How Many Amps Is 5.08 kW at 24V?

5.08 kilowatts at 24V works out to roughly 211.63 amps on DC at PF 0.85. That is typical for solar arrays, battery banks, and DC industrial equipment. See the DC and alternate-phase numbers below for other circuit types.

5.08 kW at 24V, DC (PF 0.85)
211.63 Amps
5.08 kilowatts at 24V on DC ≈ 211.63 amps
AC Single Phase (PF 0.85)248.97 A
211.63

Formulas

DC: kW to Amps

I(A) = 1000 × P(kW) ÷ V(V)

1000 × 5.08 ÷ 24 = 5,079 ÷ 24 = 211.63 A

AC Single Phase (PF = 0.85)

I(A) = 1000 × P(kW) ÷ (PF × V(V))

5,079 ÷ (0.85 × 24) = 5,079 ÷ 20.4 = 248.97 A

Equipment & Circuit Sizing

Breaker Sizing

Breaker ratings are in amps, not watts, so the real install answer depends on the equipment nameplate FLA, whether the load is continuous (NEC 210.19(A) sizes the conductor and OCP at 125% of a continuous load, equivalently 80% of breaker rating), conductor ampacity and temperature rating, ambient and bundling derates, and any motor or HVAC provisions (NEC 430 / 440). At roughly 211.63A on DC at 24V, the load sits in the bracket between a 225A standard size (non-continuous) and the next size up that covers a continuous load under 210.19(A) (around 300A). The actual install pick depends on whether the load is continuous and the factors above; a conversion page can't pick a single "right" breaker from the amp draw alone.

Energy Cost

5.08 kW costs $0.86/hour at $0.17/kWh (rates last reviewed April 2026). See breakdown.

Power Factor Reference (DC)

How the line current for 5.08 kW at 24V changes with load power factor, on the same DC circuit basis the rest of the page uses. DC has no power factor; PF 1.0 represents resistive AC loads.

Load TypePF5.08 kW at 24V (DC)
Resistive (heaters, incandescent)1211.63 A
Fluorescent lamps0.95211.63 A
LED lighting0.9211.63 A
Synchronous motors0.9211.63 A
Typical mixed loads0.85211.63 A
Induction motors (full load)0.8211.63 A
Computers (without PFC)0.65211.63 A
Induction motors (no load)0.35211.63 A

AC Conversion Comparison

On DC, 5.08kW at 24V draws 211.63A. AC single-phase at PF 0.85 pulls 248.97A because reactive current is added on top of the real power.

Circuit TypeFormulaResult
DC5,079 ÷ 24211.63 A
AC Single Phase (PF 0.85)5,079 ÷ (0.85 × 24)248.97 A

Other kW Values at 24V

kWDC AmpsAC 1-Phase PF 0.85
0.5 kW20.83 A24.51 A
0.75 kW31.25 A36.76 A
1 kW41.67 A49.02 A
1.5 kW62.5 A73.53 A
2 kW83.33 A98.04 A
2.5 kW104.17 A122.55 A
3 kW125 A147.06 A
3.5 kW145.83 A171.57 A
4 kW166.67 A196.08 A
5 kW208.33 A245.1 A
6 kW250 A294.12 A
7.5 kW312.5 A367.65 A
8 kW333.33 A392.16 A
10 kW416.67 A490.2 A
12 kW500 A588.24 A

Frequently Asked Questions

5.08 kW at 24V draws about 211.63 amps on DC. Alternate cases at the same voltage: 248.97A on AC single-phase.
Industrial equipment operates at higher power levels. 5.08 kW is easier to express than 5,079W. The math is identical, just scaled by 1000.
On AC single-phase, current scales inversely with power factor. At PF 1.0 (pure resistive, like a heater), 5.08 kW at 24V draws 211.63A. At PF 0.80 (typical induction motor), the same real power draws 264.53A. The extra current is reactive and does no real work, but still flows through the wire and the breaker.
This is a sizing question, not a conversion question, and there is no single correct answer from a page like this. Breaker selection depends on the equipment nameplate FLA, whether the load is continuous (NEC 210.19(A) applies the 125% continuous-load rule), the conductor ampacity and temperature rating, any NEC 430/440 motor or HVAC provisions, and local code interpretation. Use the nameplate and a licensed electrician for the real install value; use this page only for the current-draw estimate that feeds into that process.
5.08 kW equals 5,079 watts. Multiply kilowatts by 1000.
This calculator provides estimates for reference purposes only. Always consult a licensed electrician and verify compliance with the National Electrical Code (NEC) and local electrical codes before performing any electrical work.