Inputs
Results
Single-phaseElectrical
Energy & Cost
Example Appliances
| Appliance | Typical Watts | Hours / day | kWh / day | Cost / day | Remove |
|---|---|---|---|---|---|
| LED Bulb 9–10W | 10 | 5 | 0.05 | 0.01 | |
| Ceiling Fan | 70 | 8 | 0.56 | 0.14 | |
| Refrigerator | 150 | 12 | 1.80 | 0.45 | |
| Microwave | 1100 | 0.5 | 0.55 | 0.14 | |
| Air Conditioner 1.5 Ton | 1500 | 5 | 7.50 | 1.88 | |
| Laptop | 65 | 6 | 0.39 | 0.10 | |
| TV 42" | 100 | 4 | 0.40 | 0.10 |
Edit watts and hours directly. Costs use the current rate above.
Formula Used
Single‑phase: P(W) = V × I × PF
Three‑phase: P(W) = √3 × V × I × PF
Energy: kWh = (P(W) ÷ 1000) × hours
Cost: Cost = kWh × rate
Power factor (PF) accounts for phase difference of voltage and current in AC systems. Resistive devices like heaters have PF≈1, motor-driven devices often have PF<1.
How to Use
- Select System Type and set Power Factor (1 for purely resistive loads).
- Choose whether you know Amps or Watts.
- Enter Voltage and the known quantity.
- Provide usage hours per day, days per month, and your electricity rate.
- Press Calculate. Results show watts, amps, kWh, and costs.
- Use the example table to estimate typical usage across devices; edit values inline.
- Export your results as CSV or PDF for reporting.
FAQs
If unknown, start with 1 for heaters and incandescent lamps. For motors or compressors, typical values range 0.7–0.95. Nameplates or manuals often include PF.
Watts represent real power, while volt‑amperes represent apparent power. With PF<1, Watts = VA × PF. This tool estimates real power and energy cost.
Yes. Enter your local voltage. The formulas are independent of region provided the correct voltage and power factor are used.
Select three‑phase. Provide line‑to‑line voltage and per‑phase current. The calculator applies the √3 factor automatically.
Yes. Nameplate ratings and modes vary widely. Always prefer the device nameplate or measured values for best accuracy.
Starting surges raise instantaneous power briefly but have little effect on overall energy cost compared to steady‑state consumption.
Accuracy depends on correct inputs and realistic usage assumptions. For the most precise results, measure with a plug‑in energy meter and use its readings here.
Quick Tips
- Lower PF increases current for the same watts.
- Reducing daily hours lowers both kWh and cost linearly.
- Modern inverter ACs often consume less energy per hour than fixed‑speed units.
- Standby loads add up—unplug or use smart switches.