Single phase current:
formula and worked example

Why the amps matter more than the watts

Cables, breakers, plugs and switches are all rated in amps, not watts. A load's watts tell you how much work it does, but its current tells you how much heat it creates in the wiring on the way. Knowing the current is therefore the first step in choosing a cable size or a protective device.

For a simple heater the arithmetic is watts divided by volts. For motors, compressors and other inductive loads, the power factor enters because part of the current does no useful work, yet still flows through the wires and heats them.

The same logic explains why the supply company and the electrician both talk about amps. Two appliances of the same wattage can place different demands on the wiring if one has a poor power factor, and it is the current that decides which cable is acceptable.

The formula and its three inputs

• Power: the real power in watts that the load consumes, not the apparent power in volt-amperes.
• Voltage: the supply voltage at the load, for example 230 V.
• Power factor: a number between 0 and 1. A resistive heater is near 1, while motors and some electronics are lower.

If a nameplate gives apparent power in volt-amperes, divide by voltage directly and no power factor is needed, because the power factor is already inside that figure.

Real power in watts is what appears on most appliance labels, but for motors check whether the stated rating is the mechanical output or the electrical input. A motor delivering 1.5 kW at the shaft draws more than 1.5 kW from the supply because of its own losses.

Worked example: 1,800 W, 230 V, power factor 0.9

An appliance consumes 1,800 W from a 230 V single-phase supply at a power factor of 0.9.

1. 1Real power: 1,800 W.
2. 2Voltage × power factor: 230 × 0.9 = 207.
3. 3Divide: 1,800 ÷ 207 = 8.696.
4. 4The load draws about 8.7 A.

If the power factor were 1.0, the current would be 1,800 ÷ 230 = 7.83 A. The 0.9 power factor adds close to an amp for the same real power.

The apparent power here is 8.696 × 230 = 2,000 volt-amperes. That is the figure a supply or an inverter must be able to provide, and it is higher than the 1,800 W the load actually uses.

Turning amps into cable and breaker decisions

Once you have the running current, you compare it with the ratings of the circuit. The cable must carry the current continuously without overheating, and the protective device must be rated above the running current but below the cable's capacity.

• Running current: from this formula, based on real operating conditions.
• Margin: designs add allowance for ambient temperature, bundling and future load, following the standard that applies where you work.
• Starting current: motors draw much more when they start, which affects breaker type and voltage drop.

Specific ratings and derating rules vary by country and code, so use the applicable standard or a qualified electrician for the final selection.

The calculated current is a continuous value. Short bursts, such as motor starting, are handled by the characteristic of the breaker rather than by upsizing everything, which is why the type of protective device matters as much as its rating.

How the load and power factor move the current

A 10% increase in power takes the current from 8.696 to 9.565 A. Power factor has an even stronger effect at the low end: at 0.7 the same 1,800 W would draw about 11.2 A, almost 30% more than at 0.9. That is why a motor with a poor power factor needs thicker cable than its wattage alone suggests.

Voltage works in the opposite direction. A 10% fall in supply voltage at constant power raises the current by roughly 11%, which is why heavily loaded circuits at the end of a long run can run warmer than the calculation at nominal voltage suggests.

Mistakes in single-phase current estimates

• Using apparent power in volt-amperes and then dividing by power factor again, which double-counts it.
• Forgetting power factor and sizing from watts divided by volts, which undershoots for motors.
• Using the motor's mechanical output rating where its electrical input is needed.
• Assuming 230 V holds at the far end of a long cable run, where voltage drops.
• Applying a single-phase formula to a three-phase load.

A short scenario

A workshop adds a 1.8 kW machine to a circuit that already carries 10 A. The calculation shows a further 8.7 A, bringing the total near 19 A. If the circuit is a 16 A branch, it is overloaded even though each load alone looked acceptable. Adding currents, not watts, is what exposes the problem.

• List each load on the circuit with its power and power factor.
• Calculate each load's current and add them.
• Compare the total with the breaker and cable rating.
• Allow for starting surges of any motors.
• Check the voltage drop over the run.