---
title: "Annual charge and pf improvement for 1200 kVA load"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:pf-improvement-1200kva-annual-charge"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/pf-improvement-1200kva-annual-charge
site: BCT Engineering
---

# Annual charge and pf improvement for 1200 kVA load

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2082 Baisakh, 8 marks
- Video: [How to Solve Electrical Energy Cost and Max Demand Problems (Electrical Power PE Exam)](https://www.youtube.com/watch?v=tOd9cdGBTwQ) — Electrical PE Review
- 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/pf-improvement-1200kva-annual-charge
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6

## Question

The load on a certain installation may be considered constant at 1,200 kVA, 0.75 lagging power factor for 3,000 hours per annum. The tariff is Rs 5,000 per kVA maximum demand plus Rs 5 per kWh.
i) Determine the annual charge of energy.
ii) Power Factor improving apparatus is installed to improve the pf to 0.95 lagging. Determine the kVAR require and the annual charge of energy if the power factor improving apparatus costs Rs 8,000 per kVAR, annual interest and depreciation charges are 10% of the capital cost.

## Short answer

i) P=900 kW, MD charge 1200×5000, energy 900×3000×5, total Rs 19,500,000. ii) Qc=900(0.8819−0.3287)=497.9 kVAR; new MD 947.37 kVA, total annual charge including corrector cost Rs 18,635,170.

## Answer

**Given:** $S = 1200$ kVA at pf 0.75 lag, 3000 h/year, tariff Rs 5000/kVA of maximum demand per year + Rs 5/kWh; pf corrector cost Rs 8000/kVAR, annual interest and depreciation 10%.

#### i) Annual charge at pf 0.75
$$
\begin{aligned}
P &= 1200 \times 0.75 = 900\ \text{kW} \\
\text{MD charge} &= 1200 \times 5000 = \text{Rs } 6\,000\,000 \\
\text{Energy} &= 900 \times 3000 = 2\,700\,000\ \text{kWh} \\
\text{Energy charge} &= 2\,700\,000 \times 5 = \text{Rs } 13\,500\,000 \\
\text{Annual charge} &= \text{Rs } 19\,500\,000
\end{aligned}
$$

#### ii) After raising pf to 0.95
kVAR required:
$$
\begin{aligned}
Q_C &= P(\tan\phi_1 - \tan\phi_2) = 900(0.8819 - 0.3287) = 497.9\ \text{kVAR}
\end{aligned}
$$
New maximum demand:
$$
S_2 = \frac{900}{0.95} = 947.37\ \text{kVA}
$$
Annual charges:
$$
\begin{aligned}
\text{MD charge} &= 947.37 \times 5000 = \text{Rs } 4\,736\,842 \\
\text{Energy charge} &= \text{Rs } 13\,500\,000 \ (\text{unchanged}) \\
\text{Capital cost} &= 497.9 \times 8000 = \text{Rs } 3\,983\,278 \\
\text{Interest + depreciation} &= 0.10 \times 3\,983\,278 = \text{Rs } 398\,328 \\
\text{Total} &= 4\,736\,842 + 13\,500\,000 + 398\,328 = \text{Rs } 18\,635\,170
\end{aligned}
$$

| Item | pf 0.75 (Rs) | pf 0.95 (Rs) |
|---|---|---|
| MD charge | 6,000,000 | 4,736,842 |
| Energy charge | 13,500,000 | 13,500,000 |
| Corrector charges | — | 398,328 |
| Total | 19,500,000 | 18,635,170 |

**Answer:** (i) Rs 19,500,000 per year; (ii) 497.9 kVAR needed, new annual charge Rs 18,635,170, an annual saving of about Rs 864,830.

## Check yourself

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

1. What are the components of the annual charge at pf 0.75?
   A. MD charge 1200 x 5000 plus energy 900 kW x 3000 h x Rs 5, total Rs 19,500,000
   B. MD on 900 kW only, total Rs 18,000,000
   C. Energy on 1200 kVA x 3000 h
   D. Only energy charge Rs 13,500,000
   Answer: A. MD is charged on kVA (Rs 6,000,000) and energy on kWh (2.7 million kWh x 5). (Key point: Annual charge at 0.75)

2. Which gives the kVAR required?
   A. Q_C = 1200(0.8819 - 0.3287)
   B. Q_C = 900(0.8819 - 0.3287) = 497.9 kVAR
   C. Q_C = 900(0.95 - 0.75) = 180 kVAR
   D. Q_C = 947.37 - 900
   Answer: B. Correction is based on kW and the change in tan phi. (Key point: kVAR required)

3. What is the new maximum demand after improvement to 0.95?
   A. 1200 x 0.95 = 1140 kVA
   B. 900 x 0.95 = 855 kVA
   C. 900/0.95 = 947.37 kVA
   D. 1200 kVA, unchanged
   Answer: C. kW stays the same, so kVA = kW / new pf. (Key point: New maximum demand)

4. Why does the total annual charge fall to Rs 18,635,170?
   A. The energy charge falls because kWh is reduced
   B. The capacitor cost is not counted
   C. The kW load decreases
   D. MD charge drops while energy charge is unchanged, and the extra 10% annual cost of apparatus is less than the MD saving
   Answer: D. MD saving of about Rs 1.26 million exceeds interest and depreciation of Rs 398,328. (Key point: Why total falls)

## Close variants set in other exams

- [Annual charge and pf improvement (Rs 65/kVA tariff)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Apf-improvement-1200kva-65-tariff) — Utilization of Electrical Energy, chapter 6, asked 3×
- [Annual charge and pf improvement (Rs 1300/kVA tariff)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Apf-improvement-1200kva-1300-tariff) — Utilization of Electrical Energy, chapter 6, asked 3×

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