kW to Amps Calculator
The kW to amps conversion answers the fundamental question every electrical engineer faces when designing power systems: how much current will this equipment draw from the supply?...
Formula
Source: NEC Article 430, IEEE 141 (Red Book), NEMA MG 1 | Last reviewed: July 26, 2026
Examples
75 kW
= 106 Amps
- voltage = 480
- pf = 0.85
- phase_factor = 1.732
75 kW 480V 3-phase motor at 0.85 PF draws 106 A
7.5 kW
= 24.5 Amps
- voltage = 208
- pf = 0.85
- phase_factor = 1.732
7.5 kW 208V 3-phase motor draws 24.5 A
2 kW
= 8.33 Amps
- voltage = 240
- pf = 1
- phase_factor = 1
2 kW 240V single-phase heater draws 8.33 A
100 kW
= 16.34 Amps
- voltage = 4160
- pf = 0.85
- phase_factor = 1.732
100 kW 4160V medium-voltage motor draws 16.3 A
Quick Reference Table
| kW | HP (approx) | Amps | NEC 430.250 FLA |
|---|---|---|---|
| 0.75 | 1 | 1.06 | 1.4 A |
| 2.2 | 3 | 3.1 | 4.8 A |
| 4 | 5 | 5.7 | 7.6 A |
| 7.5 | 10 | 10.6 | 14 A |
| 15 | 20 | 21.2 | 27 A |
| 22 | 30 | 31.1 | 40 A |
| 37 | 50 | 52.3 | 65 A |
| 55 | 75 | 77.8 | 96 A |
| 75 | 100 | 106 | 124 A |
| 90 | 125 | 127.2 | 156 A |
| 132 | 175 | 186.6 | 205 A |
| 200 | 250 | 282.8 | 302 A |
| 315 | 400 | 445.4 | 477 A |
| kW | Amps | Typical Application |
|---|---|---|
| 1 | 4.17 | Small space heater (150 sq ft) |
| 2 | 8.33 | Residential water heater element |
| 3 | 12.5 | Electric dryer heating element |
| 4.5 | 18.75 | Residential cooktop/oven |
| 5 | 20.8 | Large electric furnace element |
| 7.5 | 31.25 | Commercial water heater |
| 10 | 41.7 | Industrial process heater, small |
| 15 | 62.5 | Duct heater, commercial AHU |
Popular Conversions
Quick answers for the most-searched kW to Amps values.
7.5 kW to amps 3 phase
7.5 kW = 10.6 Amps
10 HP motor equivalent. 7.5 kW at 480V 3-phase draws 10.6 A (nameplate calc) or 14 A per NEC Table 430.250. One of the most common small industrial motor sizes, used for conveyors, pumps, and fans.
15 kW to amps 3 phase
15 kW = 21.2 Amps
20 HP motor equivalent. 15 kW at 480V draws 21.2 A calculated, 27 A per NEC table. Common for HVAC compressors, hydraulic power units, and machine tool spindles.
75 kW to amps 480V
75 kW = 106 Amps
100 HP motor equivalent -- the benchmark industrial motor size. 75 kW at 480V 3-phase draws 106 A calculated, 124 A per NEC Table 430.250. Used for large pumps, air compressors, and process machinery.
200 kW to amps 480V 3 phase
200 kW = 282.8 Amps
Approximately 250 HP. 200 kW at 480V draws 283 A calculated, 302 A per NEC. Common for large chillers, 500+ ton cooling tower fans, and industrial air compressors. Requires 400 A frame breaker and parallel conductors or large single conductors.
1 kW to amps 240V single phase
1 kW = 4.17 Amps
The simplest residential reference: 1 kW at 240V single-phase = 4.17 A. A typical 4.5 kW electric water heater draws 18.75 A, fitting on a 30 A 2-pole breaker. Used for sizing branch circuits for fixed appliances per NEC 422.
55 kW to amps 3 phase
55 kW = 77.8 Amps
75 HP motor -- the most common medium industrial motor. 55 kW at 480V draws 77.8 A calculated, 96 A per NEC Table 430.250. Frequently encountered in water treatment plants, manufacturing conveyors, and building chilled water pumps.
Where is this used?
