To calculate amperage from wattage, divide the wattage by the voltage. In a direct current (DC) circuit, the formula is I = P ÷ V. In a single-phase alternating current (AC) circuit, you must also account for the power factor (PF), making the formula I = P ÷ (V × PF). For example, a 1200W space heater on a 120V AC circuit with a power factor of 1.0 draws exactly 10 amps.
While the math is straightforward, applying it incorrectly on the workbench or jobsite leads to undersized wire, tripped breakers, or melted terminal lugs. Below is the complete derivation, unit tracking, and the exact assumptions you need to make before sizing your conductors and overcurrent protection.
The Core Equation: Wattage, Voltage, and Amperage
Electrical power is the rate at which work is done in a circuit. To find the current (amperage), we use Watt's Law. The fundamental relationship changes slightly depending on whether you are working with DC or AC power.
| Symbol | Quantity | Unit | Definition & Assumptions |
|---|---|---|---|
| P | Real Power | Watts (W) | The actual work performed. Assumes steady-state load; does not account for startup surge currents (like motor LRA). |
| V | Voltage | Volts (V) | Electrical potential difference. For AC, this must be the RMS (Root Mean Square) voltage, not the peak voltage. |
| I | Current | Amperes (A) | The flow of electrical charge. This is the value we are solving for to size wire and breakers. |
| PF | Power Factor | Dimensionless (0 to 1) | The ratio of real power to apparent power. Always 1.0 for DC and purely resistive AC loads (heaters, incandescent bulbs). Typically 0.70 to 0.90 for inductive AC loads (motors, compressors). |
When the Formula Applies (and When It Doesn't)
These equations apply to single-phase AC and DC circuits under steady-state conditions. They assume the voltage provided is RMS. If you are working with three-phase AC power, the formula requires a multiplier: I = P ÷ (V × PF × √3). Using the single-phase formula on a 3-phase motor will result in a calculated amperage that is roughly 73% higher than reality, leading to oversized, unnecessarily expensive conductors.
Rearranged Forms and Unit Traps
Depending on the known variables on your equipment's spec sheet, you will need to rearrange Watt's Law. Here is the complete list of algebraic forms for single-phase and DC systems:
- Solving for Current: I = P ÷ (V × PF)
- Solving for Voltage: V = P ÷ (I × PF)
- Solving for Power: P = V × I × PF
- Solving for Power Factor: PF = P ÷ (V × I)
Which Unit Mistakes Break the Math?
The most common reason DIYers and junior techs get wildly incorrect amperage numbers is unit mismatch. The formula only works if base units are used.
- kW vs. W: A 2.5 kW generator outputs 2500 W. If you divide 2.5 by 120V, you get 0.02A instead of the actual 20.8A.
- mA vs. A: Microcontroller datasheets list GPIO limits in milliamps (e.g., 40mA). You must convert to Amps (0.04A) before multiplying by voltage to find wattage.
- Peak vs. RMS Voltage: A 120V AC wall outlet actually peaks at ~170V. If you use 170V in your denominator, your calculated amperage will be dangerously low, leading to undersized wire.
Solved Problems: DC and AC Worked Examples
Let's run through two bench scenarios with strict unit tracking to prove the math.
Example 1: Sizing Wire for a 12V DC LED Strip
Setup: You are installing a 5-meter reel of 12V DC LED strip lights. The packaging states the total power draw is 60W. You need to know the amperage to select the correct fuse and wire gauge for the 15-foot run from the power supply to the strip.
- Identify Knowns: P = 60 W, V = 12 V, PF = 1.0 (DC circuits always have a PF of 1).
- Select Formula: I = P ÷ V
- Substitute and Track Units:
I = 60 W ÷ 12 V
I = 60 (Joules/second) ÷ 12 (Joules/Coulomb)
I = 5 Coulombs/second
I = 5 Amperes (A) - Outcome: The strip draws exactly 5A. Using standard AWG tables for chassis wiring, 16 AWG wire (rated for ~10A) and a 7.5A inline automotive fuse are appropriate.
