How Many Amps Is 8.48 kW at 240V?

8.48 kW at 240V draws about 41.57 amps on an AC single-phase circuit at PF 0.85, typical for residential water heaters, dryers, ranges, EV chargers, and HVAC equipment. Actual current varies with equipment power factor and duty cycle.

8.48 kW at 240V, AC single-phase (PF 0.85)
41.57 Amps
8.48 kilowatts at 240V on AC single-phase ≈ 41.57 amps
DC (ideal baseline)35.33 A
41.57

Formulas

DC: kW to Amps

I(A) = 1000 × P(kW) ÷ V(V)

1000 × 8.48 ÷ 240 = 8,480 ÷ 240 = 35.33 A

AC Single Phase (PF = 0.85)

I(A) = 1000 × P(kW) ÷ (PF × V(V))

8,480 ÷ (0.85 × 240) = 8,480 ÷ 204 = 41.57 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 41.57A on AC single-phase at 240V, the load sits in the bracket between a 45A standard size (non-continuous) and the next size up that covers a continuous load under 210.19(A) (around 60A). 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

8.48 kW costs $1.44/hour at $0.17/kWh (rates last reviewed April 2026). See breakdown.

Power Factor Reference (AC single-phase)

How the line current for 8.48 kW at 240V changes with load power factor, on the same AC single-phase circuit basis the rest of the page uses. DC has no power factor; PF 1.0 represents resistive AC loads.

Load TypePF8.48 kW at 240V (AC single-phase)
Resistive (heaters, incandescent)135.33 A
Fluorescent lamps0.9537.19 A
LED lighting0.939.26 A
Synchronous motors0.939.26 A
Typical mixed loads0.8541.57 A
Induction motors (full load)0.844.17 A
Computers (without PFC)0.6554.36 A
Induction motors (no load)0.35100.95 A

AC Conversion Comparison

On DC, 8.48kW at 240V draws 35.33A. AC single-phase at PF 0.85 pulls 41.57A because reactive current is added on top of the real power.

Circuit TypeFormulaResult
DC8,480 ÷ 24035.33 A
AC Single Phase (PF 0.85)8,480 ÷ (0.85 × 240)41.57 A

Other kW Values at 240V

kWAC 1-Phase PF 0.85DC Amps PF 1.0 baseline
0.75 kW3.68 A3.13 A
1 kW4.9 A4.17 A
1.5 kW7.35 A6.25 A
2 kW9.8 A8.33 A
2.5 kW12.25 A10.42 A
3 kW14.71 A12.5 A
3.5 kW17.16 A14.58 A
4 kW19.61 A16.67 A
5 kW24.51 A20.83 A
6 kW29.41 A25 A
7.5 kW36.76 A31.25 A
8 kW39.22 A33.33 A
10 kW49.02 A41.67 A
12 kW58.82 A50 A
15 kW73.53 A62.5 A

Frequently Asked Questions

8.48 kW at 240V draws about 41.57 amps on an AC single-phase circuit at PF 0.85. Alternate cases at the same voltage: 35.33A on DC.
8.48 kW equals 8,480 watts. Multiply kilowatts by 1000.
DC: Amps = (kW × 1000) ÷ Volts. AC single-phase: Amps = (kW × 1000) ÷ (Volts × PF). AC three-phase: Amps = (kW × 1000) ÷ (VoltsL-L × √3 × PF).
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.
On AC single-phase, current scales inversely with power factor. At PF 1.0 (pure resistive, like a heater), 8.48 kW at 240V draws 35.33A. At PF 0.80 (typical induction motor), the same real power draws 44.17A. The extra current is reactive and does no real work, but still flows through the wire and the breaker.
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.