(1) Circuit breaker and fuse sizing per NEC 430.52: motor branch-circuit short-circuit and ground-fault protection uses the calculated FLA multiplied by a code-specified factor.
For an inverse-time breaker on a 75 kW 480V motor drawing 106 A, the maximum breaker is 106 x 250% = 265 A, rounded to the next standard size of 300 A.
A time-delay fuse would be sized at 106 x 175% = 186 A, rounded to 200 A.
(2) Conductor sizing per NEC 310.16: minimum conductor ampacity for a continuous motor load is 125% of FLA.
For 106 A, that means 132.5 A minimum, requiring #1 AWG copper (130 A at 75C is slightly low, so #1/0 AWG at 150 A may be needed after temperature and conduit fill derating).
(3) Panel load calculations: a 480V panelboard serving (1) 75 kW pump at 106 A, (2) 37 kW fan at 52.3 A, and (3) 15 kW compressor at 21.2 A totals 179.5 A.
A 225 A main breaker panelboard provides adequate headroom.
(4) Motor starter and contactor selection: NEMA contactors are rated by amp capacity -- NEMA Size 4 (135 A at 480V) covers the 106 A motor, while NEMA Size 3 (90 A) would be undersized.
IEC contactors use AC-3 utilization categories with amp ratings matched to the calculated FLA.
(5) Current transformer selection for energy metering: CTs are typically selected so the meter operates at 50-75% of CT primary at normal load.
For a 106 A motor, a 200:5 A CT provides good resolution and accommodates 150% overload without saturation.
(6) Voltage drop verification: for a 300 ft feeder to a 75 kW motor drawing 106 A with #1/0 AWG copper (0.122 ohm/1000 ft), VD = sqrt(3) x 106 x 0.122 x 300/1000 = 6.7 V, or 1.4% of 480 V -- well within the 3% NEC recommendation.
(7) Transformer secondary protection: a 500 kVA 480V transformer's secondary full-load amps = 500,000/(480 x 1.732) = 601 A.
The secondary main breaker is typically sized at 125% = 751 A, using a standard 800 A frame.
(8) Generator loading: a 500 kW standby generator at 480V 3-phase 0.8 PF delivers 752 A.
Adding a new 55 kW load drawing 77.8 A raises total from 602 A to 679.8 A -- still within the gen-set alternator rating but requiring verification of the engine kW limit and step-load transient capability.
Real-World Usage Scenarios
Motor branch circuit design for a cooling tower fan
A design-build electrical contractor is sizing the branch circuit for a new 55 kW (75 HP) cooling tower fan motor at a commercial building. The motor operates at 480V 3-phase with 0.88 PF per the nameplate. Calculated FLA: (55,000) / (480 x 0.88 x 1.732) = 75.1 A. Per NEC 430.22, conductors are sized at 125% of FLA: 75.1 x 1.25 = 93.9 A, requiring #3 AWG copper (100 A ampacity at 75C). The branch-circuit breaker per NEC 430.52 (inverse-time): 75.1 x 250% = 187.8 A, next standard size 200 A 3-pole. The electrician uses this calculation to pull the correct conductors and install the right breaker on the first visit, avoiding a costly rework order when the electrical inspector checks the installation against the approved one-line diagram. The inspector confirms the #3 AWG conductors and 200 A breaker match the NEC tables and passes the installation without corrections.
Generator loading verification for a process pump addition
A facility manager is adding a new 37 kW process pump to an existing 500 kW standby generator at a food processing plant. The generator load schedule currently totals 320 kW of connected load with an average plant power factor of 0.83. The new pump at 480V 3-phase 0.85 PF draws (37,000) / (480 x 0.85 x 1.732) = 52.3 A. Converting to kVA for alternator loading: 37 / 0.85 = 43.5 kVA. Existing load: 320 kW / 0.83 PF = 385.5 kVA. New total kVA = 385.5 + 43.5 = 429 kVA. The 500 kW gen-set at 0.8 PF has an alternator rating of 625 kVA. The new pump pushes the alternator loading to 429/625 = 68.6%, well within limits. However, the plant engineer notes that the 37 kW pump's across-the-line starting current of 6x FLA (314 A or 261 kVA starting surge) must be evaluated for voltage dip -- a separate generator transient analysis confirms the 500 kW set can start the pump while carrying existing load with less than 15% voltage dip, meeting the NEMA MG 1 requirement for motor starting.
