss function)

3. Setting dP_L/dQc = 0 gives Qc equal to:
   A. Sum of Q_i
   B. Sum(R_i Q_i)/Sum(R_i)
   C. Sum(P_i)/Sum(R_i)
   D. Q of the last load only
   Answer: B. The optimum is the resistance-weighted average of section reactive flows. (Key point: Optimum size formula)

4. The optimum capacitor size is about:
   A. 160 kVAR
   B. 120 kVAR
   C. 285 kVAR
   D. 203 kVAR
   Answer: D. Qc = 1.2162/6 = 0.2027 p.u. = 203 kVAR, larger than the 160 kVAR at C because upstream reactive loads also benefit. (Key point: Result)

## Close variants set in other exams

- [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, 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, feeder with 500 kVA end load](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-500kva-load-c) — 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