30 kVA to Amps

Find current directly from apparent power and line voltage for single and three-phase systems.

Operating a 30 kVA industrial load requires careful amp calculation. It draws 125 Amps on standard 240V single-phase lines. For commercial three-phase power, expect 83.27 Amps at 208V and 36.08 Amps at 480V.

Quick answer

30 kVA is 125.0 A at 240 V single-phase AC. At other common voltages — 208 V: 144.2 A. Use the calculator below for any voltage or circuit type.

For three-phase systems, the current is lower because the power is distributed across three alternating current waves. This relies on the

.732$ ($\sqrt{3}$) mathematical constant factor necessary for balanced three-phase loads: Amps = (kVA × 1000) ÷ (Volts × 1.732).

Phase & VoltageMax Current (Amps)
Single-phase 120V250 A
Single-phase 240V125 A
Three-phase 208V83.27 A
Three-phase 480V36.08 A
RESULT — CURRENT
Formula: A = (kVA × 1,000) ÷ V

How do you convert kVA to amps?

For single-phase, amps = (kVA × 1000) ÷ V. A 10 kVA single-phase supply at 240 V gives (10 × 1000) ÷ 240 = 41.7 A. For three-phase, amps = (kVA × 1000) ÷ (√3 × V): a 10 kVA, 400 V three-phase supply gives 10000 ÷ (1.732 × 400) = 14.4 A.

kVA to Amps Formula

Single-phase
A = (kVA × 1,000) ÷ V

Multiply kVA by 1,000 to convert to VA (volt-amps), then divide by the line voltage. This gives the current in amps for a single-phase system. Example: 5 kVA at 120 V to (5,000) ÷ 120 = 41.67 A.

Three-phase
A = (kVA × 1,000) ÷ (V × √3)

For three-phase, divide by voltage and the square root of 3 (≈ 1.7321). The √3 factor arises from the 120° phase angle between conductors. Use line-to-line voltage (e.g. 400 V, 480 V). Example: 5 kVA at 400 V to 5,000 ÷ (400 × 1.7321) = 7.22 A.

kVA to Amps Reference Table — Single-phase

kVA Amps @ 120 V Amps @ 240 V
1 kVA 8.3333 A 4.1667 A
2 kVA 16.6667 A 8.3333 A
5 kVA 41.6667 A 20.8333 A
10 kVA 83.3333 A 41.6667 A
20 kVA 166.6667 A 83.3333 A
50 kVA 416.6667 A 208.3333 A

Single-phase vs Three-phase

Single-phase (1Ø)

One live conductor plus a neutral. Power is delivered as a single sinusoidal wave. Common in homes, small offices, and retail units up to around 10 kVA. Higher current per conductor for the same power.

A = (kVA × 1,000) ÷ V
Three-phase (3Ø)

Three live conductors, each offset 120° apart. Delivers three times the power at the same current, making it ideal for motors, data centres, and industrial sites. Far more efficient for loads above 10 kVA.

A = (kVA × 1,000) ÷ (V × √3)

In practice: a 30 kVA workshop supply

Thirty kVA at 208 V three-phase works out to 83.3 A per phase, enough for a workshop running several machine tools alongside a compressor. Because motor loads pull a lagging power factor, real available power is often nearer 24 kW than 30 kW. Power factor correction capacitors frequently pay for themselves at this scale, both in recovered capacity and in avoided utility penalties.

Frequently Asked Questions

How do I convert kVA to amps?

Multiply kVA by 1,000 and divide by the voltage. For single-phase: A = (kVA × 1,000) ÷ V. For three-phase: A = (kVA × 1,000) ÷ (V × √3). For example, 10 kVA at 240 V single-phase gives (10 × 1,000) ÷ 240 = 41.67 A.

What is the formula to convert kVA to amps?
  • Single-phase: A = (kVA × 1,000) ÷ V
  • Three-phase: A = (kVA × 1,000) ÷ (V × √3), where √3 ≈ 1.7321

Note that kVA is apparent power — it does not account for power factor, unlike kW (real power). Use kVA when sizing cables, fuses, and switchgear.

How many amps is 5 kVA at 240V?
  • Single-phase at 240 V: (5 × 1,000) ÷ 240 = 20.83 A
  • Three-phase at 240 V (L-L): (5 × 1,000) ÷ (240 × 1.7321) = 12.03 A

Three-phase draws less current per conductor because load is shared across three wires.

What is the difference between single-phase and three-phase?

Single-phase power uses one live conductor and a neutral, delivering power as a single sinusoidal wave. It is standard in homes and small offices. Three-phase power uses three live conductors, each carrying a wave offset by 120°. For the same kVA, three-phase requires significantly less current per conductor, making it more efficient for large motors, industrial equipment, and commercial buildings.