Horsepower Calculator — Direct Current Power Supply
Determine the useful horsepower of a direct-current load from DC voltage, current, and efficiency. Enter any three variables to solve for the fourth.
US + metric
Formula shown
Shareable results
Horsepower — Direct Current
hp
V DC
A DC
EFF
Horsepower1.09hp(0.816 kW)
Electrical input (E × I)
960 W
Estimated load losses
144 W
Useful output
816 W
Formula & method
HP = E · I · EFF / 745.7
For a direct-current supply, electrical input power is voltage times current. Multiply by load efficiency for useful output, then divide by 745.7 W per mechanical horsepower (often rounded to 746). Metric output is useful watts divided by 1000. Efficiency is entered as a decimal from 0 to 1; there is no AC power-factor or phase multiplier.
Assumes a steady direct-current supply and a load efficiency that applies at the entered operating point. For preliminary and educational use; verify against the equipment nameplate, manufacturer data, and applicable electrical codes.
This calculator determines the useful horsepower of a load supplied by direct current. It ties together DC voltage, DC current, load efficiency, and horsepower through one relationship. Enter any three and it returns the fourth, so you can estimate output horsepower, current draw, required voltage, or operating-point efficiency.
The formula
HP = E · I · EFF / 745.7. Voltage times current is the DC electrical input in watts. Multiplying by efficiency gives the useful output power, and dividing by 745.7 watts per mechanical horsepower converts the result to hp (some references round the divisor to 746). Switch to metric and the output is useful watts divided by 1000, in kilowatts. Enter efficiency as a decimal between 0 and 1 — for example, 0.85 rather than 85.
Solve in any direction
Horsepower — the useful output implied by measured DC voltage, current, and load efficiency.
Current — the DC amperage needed to produce a known output at the specified voltage and efficiency.
Voltage — the DC supply voltage required for a known output and current.
Efficiency — useful output divided by electrical input; a result above 1.0 identifies inconsistent inputs.
Direct-current assumption
This method assumes a steady DC supply at one operating point. Unlike an AC calculation, it has no power-factor or phase multiplier. For a DC motor, use the measured or manufacturer-specified efficiency at the relevant speed and load; efficiency changes across the motor’s operating range. For AC supplies, use the single-phase or three-phase horsepower calculator.
Worked example
A DC load drawing 20 A from a 48 V supply receives 960 W. At 0.85 efficiency, its useful output is 960 × 0.85 = 816 W, so HP = 816 / 745.7 ≈ 1.09 hp (0.816 kW). The estimated conversion loss at that operating point is 144 W.
DC electrical power relation (P = E × I)NIST SP 811 (mechanical horsepower conversion)IEC 60034-2-1 (motor efficiency framework)
Verified
Constants and formula two-source checked (PumpCalcs engineering review, 2026-08-25).
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