2 kW electric heater, 4 h/day for a month
The classic "why is my bill so high" calculation.
240 kWh — 60.00 at 0.25/kWh
Electrical Power & Energy Cost Calculator
Power from any two of voltage, current and resistance — then energy in kWh and what it costs to run.
Inputs
Calculate from
V
A
Accepts 500m for 500 mA.
In your own currency — the result comes back in the same units.
Results
Power1.15 kWP = V × I = I²R = V²/R
- Voltage
- 230 V
- Current
- 5 A
- Resistance
- 46 Ω
- Energy used
- 138E = P × h / 1000 (kWh)
- Cost to run
- 34.5
Running 4 h/day, that is 4.60 kWh per day and roughly 1.68e+3 kWh — about 420 per year at this tariff.
If this is a resistor, 1.15e+3 W means a wirewound or chassis-mount part — well beyond the 0.25 W of a standard through-hole resistor.
Diagram
Worked examples
Always-on router at 12 W
Small loads look negligible until you multiply by 8,760 hours a year.
105 kWh a year — 26.28
Resistor dissipation: 12 V across 100 Ω
The component-level use — checking what a resistor has to survive.
1.44 W — needs a 2 W or 3 W part
Motor drawing 8 A at 400 V three-phase (per phase)
Per-phase power for an industrial load.
3.2 kW per phase, 563 kWh a month
Power, energy, and the difference between them
When to use this: checking what an appliance costs to run, sizing a resistor or heatsink, working out the current a load will draw, or deciding whether a standby load is worth eliminating.
Power is the rate at which energy is converted, measured in watts. A 2 kW heater converts 2,000 joules of electrical energy into heat every second, whenever it is on.
Energy is power accumulated over time, measured in kilowatt-hours. That same heater running for four hours uses 8 kWh. Power is the speedometer; energy is the odometer. Your meter counts energy, and that is what appears on the bill.
This distinction is where most confusion about electricity costs comes from. A high-power appliance used briefly can cost far less than a low-power one left on permanently.
The three power formulas
| Formula | Use when you know | Typical situation |
|---|---|---|
P = V × I | Voltage and current | Measured with a meter, or from a rating plate |
P = I² × R | Current and resistance | Heat in a cable, shunt or winding |
P = V² / R | Voltage and resistance | Dissipation in a resistor across a known supply |
These are one relationship, not three. Substituting Ohm’s law into P = V × I produces the other two, so they always agree. If two of them give different answers for the same circuit, a value has been entered wrong.
Which one you reach for depends on what is causing the heat. For a cable or a winding, the current is fixed by the load and I²R is the natural form — and its squared term is why doubling the current quadruples the heating. For a resistor across a fixed supply, V²/R is more direct.
Working out running cost
Three steps, and the only one people get wrong is the first:
- Power in kilowatts. Divide watts by 1,000. A 2,000 W heater is 2 kW.
- Energy in kWh. Multiply by hours of use. 2 kW × 4 h = 8 kWh.
- Cost. Multiply by your tariff per kWh.
The trap is reading the rating plate as continuous consumption. A 2 kW heater with a thermostat does not draw 2 kW continuously — it cycles, and might average a third of that. A fridge rated at 150 W runs perhaps a quarter of the time. Rating plates give peak draw, not average consumption.
What continuous small loads actually cost
| Device | Power | kWh / year | Cost at 0.25/kWh |
|---|---|---|---|
| Router / modem | 12 W | 105 | 26 |
| Set-top box on standby | 15 W | 131 | 33 |
| Desktop PC idling | 60 W | 526 | 131 |
| Old plasma TV on standby | 25 W | 219 | 55 |
| Phone charger, nothing plugged in | 0.1 W | 0.9 | 0.22 |
The last row is worth noting. Unplugging chargers is frequently recommended and saves essentially nothing — a modern charger with no load draws a fraction of a watt. The devices worth attacking are the ones drawing tens of watts around the clock.
AC power and power factor
Everything above assumes voltage and current are in phase, which is true for resistive loads: heaters, incandescent lamps, kettles, toasters. For those, RMS voltage times RMS current gives real power in watts, and the formulas work unchanged.
Motors, transformers and switch-mode supplies are different. Their current lags or leads the voltage, and only the in-phase component does useful work:
P (watts) = V × I × cos φS (volt-amps) = V × I
cos φ is the power factor. A motor with a power factor of 0.8 drawing 10 A at 230 V has an apparent power of 2,300 VA but a real power of only 1,840 W. You are billed for the watts, but the cable, breaker and transformer all have to carry the full 10 A — which is why large industrial installations pay penalties for poor power factor and fit correction capacitors.
