To convert voltage and amps to watts for a standard US 120V, 15-amp branch circuit, the direct answer is 1,800 watts (assuming a purely resistive load with a power factor of 1.0). The foundational formula used is W = V × A, which substitutes directly as 120V × 15A = 1,800W. However, treating this single-voltage calculation as a universal rule will lead to undersized breakers, tripped GFCIs, and melted wire insulation when you step outside basic DC circuits or purely resistive AC heaters. The real-world math shifts dramatically based on your phase configuration and the inductive nature of your load.
The Core Formula and the Power Factor Trap
In direct current (DC) circuits, or alternating current (AC) circuits powering purely resistive loads like incandescent bulbs or space heaters, the conversion is straightforward. Real power (Watts) equals Voltage multiplied by Current. The assumption that fixes the answer here is a Power Factor (PF) of 1.0, meaning the voltage and current waveforms are perfectly in phase.
But when you introduce inductive or capacitive loads—like AC compressor motors, transformers, or cheap switching power supplies—the current waveform lags or leads the voltage waveform. This phase shift creates a gap between Apparent Power (Volt-Amps, or VA) and Real Power (Watts). The corrected AC formula becomes:
This is exactly when the basic conversion becomes mathematically meaningless. If you are measuring an inductive load with a standard clamp meter, you are only reading RMS current. Multiplying that raw amp reading by the line voltage gives you VA, not Watts. Without knowing the specific power factor of the device, converting those raw amps to real watts is impossible. To get a true wattage reading on a complex AC load, you must use a true power meter (like a Kill A Watt or a Fluke 435 power quality analyzer) that samples the phase angle shift directly.
How the Math Shifts: 120V vs. 230V vs. 3-Phase
Voltage and phase architecture dictate how much current is required to deliver a specific wattage. Assuming a constant 1,800W target load with a 1.0 PF, here is how the amperage shifts across global standards:
- 120V Single-Phase (US/Canada Standard): Uses the standard W = V × A formula. To deliver 1,800W, the circuit draws 15A. This is the maximum continuous draw for a standard 15A duplex receptacle.
- 230V Single-Phase (EU/UK/AU Standard): The formula remains identical, but the higher voltage halves the required current. To deliver 1,800W at 230V, the circuit draws only 7.8A. This is why European wiring for high-wattage appliances can often use smaller cross-sectional conductors than their US counterparts.
- 480V 3-Phase (Industrial/Commercial): The formula shifts entirely to account for the three overlapping sine waves. The multiplier becomes the square root of 3 (approximately 1.732). The formula is W = 1.732 × V(L-L) × A × PF. If you have a 480V 3-phase motor drawing 20A with a 0.90 PF, the real power is 1.732 × 480 × 20 × 0.90, which equals 14,964W (or roughly 15kW). For a deep dive into the vector math behind this, reference the All About Circuits textbook chapter on reactive AC power.
Quick Reference: 120V Circuit Neighboring Values (±20% Range)
When sizing branch circuits, the National Electrical Code (NEC 210.20) requires that continuous loads (those running for 3 hours or more) be derated to 80% of the breaker's capacity. Therefore, on a standard 15A breaker, your maximum continuous current is 12A. The table below maps the real power output for a standard 120V nominal circuit across a ±20% range of a 15A baseline (12A to 18A), comparing a purely resistive load against a typical inductive motor load.
| Current (Amps) | Voltage (Nominal) | Power Factor (Assumed) | Real Power (Watts) | Apparent Power (VA) |
|---|---|---|---|---|
| 12.0A (Continuous Max) | 120V | 1.0 (Resistive) | 1,440W | 1,440 VA |
| 12.0A | 120V | 0.80 (Inductive) | 1,152W | 1,440 VA |
| 13.0A | 120V | 1.0 (Resistive) | 1,560W | 1,560 VA |
| 14.0A | 120V | 1.0 (Resistive) | 1,680W | 1,680 VA |
| 15.0A (Breaker Limit) | 120V | 1.0 (Resistive) | 1,800W | 1,800 VA |
| 15.0A | 120V | 0.80 (Inductive) | 1,440W | 1,800 VA |
| 16.0A | 120V | 1.0 (Resistive) | 1,920W | 1,920 VA |
| 17.0A | 120V | 1.0 (Resistive) | 2,040W | 2,040 VA |
| 18.0A (+20% Overload) | 120V | 1.0 (Resistive) | 2,160W | 2,160 VA |
Note: Running a 15A breaker at 18A will cause the thermal trip mechanism to open, typically within 15 to 45 minutes depending on ambient panel temperature.
Frequently Asked Questions
How do I convert DC voltage and amps to watts on a solar array?
For DC systems, the power factor is always 1.0, so you simply multiply the voltage by the current (W = V × A). However, when calculating solar array output, you must use the panel's Maximum Power Voltage (Vmp) and Maximum Power Current (Imp) from the spec sheet, not the Open Circuit Voltage (Voc). For example, a 48V nominal LiFePO4 battery bank pulling 15A from a charge controller is generating exactly 720W of real power.
Why is my AC wattage lower than volts times amps on my motor?
This discrepancy is caused by the power factor. AC motors are highly inductive loads, meaning the magnetic fields required to turn the rotor cause the current waveform to lag behind the voltage waveform. A 120V table saw motor drawing 12A on your clamp meter might only consume 1,150W of real power if its power factor is 0.80. The remaining 290 VA is "reactive power" that bounces back and forth between the motor and the grid, doing no real mechanical work but still generating heat in your supply wires.
Can I convert watts back to amps if I only know the voltage?
You can only do this accurately for DC circuits or purely resistive AC loads using the formula A = W / V. If you are dealing with an inductive AC load (like an HVAC compressor or a large refrigerator) and you do not know the power factor, calculating the amperage from the wattage rating on the nameplate will give you a dangerously low number. Always use the Rated Load Amps (RLA) or Full Load Amps (FLA) printed directly on the motor nameplate for breaker and wire sizing, rather than attempting to reverse-engineer the math from the wattage.
