Energy Academy
Level 1/Electrical Science Refresher
Three-Phase Systems6 / 10

Power in Three-Phase Systems

How to calculate power, and the advantages over single-phase.

9 min read · Jacob Willis, Net Zero Lead · Last reviewed July 2026

One formula turns a clamp meter reading at a panel into the power a machine is actually drawing, and it is the most-used piece of arithmetic in practical electrical energy work. This lesson builds the three-phase power calculation from the pieces the course has already covered, then applies it the way an energy manager actually would: estimating a motor's real consumption from a current reading, without trusting the nameplate.

The formula

For a balanced three-phase load, real power is:

P = √3 × VL × IL × PF

where VL is the line-to-line voltage (400 V in the UK), IL is the current in each line, PF is the power factor from the earlier lesson, and √3 ≈ 1.732 is the same geometric factor that related 400 V to 230 V. Everything from the single-phase story carries over; the √3 simply accounts for three staggered phases sharing the work.

Worked example — from clamp reading to kilowatts
Given
  • A three-phase supply at 400 V
  • A clamp meter reads 100 A on each line
  • The load's power factor is 0.85
Find
The real power being drawn.

The energy manager's version: what does this motor cost to run?

Nameplates state what a motor can deliver, not what it is drawing right now. Real consumption depends on load, and the way to know it is to measure the current and work backwards. Two adjustments matter. First, a motor's nameplate rating is its mechanical output; the electrical input is higher by its inefficiency. Second, the running current tells you where it is operating.

Worked example — a 15 kW motor, checked properly
Given
  • Nameplate: 15 kW output, 400 V, power factor 0.85, efficiency 90%
  • You want its electrical draw at full load, and its annual cost at 4,000 running hours
  • Electricity £0.20/kWh
Find
The input power, the expected full-load current, and the annual cost.

That £13,360 figure is why motor efficiency and control dominate industrial electricity strategy: a machine costing four figures a year to feed justifies real attention, and the motors and drives course shows how variable-speed control turns part-load operation into large savings.

Balanced or not

The formula assumes the three line currents are equal. In practice, read all three. Similar readings let you use the average with confidence. Distinctly unequal readings on a three-phase machine suggest a supply or winding problem worth an electrician's attention, and on a mixed panel they map how single-phase load is distributed, which the previous lesson flagged as a capacity and loss issue.

Current is the measurement; power is the conclusion

Current is cheap to measure non-invasively, which makes it the workhorse observation of electrical surveys. But amps alone mislead: without voltage and power factor they overstate small heavily-reactive loads and understate large well-corrected ones. The habit that keeps estimates honest is always the same multiplication: √3 × 400 × amps × power factor, then hours, then price.

The final module of this course moves from equipment to the bill itself: what a site pays for besides kilowatt hours, and how demand and metering shape the total.

Sources and further reading