Challenge

Challenge: The Current That Looks Too High

An engineer calculates a three-phase line current as 37.5 A for a 12 kW load. Find the error before opening the answer.

Electrical & Power · Last reviewed October 15, 2026

An engineer estimates the line current of a balanced three-phase load with:

I = 12,000 ÷ (400 × 0.80) = 37.5 A

Three-phase electrical real input power: 12 kW.
Line-to-line voltage: 400 V.
Power factor: 0.80.

Your task: find the error, calculate the correct line current, and say whether motor efficiency belongs in this calculation.

Show the answer (try it first)

For a balanced three-phase load:

P = √3 × V(line-to-line) × I(line) × PF

I = 12,000 ÷ (√3 × 400 × 0.80) = 21.65 A

The original calculation left out √3, which made the current too high.

Efficiency is not needed, because the 12 kW given is electrical input power. If the problem had given the motor's mechanical output power, efficiency would be needed first to find the electrical input.

Check your reasoning: did you fix the power definition and the voltage reference before choosing a formula?

More practice

Variation 1: a balanced 30 kW load on 480 V with a power factor of 0.90.

I = 30,000 ÷ (√3 × 480 × 0.90) = 40.1 A

Variation 2: a motor delivers 15 kW of mechanical output at 90% efficiency on a 400 V supply at a power factor of 0.85. Efficiency is needed here because the 15 kW is output power.

Electrical input = 15 ÷ 0.90 = 16.67 kW.
I = 16,670 ÷ (√3 × 400 × 0.85) = 28.3 A.

Quick shortcut

At unity power factor, a three-phase load draws about 1.2 A per kW at 480 V, 1.44 A per kW at 400 V and 2.78 A per kW at 208 V. Divide by the power factor to correct it. This is a fast way to check that an answer is the right size.

Related tools and labs

Calculators, labs and simulators are part of the All-Access Pass.

Questions

When is √3 not used?

For single-phase, and when working from the phase voltage of a balanced three-phase load, where P = 3 × V(phase) × I(phase) × PF.

How do I convert kVA to amps for a three-phase load?

Divide kVA × 1000 by (√3 × line-to-line voltage). For 100 kVA at 480 V, that is 100,000 ÷ (1.732 × 480) = 120 A.

What is the difference between kW and kVA?

kW is real power that does work. kVA is apparent power, the product of voltage and current. kW = kVA × power factor.

Does it matter whether the load is wye or delta connected?

Not for the line current when you use line-to-line voltage and the total power. The wye or delta connection changes the current in each winding, not the line current formula.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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