Data center power distribution unit (PDU) capacity planning
A colocation data center is provisioning a new 200 kW IT equipment row served by a 480V 3-phase Power Distribution Unit (PDU). The IT load is predominantly server power supplies with near-unity power factor (PF = 0.98). The PDU's output breaker must be sized for the maximum continuous load: calculated amps = (200,000) / (480 x 0.98 x 1.732) = 245.3 A. Per NEC 210.20 for continuous loads (data center IT loads operate 24/7), the overcurrent device must be rated at 125% of continuous load: 245.3 x 1.25 = 306.6 A. The next standard breaker size is 350 A frame. The PDU's internal bus bar is rated 400 A, providing headroom for future capacity upgrades. The facility engineer also calculates the upstream UPS output: at the UPS inverter's rated 0.9 PF, the 200 kW IT load appears as 200 / 0.9 = 222.2 kVA, drawing 267.4 A at 480V. The dual-corded server configuration means each PDU carries half the load under normal conditions (122.7 A per PDU), but each PDU must be sized for the full 245.3 A to handle failover when one feed is lost. The kW-to-amps calculation at each stage -- PDU input, PDU output, rack PDU, and server power supply inlet -- ensures the entire power chain from utility to chip is sized correctly, with no single point of undersizing that would limit capacity.
Common Mistakes to Avoid
Applying the single-phase formula to a three-phase circuit (or vice versa)
Using I = kW/(V x PF) for a three-phase load instead of I = kW/(V x PF x 1.732) overestimates the current by 73.2%, leading to oversized (wasteful) conductors but a safe installation. The dangerous error is the reverse: applying the three-phase formula to a single-phase load underestimates current by 42% (factor of 0.577). A 15 kW 240V single-phase heater correctly draws 15,000/(240 x 1.0) = 62.5 A, but the erroneous three-phase formula gives 15,000/(240 x 1.0 x 1.732) = 36.1 A. Installing #8 AWG conductors (40 A ampacity at 60C) based on the incorrect 36.1 A would cause the conductors to overheat under the actual 62.5 A load, risking insulation damage and fire. Always verify the system phase configuration from the panel schedule or one-line diagram before applying the formula.
Omitting power factor from the calculation
Assuming PF = 1.0 for inductive loads like motors significantly understates the current. A 75 kW motor at 480V with PF = 1.0 would theoretically draw 75,000/(480 x 1.0 x 1.732) = 90.2 A. At the actual PF of 0.85, the current is 106.0 A -- a 17.5% increase. While 90.2 A conductors (#3 AWG at 100 A) might survive at 106 A, they would be undersized per NEC 430.22 which requires 125% of the actual (106 A) FLA = 132.5 A minimum. The breaker sized at 250% of 90.2 A = 225 A would not provide adequate protection for the 106 A motor's locked-rotor current. For VFD-fed motors, the drive input PF is typically 0.95 or higher due to the DC bus capacitors, and using the motor nameplate PF (0.85) for the VFD input would overestimate the line-side current -- in this case, the kW-to-amps for the VFD input should use the drive manufacturer's specified input PF, not the motor PF.
Neglecting voltage variation in kVA-limited or kW-limited generators
When adding loads to a standby generator, engineers sometimes look only at the kW rating and ignore the kVA rating of the alternator. A 500 kW generator at 0.8 PF has an alternator rated 625 kVA. Adding a new 100 kW load with PF = 0.7 draws 100/0.7 = 142.9 kVA from the alternator, not just 100 kW. At 480V 3-phase, that is 142,900/(480 x 1.732) = 171.9 A. The generator's kW engine limit (500 kW) might still have headroom, but the alternator's kVA limit (625 kVA) could be exceeded if the plant's average PF is low. This is a common cause of generator breaker trips during monthly load tests -- the facility added kW within the engine limit but exceeded the alternator's amp rating because low-PF loads (old motors, uncorrected fluorescent lighting) push the kVA demand well above the kW demand.