Example 2: Breaker Sizing for an AC Refrigerator Compressor
Setup: You are wiring a dedicated branch circuit for a commercial reach-in refrigerator. The compressor nameplate reads: 120V, 800W, PF 0.65. What is the running amperage?
- Identify Knowns: P = 800 W, V = 120 V (RMS), PF = 0.65.
- Select Formula: I = P ÷ (V × PF)
- Substitute and Track Units:
I = 800 W ÷ (120 V × 0.65)
I = 800 ÷ 78
I = 10.25 Amperes (A) - Outcome: The compressor draws 10.25A while running. If you had ignored the power factor and just divided 800 by 120, you would have calculated 6.66A. That 3.6A difference is the reactive current bouncing back and forth, which still generates heat in your wires and must be accounted for when sizing the breaker.
Bench War Story: The 2000W Inverter Wiring Disaster
Formulas are useless if you apply them to the wrong side of the equipment. A few years ago, a hobbyist brought a scorched 2000W pure sine wave inverter into the shop after it failed during a solar cabin build.
The Setup: The builder was connecting a 2000W inverter to a 12V LiFePO4 battery bank to run AC power tools. They needed to size the DC cables running from the battery busbar to the inverter's positive and negative terminals.
The Numbers: The builder looked at the inverter's faceplate, saw "2000W", and knew the output was standard US AC voltage (120V). They calculated:
I = 2000W ÷ 120V = 16.6A.
Based on 16.6A, they purchased 12 AWG THHN wire and 20A ANL fuses for the DC battery connections.
The Outcome: When they turned on a 1500W miter saw, the inverter attempted to pull the required power from the 12V battery bank. The 12 AWG wire instantly glowed hot, the insulation melted onto the terminal lug, and the voltage drop caused the inverter's low-voltage cutoff to trip, killing the saw mid-cut.
What Went Wrong: The builder calculated the amperage for the AC output side, but wired the DC input side. Power must be conserved (minus efficiency losses). To get 2000W of AC power out of a 12V DC battery, the DC amperage is massive:
I = 2000W ÷ 12V = 166.6A.
Factoring in an 85% inverter efficiency, the actual DC draw is closer to 196A. Pushing 196A through 12 AWG wire (rated for ~20A) is a severe fire hazard. The correct cable size for this run is 2/0 AWG copper with a 250A Class T fuse. Always calculate amperage using the voltage of the specific circuit segment you are wiring.
Realistic Magnitudes and NEC Sizing Rules
When you calculate amperage from wattage, you need a sanity check. What does a realistic answer look like in residential and commercial environments?
- Standard 120V / 15A Branch Circuit: Maximum theoretical wattage is 1800W (120V × 15A). However, the National Electrical Code (NEC) requires continuous loads (those running for 3 hours or more) to be derated to 80%. Therefore, a realistic continuous maximum is 1440W.
- Standard 120V / 20A Branch Circuit: Maximum theoretical is 2400W. Continuous load maximum is 1920W. This is why most high-end space heaters and hair dryers are factory-limited to 1500W or 1875W.
- 240V / 30A Dryer Circuit: Maximum theoretical is 7200W. Continuous maximum is 5760W.
If your calculation yields an amperage that exceeds the breaker rating of the circuit you intend to plug the device into, you cannot simply swap the breaker for a larger one. The wire inside the walls is sized to match the original breaker. Upgrading a 15A breaker to 20A on 14 AWG wire removes the overcurrent protection and creates a concealed fire hazard.
For deeper reading on AC power dynamics and the relationship between real, reactive, and apparent power, the All About Circuits textbook chapter on AC Power provides excellent phasor diagrams. If you are sizing battery cables for DC inverter setups similar to the war story above, consult the Blue Sea Systems Circuit Wizard and wire sizing charts, which account for voltage drop and insulation temperature ratings in high-current DC environments.