Worked examples
Sizing a resistor
A 100 Ω resistor across a 12 V supply:
P = V²/R = 144 / 100 = 1.44 W
With the usual 2× margin that calls for a 3 W part. A 0.25 W resistor here would reach several hundred degrees and fail within seconds — this calculation takes five seconds and is the most commonly skipped step in beginner designs.
Current draw of a 2 kW appliance
I = P / V = 2000 / 230 = 8.7 A
Comfortable on a 13 A plug. On a 120 V supply the same appliance would draw 16.7 A, needing a dedicated 20 A circuit — which is why kettles are noticeably slower in 120 V countries. The plug limits the power available.
Is the standby worth killing?
A device drawing 15 W continuously: 15 × 8760 / 1000 = 131 kWh a year. At 0.25 per unit that is 33 a year. A smart plug costs about the same, so it pays for itself in twelve months — and only if the device genuinely does not need to be on.
Where the formulas stop working
- Non-resistive AC loads. Without power factor you get volt-amps, not watts.
- Non-sinusoidal current. Switch-mode supplies and LED drivers draw current in short spikes. RMS values still work, but simple V × I overstates real power unless the meter is a true-RMS type measuring both waveforms.
- Three-phase. Total power is √3 × Vline × Iline × cos φ for a balanced load — not three times the single-phase figure.
- Varying loads. A rating plate gives peak draw. For anything thermostatic or duty-cycled, measure over time or use a manufacturer’s annual consumption figure.
How to use this calculator
- Pick what you know
Voltage and current, voltage and resistance, current and resistance, or just the wattage from a rating plate.
- Enter the values
Engineering notation works throughout — 500m for 500 mA, 4k7 for 4.7 kΩ.
- Add the usage pattern
Hours per day and number of days. Set hours to 0 if you only want the instantaneous power.
- Enter your tariff
Use your own currency. The cost comes back in whatever units you put in, so there is no currency conversion to get wrong.
Frequently asked questions
How do I calculate electrical power?
Power in watts is voltage times current: P = V × I. If you know resistance instead, use P = I²R or P = V²/R. All three are the same relationship rearranged, and give identical answers for the same circuit — use whichever matches the values you already have.
How do I convert watts to kWh?
Multiply the power in watts by the number of hours, then divide by 1000. A 2000 W heater running 4 hours uses 2000 × 4 / 1000 = 8 kWh. The kilowatt-hour is a unit of energy, not power — it is what your meter counts and what you are billed for.
What is the difference between watts and kilowatt-hours?
Watts measure the rate of energy use at an instant; kilowatt-hours measure the total energy used over time. A 2 kW heater always draws 2 kW while it is on, but whether that costs pennies or pounds depends entirely on how long you run it. Power is the speedometer, energy is the odometer.
Why do the three power formulas give the same answer?
Because they are the same equation with Ohm’s law substituted in. Starting from P = V × I, replacing V with I × R gives P = I²R, and replacing I with V/R gives P = V²/R. They are algebraically identical, so any disagreement means a value has been entered wrong.
Does this work for AC circuits?
For purely resistive AC loads — heaters, incandescent lamps, kettles — yes, using RMS voltage and current. For motors, transformers and switch-mode supplies, the current and voltage are out of phase and you need power factor: real power P = V × I × cos φ. Without it you get apparent power in volt-amps, which is larger than the watts you are billed for.
How much does it cost to run an appliance?
Multiply the power in kW by the hours used, then by your tariff per kWh. A 100 W device left on continuously uses 876 kWh a year — at 0.25 per unit, about 219. Standby loads are worth checking for exactly this reason: small numbers multiplied by 8,760 hours stop being small.
What is the difference between watts and volt-amps?
Watts are real power — energy actually converted to heat, light or motion. Volt-amps are apparent power, the product of RMS voltage and RMS current regardless of phase. For a resistive load they are equal. For an inductive load like a motor, the VA figure is higher, and cables and breakers must be sized for the VA even though you are billed for the watts.
How do I work out the current an appliance draws?
Divide the power rating by the supply voltage: I = P / V. A 2000 W heater on a 230 V supply draws 8.7 A. On a 120 V supply the same heater would draw 16.7 A — which is why high-power appliances need heavier wiring in 120 V countries.
Sources and further reading
Last reviewed .
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