Watts to Amps Calculator — Current Drawn by a Load on DC, Single-Phase and Three-Phase Supplies

Background.

Watts tell you how much work a device does. Amps tell you how hard the wiring has to work to deliver it. Converting between the two takes one more piece of information — the voltage — and, on an AC supply, one more still: the power factor. This calculator handles all three supply types you are likely to meet, and runs the relation in whichever direction you need.

On a DC circuit the arithmetic is as simple as it looks. Power is the product of voltage and current, P = IV, so the current is I = P ÷ V. A 60 W bulb on a 12 V system draws 5 A; a 0.5 W standby load on USB's 5 V draws 100 mA. Nothing else enters into it, because a DC supply has no phase angle and every volt-ampere the supply delivers arrives as a watt.

AC is where people get caught out. The voltage and the current both alternate, and in anything containing a coil or a capacitor they do not peak at the same instant. The cosine of the angle between them is the power factor, and the real power is P = V × I × PF. Turn that around and the current is I = P ÷ (V × PF) — which means a load with a power factor of 0.8 draws a quarter more current than its wattage alone suggests. That extra current is real. It heats the conductors, it counts against the breaker, and it is the reason motors and drives need bigger supply cable than a heater of the same wattage. Purely resistive loads — heaters, kettles, toasters, incandescent lamps — have a power factor of 1, so for those the AC calculation collapses back to the DC one.

Three-phase supplies add a factor of √3. Each phase carries current at the phase voltage, but three-phase equipment is labelled with the voltage measured between two lines, which in a balanced system is √3 times the phase voltage. The result is I = P ÷ (√3 × V × PF), the form the US Department of Energy publishes in its motor load fact sheet as Equation 1, where it defines the voltage explicitly as "RMS voltage, mean line-to-line of 3 phases". That definition is worth reading twice. On a 480Y/277 V system the line-to-line voltage is 480, not 277; on a 400Y/230 V system it is 400, not 230. Entering the line-to-neutral figure instead inflates the calculated current by 73 percent, which is the single most common error in this calculation.

The calculator also reports the apparent power in volt-amperes alongside the current, because that is the number transformers, generators and UPS units are rated in. At a power factor of 1 the two are numerically identical; below 1 the volt-amperes exceed the watts, and the gap is exactly what a kVA rating exists to describe.

Two limits matter before you act on a number. First, everything here is a steady-state figure. A motor's starting inrush can be five or more times its running current for a fraction of a second, and switch-mode supplies draw a brief charging surge at switch-on; neither is captured by a wattage-to-current conversion, and both matter when sizing a generator or a protective device. Second, loads with distorted current waveforms — variable-frequency drives, LED drivers, computer power supplies — draw more current than the fundamental-frequency arithmetic predicts, because the harmonic content adds to the rms total without adding to the watts.

Finally, and this is on the page next to the result rather than buried in an FAQ: this is physics, not compliance. Conductor ampacity, overcurrent-device rating, continuous-load derating, ambient-temperature correction, conduit fill and voltage-drop limits are all governed by the electrical code adopted in your jurisdiction, and local amendments change the answer. Use this page to understand the load and to sanity-check a design, then have a licensed electrician size and sign off the circuit before anything is installed.

What is watts to amps calculator?

A watts-to-amps conversion is the rearrangement of the electrical power equation to solve for current. Power, measured in watts, is the rate at which a device converts electrical energy into something useful. Current, measured in amperes, is the rate at which charge flows through the conductors feeding it. They are linked by the voltage, and on AC by the power factor as well.

The three forms are: I = P ÷ V for direct current, I = P ÷ (V × PF) for single-phase alternating current, and I = P ÷ (√3 × V × PF) for a balanced three-phase supply where V is the line-to-line voltage. The √3 is not a correction factor invented for three-phase work — it is the ratio between line-to-line and line-to-neutral voltage in a balanced system, and it appears because three-phase equipment is labelled with the former.

The distinction between watts and volt-amperes matters more the further the power factor falls below 1. Watts measure the energy actually converted; volt-amperes measure the product of voltage and current the supply must deliver. Conductors heat according to current, not according to watts, so a 1000 W load at a power factor of 0.5 stresses the wiring exactly as hard as a 2000 W resistive load would. That is why supply equipment is rated in kVA and why the power factor is not an optional refinement on an AC calculation.

How to use this calculator.

1. 01Choose the supply type: single-phase AC for ordinary mains, three-phase AC for industrial and commercial supplies, or DC for battery, solar and USB systems.
2. 02Choose what you are solving for — amps from watts and volts, watts from amps and volts, or volts from watts and amps.
3. 03Enter the load's power in watts, taken from its nameplate or label rather than estimated.
4. 04Enter the supply voltage. On three phase this must be the line-to-line value: 208, 400, 415 or 480 V on common systems, never the 120 V or 230 V measured to neutral.
5. 05Enter the power factor as a decimal. Use 1 for heaters, kettles and incandescent lamps; take the nameplate figure for motors and drives. On DC the field is ignored.
6. 06Read the current as the headline, and check the apparent power in volt-amperes if you are sizing a generator, transformer or UPS.
7. 07Add headroom for motor starting inrush and for harmonic-rich loads, then have a licensed electrician confirm conductor and overcurrent sizing against the code adopted where you are.

The formula.

The direct-current case is the base everything else is built on. OpenStax University Physics Volume 2, section 9.5, gives it as Equation 9.12: "P = IV". Dividing both sides by V gives I = P ÷ V. Because a DC supply has no phase angle, there is nothing else to account for; the calculator forces the power factor to 1 in DC mode and ignores whatever is in the field.

