Complete conversion table for 5 kW to Amps at common voltages for both single-phase and three-phase systems.
Single Phase (240V, PF 0.9)
I = 5000 / (240 × 0.9)
Three Phase (415V L-L, PF 0.85)
I = 5000 / (√3 × 415 × 0.85)
Power Factor = 0.9 (typical for mixed loads)
| Voltage | Region | Current (PF 0.8) | Current (PF 0.9) | Current (PF 1.0) |
|---|---|---|---|---|
| 120V | USA | 52.08 A | 46.30 A | 41.67 A |
| 220V | Asia | 28.41 A | 25.25 A | 22.73 A |
| 240V | UK/AU | 26.04 A | 23.15 A | 20.83 A |
Formula: I = P / (V × PF) = 5000W / (V × PF)
Line-to-Line voltage, Power Factor = 0.85
| Voltage (L-L) | Application | Current (PF 0.8) | Current (PF 0.85) | Current (PF 0.9) |
|---|---|---|---|---|
| 415V | 3-Phase | 8.70 A | 8.18 A | 7.73 A |
| 440V | Industrial | 8.20 A | 7.72 A | 7.29 A |
| 480V | USA 3-Phase | 7.52 A | 7.08 A | 6.68 A |
Formula: I = P / (√3 × V × PF) = 5000W / (1.732 × V × PF)
To convert 5 kW to Amps at 240V with a power factor of 0.9:
I = P / (V × PF)
I = 5000 W / (240 V × 0.9)
I = 5000 / 216
I = 23.15 A
To convert 5 kW to Amps at 415V (L-L) with a power factor of 0.85:
I = P / (√3 × V × PF)
I = 5000 W / (1.732 × 415 V × 0.85)
I = 5000 / 610.96
I = 8.18 A
Page intent
Five kilowatts is not one electrical scenario. A resistive heater, an inverter AC output, and a generator nameplate can all be described as 5 kW while requiring different assumptions about PF, efficiency, duty, and whether the rating is kW or kVA.
A near-pure resistive heater normally uses PF approximately 1.0. At 230 V single-phase, 5,000 / 230 = 21.74 A; at 240 V single-phase, 5,000 / 240 = 20.83 A; at 400 V three-phase L-L, 5,000 / (√3 × 400) = 7.22 A per line.
Thermostatic cycling and staged heating elements mean the measured current may be zero, full stage current, or a combination rather than a constant 5 kW. Continuous-load and circuit-sizing rules vary by jurisdiction; the heater nameplate and installation instructions take priority.
For the inverter AC-output side, 230 V single-phase at PF 1.0 is 21.74 A, while PF 0.95 is 22.88 A. At 415 V three-phase L-L and PF 1.0, the result is 6.96 A per line.
Inverter current limits, export limits, grid-code settings, adjustable displacement PF, and conversion efficiency determine the real operating point. The DC array voltage and current require a separate calculation, and a 5 kW array does not produce 5 kW continuously.
For 5 kW real power at 240 V single-phase and PF 0.8, current is 26.04 A. A 5 kVA apparent-power rating at 240 V is 5,000 / 240 = 20.83 A, and at PF 0.8 that 5 kVA represents only 4 kW usable real power.
Starting loads, continuous rating, voltage regulation, and the generator nameplate limit the practical load. Do not silently convert a kVA label into kW.
It may require a high-current or dedicated circuit. Check the actual voltage, continuous-load rules, heater nameplate, terminal temperature, and service capacity rather than assuming a standard outlet is suitable.
The wording matters. This page’s inverter examples are AC-output examples; the DC array side has separate voltage, current, efficiency, and MPPT limits.
No. kW is real output and kVA is apparent capacity. At PF 0.8, 5 kVA corresponds to only 4 kW real power.
A motor has PF below 1.0 and efficiency below 100%, and it can draw much more current during starting.
The displayed result is approximate current per line under a balanced-load assumption. Imbalance and harmonics can make actual phase currents differ.
Custom kW to Amps conversion
Dedicated three-phase converter
Complete reference tables
