Watts to Amps Calculator
Amps from watts and volts — the breaker-and-wire-sizing question, with the 80% circuit rule.
Updated
You need
10.00 Adraw
1,200 W at 120 V
- Current draw
- 10.00 A
- Min. breaker (80% rule)
- 12.5 A continuous
In short
How many amps is 1,200 watts?
Amps = watts ÷ volts, so 1,200 W on a 120 V circuit is exactly 10.00 A, and the same 1,200 W on a 240 V circuit is only 5.00 A. Treated as a continuous load, that 10 A needs a breaker of at least 10 ÷ 0.8 = 12.5 A, which in practice means the next standard size up, a 15 A breaker.
The formula is exact for resistive loads such as heaters and kettles; motors and switching electronics draw more current than it predicts because their power factor is below 1.
How to use the watts to amps calculator
Enter the power in watts and pick the circuit voltage, and this tool returns the current in amps plus the smallest breaker rating that would carry it as a continuous load. That is the direction that matters in practice, because almost nothing in the electrical world is rated in watts.
Breakers are rated in amps, fuses are rated in amps, extension leads and plug tops and wire gauges are all rated in amps, while the appliance you just bought advertises itself in watts.
Converting between the two is the step that turns a marketing number on a box into a number you can compare against the rating printed on the panel door, and it is the first thing anyone does before deciding whether a new load can share an existing line or needs one of its own.
10.00 A
1,200 W at 120 V
the tool default
5.00 A
The same load at 240 V
double the volts, half the amps
15 A
Breaker it needs, run continuously
the next standard size above 12.5 A
The arithmetic is amps = watts ÷ volts, and it holds exactly for direct current and for resistive alternating-current loads. Resistive means the load is essentially a heating element: a space heater, a kettle, a toaster, an immersion heater, an incandescent bulb, a hair dryer. For those, every amp of current does useful work and the power factor sits at approximately 1.
Motors, compressors, transformers, fluorescent ballasts, and a good deal of switching electronics behave differently. Their current and voltage waveforms drift out of step, so the real power in watts is less than the apparent power in volt-amps, and the actual current they pull is higher than watts ÷ volts predicts. For anything with a motor in it, treat this result as a floor and use the nameplate amps instead.
Working from the amps instead?
When the label prints current rather than power, the mirror tool multiplies amps by volts to give the wattage that circuit represents.
Open the amps to watts calculator →Two voltages cover most North American homes, and choosing the wrong one throws the answer by a factor of two. Standard receptacles and everything that plugs into them run at 120 V nominal; ranges, electric dryers, water heaters, and electric-vehicle chargers run at 240 V, which halves the current for the same power and is precisely why they are wired that way.
Europe and the United Kingdom sit at 230 V nominal, so a European appliance drawing 3,000 W pulls 13.04 A there but would pull 25.00 A on a North American 120 V circuit. Use the voltage of the circuit in front of you, not the voltage of the country the appliance was designed for. Finally, treat every figure here as a planning estimate: sizing a circuit for installation is a licensed electrician’s job, not a calculator’s.
Do
- Read the watts off the rating plate rather than off a similar model you remember.
- Pick the voltage of the circuit the appliance will actually sit on.
- Use the nameplate amps for anything with a motor or a compressor in it.
- Sum every load running at the same time before assuming the new one fits.
Don't
- Mix up the two voltages, which throws the answer by a factor of two.
- Assume a load that fits under the breaker also fits the continuous budget.
- Apply the plain division to a motor, whose power factor sits below 1.
- Read a wire gauge out of this figure, since temperature, run length and terminations all bear on it.
Standard North American breaker ratings and the power each one can actually carry, once the continuous-load margin comes off. The continuous columns are the ones to plan against for anything that runs for three hours or more; the maximum column applies only to short bursts.
| Breaker rating | Continuous limit at 80 percent | Maximum watts at 120 V | Continuous watts at 120 V | Continuous watts at 240 V |
|---|---|---|---|---|
| 15 A | 12 A | 1,800 W | 1,440 W | 2,880 W |
| 20 A | 16 A | 2,400 W | 1,920 W | 3,840 W |
| 25 A | 20 A | 3,000 W | 2,400 W | 4,800 W |
| 30 A | 24 A | 3,600 W | 2,880 W | 5,760 W |
| 35 A | 28 A | 4,200 W | 3,360 W | 6,720 W |
| 40 A | 32 A | 4,800 W | 3,840 W | 7,680 W |
| 45 A | 36 A | 5,400 W | 4,320 W | 8,640 W |
| 50 A | 40 A | 6,000 W | 4,800 W | 9,600 W |
| 60 A | 48 A | 7,200 W | 5,760 W | 11,520 W |
| 70 A | 56 A | 8,400 W | 6,720 W | 13,440 W |
| 80 A | 64 A | 9,600 W | 7,680 W | 15,360 W |
| 90 A | 72 A | 10,800 W | 8,640 W | 17,280 W |
| 100 A | 80 A | 12,000 W | 9,600 W | 19,200 W |
| 125 A | 100 A | 15,000 W | 12,000 W | 24,000 W |
The 80 percent rule, in code terms
The second output applies the continuous-load rule. The National Electrical Code defines a continuous load in Article 100 as one whose maximum current is expected to continue for three hours or more, and NEC 210.20(A) then requires the branch-circuit overcurrent device to be rated at not less than the noncontinuous load plus 125 percent of the continuous load.
NEC 210.19(A) applies the same 125 percent to the conductor. Multiplying the load by 1.25 is the same thing as limiting it to 80 percent of the breaker rating, which is why the rule is usually quoted as the 80 percent rule: a 15 A circuit takes 12 A continuously, a 20 A circuit takes 16 A, a 30 A circuit takes 24 A. The tool divides your current by 0.8 to show what that margin demands.
