---
title: "Annual saving by capacitors for 200 kW factory"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:factory-200kw-capacitor-annual-saving"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/factory-200kw-capacitor-annual-saving
site: BCT Engineering
---

# Annual saving by capacitors for 200 kW factory

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2071 Chaitra
- Video: [Capacitor Value to Save Electricity Bill 💵 @TheElectricalGuy](https://www.youtube.com/watch?v=1cNeia9D5Fk) — The Electrical Guy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/factory-200kw-capacitor-annual-saving
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6

## Question

A factory takes a load of 200kW at 0.85 pf lagging for 2500 hrs per annum. The tariff is Rs. 150 per KVA plus 5 paisa per KWH consumed. If the pf is improved to 0.9 lagging by means of capacitors costing Rs 420 per KVAR and having a power loss of 100W per KVA, Calculate the annual saving effected by their use. Allow 10% per annum for interest and depreciation.

## Short answer

With 200 kW load, demand charge saving from improving pf 0.85 to 0.9 is Rs 1,960.80. Capacitors of 27.08 kVAR cost Rs 11,375; their loss costs Rs 338.60 and interest/depreciation Rs 1,137.50. Net annual saving is about Rs 485.

## Answer

Saving comes from the reduced kVA demand charge; against it are the capacitor loss (extra kWh) and the interest and depreciation on capacitors. The capacitor kVAR is found from the 200 kW load, and the loss of 100 W per kVA of capacitor rating is taken as 0.1 kW per kVAR.

#### kVA demand before and after
$$
\begin{aligned}
S_1 &= \frac{200}{0.85} = 235.29\ \text{kVA}, \qquad S_2 = \frac{200}{0.9} = 222.22\ \text{kVA} \\
\text{Reduction} &= 235.29 - 222.22 = 13.07\ \text{kVA} \\
\text{Saving in demand charge} &= 150 \times 13.07 = \text{Rs } 1960.8
\end{aligned}
$$

#### Capacitor rating and its costs
$$
\begin{aligned}
\tan\phi_1 &= 0.6197,\quad \tan\phi_2 = 0.4843 \\
Q_C &= 200\,(0.6197 - 0.4843) = 27.08\ \text{kVAR} \\
\text{Loss} &= 0.1 \times 27.08 = 2.708\ \text{kW} \\
\text{Cost of loss} &= 2.708 \times 2500 \times 0.05 = \text{Rs } 338.6 \\
\text{Capital cost} &= 420 \times 27.08 = \text{Rs } 11{,}375 \\
\text{Interest and depreciation} &= 0.10 \times 11{,}375 = \text{Rs } 1137.5
\end{aligned}
$$

#### Net annual saving
$$
\begin{aligned}
\text{Saving} &= 1960.8 - 338.6 - 1137.5 \\
&= \text{Rs } 484.7
\end{aligned}
$$

**Answer: Annual saving ≈ Rs 485 (with 27.08 kVAR of capacitors).**

## Check yourself

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

1. What is the demand charge saving?
   A. 150 x (235.29 - 222.22) = Rs 1960.8
   B. 150 x 200
   C. 150 x 27.08
   D. 0.05 x 200 x 2500
   Answer: A. kVA falls from 200/0.85 to 200/0.9, a reduction of 13.07 kVA. (Key point: Demand charge saving)

2. How is the capacitor loss cost found?
   A. 0.1 kW per kVAR x 27.08 kVAR x 2500 h x Rs 0.05 = Rs 338.6
   B. 100 W x Rs 150
   C. 27.08 x 420
   D. Zero, since capacitors are lossless
   Answer: A. Loss of 2.708 kW over 2500 h at 5 paisa per kWh costs Rs 338.6. (Key point: Capacitor loss)

3. What is the net annual saving?
   A. 1960.8 - 338.6 - 1137.5 = Rs 484.7
   B. Rs 1960.8
   C. Rs 11,375
   D. 1960.8 + 338.6 + 1137.5
   Answer: A. Saving in demand charge minus the capacitor loss cost and interest and depreciation (10% of Rs 11,375). (Key point: Net saving)

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