---
title: "Annual charge and pf improvement (Rs 1300/kVA tariff)"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:pf-improvement-1200kva-1300-tariff"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/pf-improvement-1200kva-1300-tariff
site: BCT Engineering
---

# Annual charge and pf improvement (Rs 1300/kVA tariff)

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2080 Bhadra, 8 marks
- 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-1300-tariff
- 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 1200 KVA, 0.75 lagging power factor for 3000 hr per annum. The tariff is 1300 per KVA maximum demand plus 80 paisa per KWh. Determine:
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 required and annual new charge of energy if PF correcting apparatus cost to Rs 1200 per KVAr, annual interest and deprecation charge are 10% of capital cost.

## Short answer

i) P=1200×0.75=900 kW, MD=1200×1300, energy=900×3000×0.80 gives Rs 3,720,000. ii) Qc=900(0.8819−0.3287)=497.9 kVAR, new MD=900/0.95=947.37 kVA, total annual charge with corrector Rs 3,451,328.

## Answer

**Given:** $S = 1200$ kVA, pf 0.75 lag, 3000 h/year; tariff Rs 1300/kVA of maximum demand per year + 80 paisa (Rs 0.80)/kWh; corrector Rs 1200/kVAR; interest and depreciation 10%.

#### i) Annual charge at pf 0.75
$$
\begin{aligned}
P &= 1200 \times 0.75 = 900\ \text{kW} \\
\text{MD charge} &= 1200 \times 1300 = \text{Rs } 1\,560\,000 \\
\text{Energy charge} &= 900 \times 3000 \times 0.80 = \text{Rs } 2\,160\,000 \\
\text{Annual charge} &= \text{Rs } 3\,720\,000
\end{aligned}
$$

#### ii) kVAR required for pf 0.95
$$
\begin{aligned}
\tan\phi_1 &= \tan(\cos^{-1}0.75) = 0.8819, \quad \tan\phi_2 = \tan(\cos^{-1}0.95) = 0.3287 \\
Q_C &= 900(0.8819 - 0.3287) = 497.9\ \text{kVAR}
\end{aligned}
$$

#### New annual charge
$$
\begin{aligned}
S_2 &= \frac{900}{0.95} = 947.37\ \text{kVA} \\
\text{MD charge} &= 947.37 \times 1300 = \text{Rs } 1\,231\,579 \\
\text{Energy charge} &= \text{Rs } 2\,160\,000 \\
\text{Corrector cost} &= 497.9 \times 1200 = \text{Rs } 597\,492 \\
\text{Interest + depreciation} &= 0.10 \times 597\,492 = \text{Rs } 59\,749 \\
\text{New annual charge} &= 1\,231\,579 + 2\,160\,000 + 59\,749 = \text{Rs } 3\,451\,328
\end{aligned}
$$

| Item | pf 0.75 (Rs) | pf 0.95 (Rs) |
|---|---|---|
| MD charge | 1,560,000 | 1,231,579 |
| Energy charge | 2,160,000 | 2,160,000 |
| Corrector charges | — | 59,749 |
| Total | 3,720,000 | 3,451,328 |

**Answer:** (i) Rs 3,720,000 per year; (ii) 497.9 kVAR, new annual charge Rs 3,451,328 (saving about Rs 268,672 per year).

## Check yourself

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

1. What is the annual charge at pf 0.75?
   A. Energy at Rs 80 per kWh
   B. MD on 900 kVA
   C. Rs 2,160,000 only
   D. MD 1200 x 1300 = Rs 1,560,000 plus energy 900 x 3000 x 0.80 = Rs 2,160,000, total Rs 3,720,000
   Answer: D. 80 paisa is Rs 0.80 per kWh. (Key point: Annual charge)

2. How is the kVAR for pf 0.95 found?
   A. Q_C = 900(0.8819 - 0.3287) = 497.9 kVAR
   B. 1200 x 0.2
   C. 900 x (0.95 - 0.75)
   D. 1200(0.8819 - 0.3287)
   Answer: A. Uses kW and tan phi difference. (Key point: kVAR required)

3. What is the new MD charge?
   A. 1200 x 1300
   B. 947.37 kVA x 1300 = Rs 1,231,579
   C. 900 x 1300
   D. 1140 x 1300
   Answer: B. New kVA = 900/0.95. (Key point: New MD charge)

4. What is included in the new annual charge Rs 3,451,328?
   A. Only MD charge
   B. MD charge plus full corrector cost
   C. New MD charge, unchanged energy charge and 10% of corrector cost (Rs 59,749)
   D. Reduced energy charge
   Answer: C. Corrector cost 497.9 x 1200 = Rs 597,492; 10% is interest and depreciation. (Key point: New total)

## 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 for 1200 kVA load](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Apf-improvement-1200kva-annual-charge) — Utilization of Electrical Energy, chapter 6, asked 3×

## Related in Utilization of Electrical Energy

- [Annual saving by pf improvement for 800 kW load](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainstallation-800kw-capacitor-annual-saving) — Utilization of Electrical Energy, chapter 6
- [Annual cost and saving with pf correction (400 kVA)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aindustry-50000-units-pf-saving) — Utilization of Electrical Energy, chapter 6

## Related in other subjects

- [PF improvement of 490 kW load with demand charge](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Apf-490kw-6-6kv-demand-charge) — Industrial Power Distribution and Illumination, chapter 10, asked 2×
- [PF improvement of 1020 kVA factory load](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Afactory-1020kva-pf-improvement) — Industrial Power Distribution and Illumination, chapter 10, asked 2×