---
title: "Annual charge and pf improvement (Rs 65/kVA tariff)"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:pf-improvement-1200kva-65-tariff"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/pf-improvement-1200kva-65-tariff
site: BCT Engineering
---

# Annual charge and pf improvement (Rs 65/kVA tariff)

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2072 Chaitra, 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-65-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 1200kVA, 0.75 lagging power factor for 3000 hours per annum. The tariff is Rs 65 per kVA of maximum demand plus 2 paisa per kWH.
i) Determine the annual charge for electrical energy
ii) Power factor improving apparatus is installed to improve the power factor to 0.95 lagging. Determine the kVAR required and the new annual charge if the power factor improving apparatus costs Rs 60 per kVAR, annual interest and depreciation charges are 10% of the capital cost and the losses in the apparatus are 5% of the kVAR rating.

## Short answer

Initial annual charge is Rs 132,000. Improving pf to 0.95 requires about 489.9 kVAR including 5% apparatus loss, giving a new annual charge of Rs 121,664, a saving of about Rs 10,336.

## Answer

#### (i) Present annual charge
$$
\begin{aligned}
P &= 1200 \times 0.75 = 900\ \text{kW} \\
\text{Demand charge} &= 65 \times 1200 = \text{Rs } 78{,}000 \\
\text{Energy charge} &= 0.02 \times 900 \times 3000 = \text{Rs } 54{,}000 \\
\text{Annual charge} &= \text{Rs } 132{,}000
\end{aligned}
$$

#### (ii) After improving pf to 0.95
Let the apparatus rating be $Q_C$ kVAR. Its loss, $0.05\,Q_C$ kW, is also drawn from the supply, so the overall pf of 0.95 must include it.
$$
\begin{aligned}
\tan\phi_1 &= \tan(\cos^{-1}0.75) = 0.8819,\quad \tan\phi_2 = \tan(\cos^{-1}0.95) = 0.3287 \\
\text{Load kVAR} &= 900 \times 0.8819 = 793.73\ \text{kVAR} \\
\text{Condition: } \frac{793.73 - Q_C}{900 + 0.05\,Q_C} &= 0.3287 \\
793.73 - Q_C &= 295.82 + 0.01643\,Q_C \\
Q_C &= \frac{497.91}{1.01643} = 489.9\ \text{kVAR}
\end{aligned}
$$
New annual charge:
$$
\begin{aligned}
\text{Apparatus loss} &= 0.05 \times 489.9 = 24.5\ \text{kW} \\
\text{Total kW} &= 900 + 24.5 = 924.5\ \text{kW} \\
\text{New max demand} &= \frac{924.5}{0.95} = 973.15\ \text{kVA} \\
\text{Demand charge} &= 65 \times 973.15 = \text{Rs } 63{,}255 \\
\text{Energy charge} &= 0.02 \times 924.5 \times 3000 = \text{Rs } 55{,}470 \\
\text{Capital cost} &= 60 \times 489.9 = \text{Rs } 29{,}392 \\
\text{Interest and depreciation} &= 0.10 \times 29{,}392 = \text{Rs } 2{,}939 \\
\text{New annual charge} &= 63{,}255 + 55{,}470 + 2{,}939 = \text{Rs } 121{,}664
\end{aligned}
$$

| Item | Before (Rs) | After (Rs) |
|---|---|---|
| Demand charge | 78,000 | 63,255 |
| Energy charge | 54,000 | 55,470 |
| Apparatus charges | 0 | 2,939 |
| Total | 132,000 | 121,664 |

(If the apparatus loss is ignored when sizing, $Q_C = 497.9$ kVAR; but then the kVA demand including the 24.9 kW loss is about 971 kVA, not 947.4 kVA.)

**Answer: (i) Rs 132,000 per year; (ii) kVAR required ≈ 489.9 kVAR, new annual charge ≈ Rs 121,664 (saving about Rs 10,336 per year).**

## Check yourself

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

1. What is the present annual charge?
   A. Rs 132,000 (78,000 demand + 54,000 energy)
   B. Rs 78,000
   C. Rs 54,000
   D. Rs 120,060
   Answer: A. Demand = 65 x 1200 = 78,000; energy = 0.02 x 900 x 3000 = 54,000. (Key point: Present annual charge)

2. What is the new maximum demand after improving to 0.95?
   A. 1200 kVA
   B. 900/0.95 = 947.37 kVA
   C. 900 x 0.95 = 855 kVA
   D. 1200 x 0.95 = 1140 kVA
   Answer: B. kVA = kW / pf = 900/0.95. (Key point: New maximum demand)

3. How does the 5% apparatus loss affect the new charge?
   A. It adds 24.9 kW to the energy charge
   B. It reduces the demand charge
   C. It is ignored
   D. It reduces capital cost
   Answer: A. Loss = 0.05 x 497.9 = 24.9 kW, so energy is on 924.9 kW, giving Rs 55,494; total new charge Rs 120,060. (Key point: Apparatus loss effect)

## Close variants set in other exams

- [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×
- [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×

## 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

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- [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×