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

# Capacitor size for maximum loss reduction (variant)

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

- Asked in: 2075 Bhadra, 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-11kv-network-b
- 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 the maximum loss reduction. For simplifying the 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+j1, A to node B through 1+j1, B to node C through 1+j2. Loads through distribution transformers: 100 kVA at 0.8 p.f. lag at A, 50 kVA at 0.85 p.f. lag at B, 100 kVA at 0.8 p.f. lag at C. The capacitor is placed at node C.]

## Short answer

The optimum capacitor size at node C is the resistance-weighted average of reactive flows: (146.34+86.34+60)/3 = 97.56 kVAR. Use 97.6 kVAR (≈100 kVAR standard).

## 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 --(1+j1)-- A --(1+j1)-- B --(1+j2)-- C
              |            |            |
          100 kVA       50 kVA      100 kVA
          0.8 lag      0.85 lag     0.8 lag   [Qc]
```

#### Reactive load at each node
| Node | S (kVA) | p.f. | $\sin\phi$ | Q (kVAR) |
|---|---|---|---|---|
| A | 100 | 0.8 | 0.6 | 60.00 |
| B | 50 | 0.85 | 0.5268 | 26.34 |
| C | 100 | 0.8 | 0.6 | 60.00 |

#### Reactive flow in each section
| Section | R (p.u.) | Q flow (kVAR) | R × Q |
|---|---|---|---|
| S–A | 1 | 60 + 26.34 + 60 = 146.34 | 146.34 |
| A–B | 1 | 26.34 + 60 = 86.34 | 86.34 |
| B–C | 1 | 60.00 | 60.00 |
| Total | 3 | | 292.68 |

#### Optimum capacitor
$$
Q_c = \frac{146.34+86.34+60}{1+1+1} = \frac{292.68}{3} = 97.56\ \text{kVAR}
$$
In p.u. on 1000 kVA: $Q_c = 0.0976$ p.u.

Check of the loss reduction (p.u. on 1000 kVA):
$$
\begin{aligned}
\Delta P_L &= \sum R_i(2Q_iQ_c - Q_c^2)\\
&= 2(0.09756)(0.29268) - 3(0.09756)^2 = 0.02855\ \text{p.u.} \approx 28.6\ \text{kW}
\end{aligned}
$$
The capacitor is larger than the 60 kVAR load at C because it also relieves the heavier flows in S–A and A–B. In practice the nearest standard bank (about 100 kVAR) would be used.

**Answer: Optimum capacitor at node C ≈ 97.6 kVAR (≈ 100 kVAR standard).**

## Check yourself

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

1. The optimum single capacitor equals:
   A. Sum of Q loads at the node
   B. Sum(Ri Qi)/Sum(Ri), resistance-weighted average of reactive flows
   C. The total kVA
   D. Half the end load
   Answer: B. Setting d(loss reduction)/dQc = 0 gives that. (Key point: Formula)

2. The reactive flow in S-A is:
   A. 60 kVAR
   B. 86.34 kVAR
   C. 146.34 kVAR
   D. 26.34 kVAR
   Answer: C. 60 + 26.34 + 60 = 146.34. (Key point: Section flows)

3. The optimum capacitor is:
   A. 60 kVAR
   B. 97.6 kVAR
   C. 146 kVAR
   D. 292.7 kVAR
   Answer: B. 292.68/3 = 97.56 kVAR. (Key point: Result)

4. Why is Qc larger than the 60 kVAR at C?
   A. It also relieves heavier flows in S-A and A-B
   B. Capacitors are oversized
   C. Voltage is above 1 p.u.
   D. Because of load at A only
   Answer: A. It reduces flow in all sections between the source and C. (Key point: Why larger)

## 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](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×
- [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