Read it: A fifth of every breaker rating is spoken for before any load is plugged in, which is why a device that fits the breaker on paper can still be over the continuous budget.
Computed as rating × 0.8, the inverse of the 125 percent NEC 210.20(A) requires.
The formula, worked line by line
Every question in this family comes out of one relationship, P = V × I: power in watts equals voltage in volts multiplied by current in amperes. Rearranged to solve for current it becomes amps = watts ÷ volts, which is the form this tool uses.
It is exact for direct current, and exact for alternating current whenever the load is purely resistive, because in that case the current rises and falls in lockstep with the voltage and every volt-ampere the supply delivers becomes a watt of real work.
For alternating current in general the honest version carries a fourth term: P = V × I × PF, where PF is the power factor, the fraction of the apparent power that does useful work. A resistive load sits at a power factor of about 1 and the term disappears.
A motor, a compressor, or a switch-mode supply sits well below 1, so for the same watts it draws more amps than the simple division suggests. Three-phase work adds another factor again: for a balanced three-phase load, P = √3 × V × I × PF, with V the line-to-line voltage.
amps = watts ÷ volts
amps = watts ÷ (volts × PF) (single-phase AC, real loads)
amps = watts ÷ (√3 × volts × PF) (balanced three-phase)
minimum breaker = amps ÷ 0.8 (continuous load, NEC 210.20(A))- Load
- 1,200 W
- Circuit voltage
- ÷ 120 V
- Current drawn
- 10.00 A
- Continuous margin
- 10 ÷ 0.8 = 12.5 A
- Minimum breaker
- 12.5 A, so the next standard size, 15 A
Move the same 1,200 W to a 240 V circuit and it draws 1,200 ÷ 240 = 5.00 A, needing only 6.25 A of continuous-rated breaker. On a European 230 V supply the same appliance draws 1,200 ÷ 230 = 5.22 A.
One caveat worth stating plainly, because it is where this formula most often misleads. NEC 240.4(D) caps the overcurrent protection on small copper conductors at 15 A for 14 AWG, 20 A for 12 AWG, and 30 A for 10 AWG, regardless of what an ampacity table might otherwise permit, and aluminium is limited to 15 A at 12 AWG and 25 A at 10 AWG.
Those caps mean the amps figure here never tells you a wire size by itself: conductor selection also depends on ambient temperature, how many conductors share a raceway, the length of the run, and the termination ratings. This tool answers the arithmetic question. Sizing an actual circuit is a licensed electrician’s call.
Questions people ask
How many amps is 1,500 watts?
On a 120 V circuit, 1,500 watts is 1,500 ÷ 120 = 12.50 amps. That sits under a 15 A breaker but above the 1,440 W the same circuit may carry continuously, so a 1,500 W heater running all evening is already at the limit with nothing else sharing the line. On a 240 V circuit the same 1,500 W is only 6.25 amps, and on a European 230 V supply it is 6.52 amps. The voltage you choose changes the answer, so match it to the circuit rather than to the appliance’s country of origin.
What is the 80 percent rule for circuits?
It is the plain-language form of a code requirement. NEC Article 100 defines a continuous load as one whose maximum current is expected to last three hours or more, and NEC 210.20(A) requires the overcurrent device to be rated at the noncontinuous load plus 125 percent of the continuous load, with NEC 210.19(A) applying the same 125 percent to the conductor. Multiplying a load by 1.25 is the same as limiting it to 80 percent of the rating, which gives the familiar ladder: 12 A on a 15 A circuit, 16 A on 20 A, 24 A on 30 A, 40 A on 50 A.
Why do dryers and ranges use 240 volts instead of 120?
Because doubling the voltage halves the current for the same power. A 4,800 W load is 40.00 amps at 120 V but only 20.00 amps at 240 V. Less current means a smaller conductor, a smaller breaker, and less energy wasted heating the wire on the way, since resistive loss climbs with the square of the current. That is why clothes dryers, electric ranges, water heaters, and electric-vehicle chargers are wired for 240 V, and why long direct-current runs in solar and battery systems push their voltage up for exactly the same reason.
Is amps = watts ÷ volts always exact?
For direct current and for resistive alternating-current loads such as heaters, kettles, toasters, and incandescent bulbs, yes. For motors, compressors, and many switching power supplies it underestimates the current, because their power factor is below 1: the real power in watts is less than the apparent power in volt-amps, so the same watts arrive with more amps behind them. For those loads use the nameplate amperage and treat this figure as a lower bound. Balanced three-phase equipment needs a further √3 in the denominator.
Can I use this to pick a wire gauge?
No, and it is worth being clear about why. NEC 240.4(D) caps small copper conductors at 15 A for 14 AWG, 20 A for 12 AWG, and 30 A for 10 AWG, but the correct conductor for a real installation also depends on ambient temperature, how many current-carrying conductors share the raceway, the length of the run and its voltage drop, the insulation type, and the temperature rating of the terminations at both ends. This tool gives you the current a load draws, which is one input among several. Wire and breaker selection for an actual circuit belongs to a licensed electrician.
Sources
Where the constants and formulas on this page come from. Each line names the figure it backs.
That 240.4(D) caps overcurrent protection on small copper conductors at 15 A for 14 AWG, 20 A for 12 AWG and 30 A for 10 AWG, and that Article 100 with 210.19(A) sets the three-hour continuous load and its 125 percent factor.
NFPA 70, National Electrical Code — NFPA, Free registered access; article numbering is edition-sensitive
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