Industry Standards Referenced
Frequently Asked Questions
How do you convert kW to amps?
For 3-phase: Amps = (kW x 1000) / (V x PF x 1.732). For single-phase/DC: Amps = (kW x 1000) / (V x PF). Example: 75 kW, 480V 3-phase, 0.85 PF = (75 x 1000) / (480 x 0.85 x 1.732) = 106 A. Always verify whether your system is single-phase or three-phase -- the result differs by a factor of sqrt(3) (1.732). For DC systems, set PF = 1.0 and phase factor = 1.
How many amps does a 7.5 kW motor draw?
At 480V 3-phase 0.85 PF: (7.5 x 1000) / (480 x 0.85 x 1.732) = 10.6 A. At 208V 3-phase: 24.5 A. At 240V single-phase: 36.8 A. The current depends strongly on voltage -- always check the motor nameplate for the actual full-load amps (FLA), which accounts for motor efficiency and may differ slightly from the theoretical kW-to-amps calculation. NEC Table 430.250 lists 14 A for a 10 HP (7.5 kW) motor at 460V, reflecting conservative efficiency assumptions.
What is the sqrt(3) (1.732) factor for 3-phase?
In a balanced three-phase system, the total power P = sqrt(3) x V_L-L x I_L x PF, where V_L-L is line-to-line voltage and I_L is line current. Solving for current: I = P / (sqrt(3) x V x PF). The sqrt(3) = 1.732 is the geometric consequence of the 120-degree phase displacement between the three sinusoidal line voltages. Using 1 instead of 1.732 (i.e., the single-phase formula for a three-phase system) overestimates the current by 73.2%, leading to oversized (but safe) conductors. The reverse error -- using the three-phase formula for a single-phase load -- underestimates current by 42% (1/1.732 = 0.577), which is dangerous and would result in severely undersized breakers and conductors.
Can I use this calculator on a DC system?
Yes. For DC circuits, set phase_factor = 1 (single-phase) and PF = 1.0. For a purely resistive DC load, Amps = (kW x 1000) / Voltage. Example: a 2 kW 48V DC load draws (2 x 1000) / 48 = 41.7 A. DC systems have no power factor (no reactive component) and no phase factor -- the calculation is simply Watts / Volts = Amps. This applies to battery storage systems, solar PV DC circuits, DC motor drives, and telecom 48V DC power plants.
Why is the NEC table FLA different from the calculated amps?
NEC Table 430.250 lists full-load currents for three-phase AC motors that include conservative assumptions for motor efficiency and power factor. The NEC values are generally 10-20% higher than the theoretical kW-to-amps calculation, providing a safety margin for conductor and overcurrent protection sizing that accounts for manufacturing tolerances and future motor replacement. Always use the motor nameplate FLA for overload protection (NEC 430.32) and the NEC table FLA for conductor and short-circuit protection sizing (NEC 430.22 and 430.52). The kW-to-amps calculator gives you the theoretical current -- the starting point for understanding the load physics before applying code-mandated safety factors.
What happens to amps when voltage drops below nominal?
For a constant-power load like a motor driving a pump or fan, current increases as voltage decreases: I = P / (V x PF x sqrt(3)). If a 480V nominal system sag to 440V (8.3% drop), a 75 kW motor's current rises from 106 A to (75,000) / (440 x 0.85 x 1.732) = 115.6 A -- a 9% increase. This higher current produces additional I^2R heating in both the motor windings and the supply conductors. For induction motors, the relationship is more complex at severe undervoltage because slip increases and efficiency degrades, causing current to rise faster than the inverse-voltage relationship predicts. NEMA MG 1 specifies that motors shall operate successfully at +/-10% of rated voltage, but sustained operation at -10% voltage may reduce motor life by 50% due to increased winding temperature. This is why voltage drop calculations from the service to the motor terminals are critical -- a motor rated for 40C ambient at nominal voltage may overheat at 440V.
Reviewed for accuracy
Reviewed against NEC 2023 Article 430 motor circuit requirements and IEEE 141 power distribution standards · Last reviewed: July 26, 2026
All calculations are for reference only. Always verify with manufacturer data and a qualified engineer for critical applications. Learn about our editorial process.