Electric current (I) is the flow of electric charge in a circuit. You can calculate it using three equations, depending on which quantities you know:
A circuit has a voltage of 12 V and a resistance of 6 Ω. What is the current?
Solution: I = 12 / 6 = 2 A
A device uses 60 W of power at 12 V. What is the current?
Solution: I = 60 / 12 = 5 A
A resistor has a resistance of 16 Ω and dissipates 64 W. What is the current?
Solution: I = √(64 / 16) = √4 = 2 A
A 230V device with 2000W power and PF=0.8. What is the current?
Solution: I = 2000 / (230 × 0.8) = 10.87 A
A 400V three-phase system, 5000W load, PF=0.9. What is the current?
Solution: I = 5000 / (√3 × 400 × 0.9) = 8.02 A
All three tabs on this page answer the same question — how many amperes flow — but they use different formulas because the relationship between power, voltage and current is not the same in each system.
In a DC circuit, voltage and current are steady and in step, so power is simply P=V×I and current follows directly as I=P/V or I=V/R. There is no phase angle to account for and no distinction between real and apparent power.
In an AC circuit the current may not peak at the same instant as the voltage. Inductive loads such as motors and transformers make the current lag; capacitive loads make it lead. The power factor (PF) is the cosine of that phase angle, and it is the fraction of the apparent power (volt-amperes) that actually does useful work (watts).
This is why the single-phase formula is I=P/(V×PF) rather than I=P/V. A load with a poor power factor draws more current for the same real power, because part of that current is simply sloshing energy back and forth between the source and the load's magnetic or electric field without delivering any net work.
In a balanced three-phase system, total real power is P=√3×VLL×IL×PF, where VLL is the line-to-line voltage and IL the line current. The √3 is not a fudge factor: line-to-line voltage is √3 times the line-to-neutral voltage because the three phases are 120° apart, and combining the three phase powers while expressing everything in line quantities leaves that factor behind.
The practical upshot is that three-phase delivers the same power at a lower line current than single-phase at the same voltage — which means thinner conductors, smaller switchgear and lower I2R losses. That is the main reason industrial equipment is three-phase.
| System | Current from power | Current from Ohm's law |
|---|---|---|
| DC | I = P / V | I = V / R |
| AC single-phase | I = P / (V × PF) | I = V / Z |
| AC three-phase | I = P / (√3 × VLL × PF) | I = Vphase / Z |
Power factor is easiest to understand as a current penalty. Take a 5 kW load on a 400 V three-phase supply and vary nothing but the power factor:
| Power factor | Line current | Relative I2R loss |
|---|---|---|
| 1.00 (purely resistive) | 7.22 A | 1.00× |
| 0.95 | 7.60 A | 1.11× |
| 0.80 | 9.02 A | 1.56× |
| 0.60 | 12.03 A | 2.78× |
The real power delivered is identical in every row — 5 kW — yet the cable at PF 0.60 carries two-thirds more current than at unity and dissipates roughly 2.5 times the copper loss of the PF 0.95 case. That extra current must still be carried by the conductors, the breaker and the transformer, which is why utilities penalise poor power factor and why capacitor banks are fitted to correct it.
One detail worth stating plainly: AC currents from these formulas are RMS values, the figure that produces the same heating as an equal DC current. For a sine wave the peak is √2 times higher, so a 10 A RMS current actually peaks at about 14.14 A. Semiconductor ratings, surge ratings and insulation stress all care about that peak, even though conductor sizing and meter readings are based on the RMS value.
For DC, I = V/R (Ohm's law) or I = P/V. For AC single-phase, I = P/(V×PF). For balanced AC three-phase, I = P/(√3×VLL×PF), where VLL is the line-to-line voltage and PF the power factor.
Because the three phases are 120° apart, the line-to-line voltage is √3 times the line-to-neutral voltage. When the total power of all three phases is written in terms of line voltage and line current, that factor remains, giving P = √3×VLL×IL×PF.
It raises it. Since I = P/(V×PF), halving the power factor doubles the current for the same real power. A 5 kW load at 400 V three-phase draws 7.22 A at unity power factor but 12.03 A at PF 0.60 — about 2.5 times the copper loss for identical useful output.
RMS. The RMS value is the equivalent DC current that would produce the same heating, and it is what meters read and what conductor sizing is based on. For a sine wave the peak is √2 times the RMS value, so 10 A RMS peaks at about 14.14 A — the figure that matters for semiconductor and insulation ratings.
No. A motor nameplate states mechanical shaft output, but the current formula needs electrical input power. Divide the rated output by the efficiency first. For a 7.5 kW motor at 90% efficiency the electrical input is about 8,333 W, and using 7,500 W instead would under-estimate the current by roughly 10%.
Use I = √(P/R), which comes from combining P = I2R with Ohm's law. For a 16 Ω resistor dissipating 64 W, I = √(64/16) = √4 = 2 A.
For the same real power and voltage, the √3 factor means the three-phase line current is lower by about 42% compared with the single-phase equivalent. Lower current allows smaller conductors, smaller switchgear and lower I2R losses, which is why industrial loads are almost always three-phase.
Use 1.0 only for genuinely resistive loads such as heaters and incandescent lamps. Induction motors typically run between 0.80 and 0.90 at full load and considerably worse when lightly loaded. Modern electronic supplies with active power-factor correction usually reach 0.95 or better. When sizing protection, assume the worst realistic value rather than the best.
Current. Conductor cross-section is chosen from the current it must carry without exceeding its temperature limit, together with the acceptable voltage drop over the run length. Voltage governs the insulation rating, not the copper area — which is why a poor power factor, by raising current at unchanged voltage, directly forces a larger cable.
Common reasons are the real supply voltage differing from nominal, the actual power factor differing from the assumed one, non-linear loads drawing harmonic current that raises true RMS above the fundamental, and nameplate ratings describing full load while the equipment runs partly loaded. A true-RMS meter is essential when the load is electronic rather than resistive.
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