On alternating current the same product holds, but between root-mean-square quantities and with the phase angle carried explicitly. Section 15.4 of the same book gives Equation 15.12 as "Pave = (1/2)I₀V₀cos(ϕ)" and Equation 15.13 as "Pave = (1/2)I₀V₀ = IrmsVrms = Irms²R", and defines the rms values as "Irms = (1/√2)I₀ and Vrms = (1/√2)V₀". The section names the factor directly: "In engineering applications, cos(ϕ) is known as the power factor, which is the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase." Putting those together, P = V_rms × I_rms × PF, so I = P ÷ (V × PF).

For a balanced three-phase supply the US Department of Energy publishes the relation directly. Fact sheet DOE/GO-10097-517, Determining Electric Motor Load and Efficiency, gives Equation 1 as "Pi = V x I x PF x √3 / 1000", with V defined as "RMS voltage, mean line-to-line of 3 phases", I as "RMS current, mean of 3 phases" and PF as "Power factor as a decimal". Rearranged for current, I = P ÷ (√3 × V × PF). The √3 arises because each phase delivers V_phase × I_line, giving 3 × V_phase × I_line in total, and substituting V_phase = V_line-to-line ÷ √3 turns the 3 into a √3.

This calculator carries √3 as 1.7320508075688772935… at forty significant digits and rounds only once, at the very end. The difference is measurable: the DOE fact sheet's own worked example measures a 40-hp motor at 469.7 V, 37 A and power factor 0.763 and prints 22.9 kW, whereas the exact √3 gives 22.9671681259 kW. Substituting the schoolbook √3 ≈ 1.73 reproduces the printed figure at 22.94 kW, so the gap is the source's 1997 hand-calculation rounding rather than a disagreement about the physics. Both values are recorded in the citation below, this page uses the exact one, and a test asserts 22.9671681259 while explicitly rejecting 22.94. Run the other way — the direction this page is actually used in — 22 967.1681259 W at 469.7 V and power factor 0.763 returns exactly 37 A, closing the loop on the source's own measurement.

The apparent power reported alongside the current is S = k × V × I, which is algebraically identical to P ÷ PF. It is the quantity transformers, generators and UPS units are rated in, and it exceeds the watts by exactly the reciprocal of the power factor. Every mode on the page is a closed-form rearrangement of the same single relation; nothing is solved numerically and nothing is approximated.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), University Physics Volume 2, §9.5 "Electrical Energy and Power". Primary source for the DC form. Equation 9.12 gives "P = IV" and Equation 9.13 gives "P = I²R = V²/R"; the section states "The energy unit on electric bills is the kilowatt-hour (kW·h), consistent with the relationship E = Pt" and "1 kW·h = 3.6 × 10⁶ J". Access: open. Retrieved 2026-07-29.
2. [2]OpenStax (Rice University), University Physics Volume 2, §15.4 "Power in an AC Circuit". Primary source for the AC form and the power factor. Equation 15.12 gives "Pave = (1/2)I₀V₀cos(ϕ)"; Equation 15.13 gives "Pave = (1/2)I₀V₀ = IrmsVrms = Irms²R"; the rms values are defined as "Irms = (1/√2)I₀ and Vrms = (1/√2)V₀"; and the section states "In engineering applications, cos(ϕ) is known as the power factor, which is the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase", plus "For a resistor, ϕ = 0". Access: open. Retrieved 2026-07-29.
3. [3]US Department of Energy, Office of Energy Efficiency and Renewable Energy, fact sheet DOE/GO-10097-517, "Determining Electric Motor Load and Efficiency", January 1997 (OSTI report numbers DOE/GO--10097-517; 319). Primary source for the three-phase form. Equation 1 reads "Pi = V x I x PF x √3 / 1000" with "V = RMS voltage, mean line-to-line of 3 phases", "I = RMS current, mean of 3 phases" and "PF = Power factor as a decimal". RECORDED CONFLICT: the same document's worked example prints "469.7 x 37 x 0.763 x √3 / 1000 = 22.9 kW", while the exact √3 gives 22.9671681259 kW; the printed figure is reproduced by √3 ≈ 1.73, which yields 22.94 kW. This page uses the exact √3, a test asserts 22.9671681259 and explicitly rejects 22.94, and the inverse returns the source's own 37 A exactly. Access: open. Retrieved and text-verified 2026-07-29.
4. [4]International Electrotechnical Commission, IEC 60050 International Electrotechnical Vocabulary, part 131, entry 131-11-41 "apparent power". Cited by IEV reference number for the standardised definition of apparent power and its unit, the volt-ampere. ACCESS: GATED — electropedia.org returned HTTP 403 to the retrieval used for this page, so nothing is quoted from the entry and no figure on this page rests on it. Declared as unopened rather than presented as verified. Retrieval attempted 2026-07-29.
5. [5]NIST, Special Publication 811, "Guide for the Use of the International System of Units (SI)", 2008 edition (DOI 10.6028/NIST.SP.811e2008). Used only for unit symbols and typography — W, V, A, VA, mA — not for any electrical relation. NIST's own landing page notes that SP 811 "has not yet been updated to reflect the changes in the SI that came into effect on May 20, 2019"; none of those changes affect the symbols used here. Access: open. Retrieved 2026-07-29.
6. [6]OpenStax (Rice University), University Physics Volume 2, §15.2 "Simple AC Circuits", consulted as a secondary check on the phase relationships behind the power factor — the resistor's zero phase angle, and the ±90° angles of the inductor and capacitor that make a real load's power factor less than 1. Access: open. Retrieved 2026-07-29.