Convert AC tons or BTU/h to amps at your supply voltage, single- or three-phase, using a typical efficiency and power factor or your own values.
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- LRA to Amps Calculator (Starting and Running Current)
Tons to Amps Formula
The following formula is used to calculate the electrical current in amps from tons of refrigeration for a single-phase system.
Variables:
- I is the electrical current (amps)
- TR is the tons of refrigeration
- V is RMS supply voltage (volts); use line-to-line RMS voltage for balanced three-phase
- COP is cooling watts divided by electrical input watts; this ratio may exceed1
- PF is the power factor (as a decimal)
Convert refrigeration tons to cooling watts using approximately3516.853 W/TR, divide by cooling COP to estimate electrical real power, then divide by RMS voltage and power factor. Balanced three-phase also divides by √3. COP and electrical data must cover the same equipment and operating conditions.
Three-Phase Tons to Amps Formula
For three-phase electrical systems, the formula requires an additional factor of the square root of 3 (approximately 1.732) in the denominator to account for the way power is distributed across three conductors:
For balanced three-phase, total real power is √3 × line-to-line RMS voltage × line current × power factor. The √3 factor follows the three-phase voltage relationships. This model does not infer unbalanced currents, starting current or a wire size.
| Tons (TR) | Amps (1-Phase, 240V) | Amps (3-Phase, 460V) |
|---|---|---|
| 1 | 5.1 | 1.5 |
| 1.5 | 7.6 | 2.3 |
| 2 | 10.2 | 3.1 |
| 2.5 | 12.7 | 3.8 |
| 3 | 15.3 | 4.6 |
| 3.5 | 17.8 | 5.4 |
| 4 | 20.4 | 6.1 |
| 5 | 25.4 | 7.7 |
| 7.5 | 38.2 | 11.5 |
| 10 | 50.9 | 15.3 |
| 12.5 | 63.6 | 19.2 |
| 15 | 76.3 | 23.0 |
| 20 | 101.8 | 30.7 |
| 25 | 127.2 | 38.3 |
| Single-phase calculated at 240V. Three-phase calculated at 460V with a √3 factor. Both use the calculator’s typical defaults: cooling COP 3.2 (about EER 11) and power factor 0.90. Actual amp draw varies by equipment. Always reference the unit nameplate. | ||
What is a Ton of Refrigeration?
A ton of refrigeration (TR) is a unit of cooling capacity rooted in the ice harvesting era. It represents the amount of heat absorption needed to melt one short ton (2,000 lb) of ice at 32 degrees F over a 24-hour period. Because the latent heat of fusion of water is 144 BTU per pound, one ton of refrigeration equals 2,000 x 144 / 24 = 12,000 BTU per hour, or 3,516.85 watts.
This unit remains the standard in the United States, Japan, and parts of the Middle East for rating air conditioning and refrigeration equipment. Most of the rest of the world uses kilowatts of cooling (kWc) instead. A quick conversion: 1 TR = 3.517 kW of cooling capacity.
Efficiency, SEER, and EER in the Formula
Use cooling COP: cooling output divided by electrical input in identical power units. Unlike motor efficiency, cooling COP is not capped at1 because refrigeration transfers heat. This corrects the former0–1 efficiency label in this calculator.
EER is cooling Btu/h per electrical watt at specified conditions. Convert EER to cooling COP by dividing by approximately3.412142; EER12 corresponds to COP about3.517. Do not divide EER by a theoretical maximum or substitute motor efficiency.
SEER and SEER2 describe seasonal performance. They do not directly determine current at one operating point, and no fixed SEER-to-SEER2 equivalence is assumed here. Use matching cooling COP or steady-state EER and electrical measurements.
| Tons (TR) | 10 EER | 14 EER | 16 EER | 20 EER |
|---|---|---|---|---|
| 1.5 | 8.8 | 6.3 | 5.5 | 4.4 |
| 2 | 11.8 | 8.4 | 7.4 | 5.9 |
| 2.5 | 14.7 | 10.5 | 9.2 | 7.4 |
| 3 | 17.6 | 12.6 | 11.0 | 8.8 |
| 3.5 | 20.6 | 14.7 | 12.9 | 10.3 |
| 4 | 23.5 | 16.8 | 14.7 | 11.8 |
| 5 | 29.4 | 21.0 | 18.4 | 14.7 |
| Illustrative operating-point arithmetic: electrical watts = TR ×12000/EER; amps = watts/(V ×PF). EER is supplied steady-state performance, not SEER. These figures do not predict a specific unit or size its circuit. | ||||
Power Factor in HVAC Systems
Power factor is real electrical power divided by apparent power. For sinusoidal conditions it includes voltage-current phase displacement; nonlinear loads can also have distortion. Use measured or manufacturer data for the same operating point; do not assume a universal compressor value.
For fixed real input power and voltage, current is inversely proportional to power factor. A change from0.95 to0.80 increases modeled current by18.75%. A cooling-capacity comparison also requires the same COP and equipment scope.
When Tonnage and Current Are Proportional
With voltage, COP, power factor and phase fixed, doubling cooling tonnage doubles the modeled current. Real units can have different performance and operating conditions, so a fixed amps-per-ton rule does not establish their actual current.
Operating conditions and variable-speed controls can change cooling output, electrical input and power factor. Use measured or published matching data; this calculator does not predict those changes from outdoor temperature alone.
NEC Electrical Sizing for Air Conditioning Circuits
This running-current estimate cannot select conductors, breakers or fuses. Use the specific equipment nameplate Minimum Circuit Ampacity (MCA), Maximum Overcurrent Protection (MOCP), manufacturer instructions and applicable local requirements with qualified electrical design.
MCA and MOCP serve different purposes and cannot be inferred safely from tonnage alone. The retained historical table below has no verified equipment basis and is not installation guidance. Wire and protection selection also depend on installation conditions and applicable requirements.
| AC Size (Tons) | Typical MCA Range (A) | Typical MOCP (A) | Common Wire Size (AWG) |
|---|---|---|---|
| 1.5 | 10 – 13 | 15 – 20 | 14 or 12 |
| 2 | 13 – 17 | 20 – 25 | 12 or 10 |
| 2.5 | 15 – 19 | 20 – 30 | 12 or 10 |
| 3 | 17 – 23 | 25 – 35 | 10 |
| 3.5 | 19 – 26 | 30 – 40 | 10 or 8 |
| 4 | 22 – 30 | 35 – 45 | 10 or 8 |
| 5 | 27 – 37 | 40 – 60 | 8 or 6 |
| Historical values retained for context only; equipment sources and installation conditions were not supplied. They are not verified MCA/MOCP ratings or wire recommendations. Use the actual equipment nameplate and qualified electrical design. | |||
Industry Rules of Thumb
A generic horsepower-per-ton assumption does not establish electrical input. Motor shaft output, motor input and cooling output are different quantities. Use known cooling COP and power factor or actual electrical data instead of assuming a universal amps-per-ton ratio.
If apparent power is independently known, single-phase current equals VA divided by RMS voltage. For example, an explicitly assumed1.5kVA/TR would give6.25A/TR at240V; that assumption is not a verified equipment rule and cannot size wiring or protection.
Refrigerant Type and Amp Draw
Refrigerant affects cycle performance, but refrigerant identity or pressure alone does not establish current. Cooling COP, equipment design, operating conditions, voltage and power factor must be considered together.
This calculator has no refrigerant-specific current adjustment. Compare manufacturer electrical and cooling data at matching conditions; no fixed percentage difference between refrigerants is assumed.
