How to Convert HP to Amps & Volts

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Electrical power, in physical terms, is a function of the current flowing through a system and the voltage (potential difference) of that system. In fact, the power is simply the product of these two quantities:

P = (V)(I)

Where P is the power in watts (or joules per second), V is the potential difference in volts, and I is the current in amperes. Power can also be expressed in volt-amperes and horsepower (HP_, with the latter often used in everyday engines such as those in motor vehicles. 1 HP is equal to 746 watts.

Other factors affect the true power output of en electrical system, notably the phase of the circuit and its efficiency.

If you get the power of a system in HP and the current in amps, you can calculate the volts; if you know the power and the number of volts, you can determine the current in amps; and if you have amps and volts, you can convert to horsepower.

Assume you are working with a 30-HP circuit that draws 800 amps of current. Before you determine the voltage, you must transform the basic power equation above to a more specific one involving multiplicative coefficients, if necessary.

Step 1: Convert Horsepower to Watts

Since amps and volts are standard units, but HP is not, you need the power in watts to solve the equation. Since 1 HP = 746 W, the wattage in this example is (746)(30) = 22,380 W.

Step 2: Is the System a Three-Phase System?

If yes, introduce a correction factor of 1.728, which is the square root of 3, into the basic power equation above, so that P = (1.728)(V)(A). Assume your 22,380-watt circuit is a three-phase system:

22,380 = (1.728)(V)(800)

Step 3: What is the Efficiency?

Efficiency is a measure of how much current and voltage is converted to useful power and is expressed as decimal number. Assume for this problem the efficiency of the circuit is 0.45. This also factors into the original equation, so you now have:

22,380 = (0.45)(1.728)(V)(800)

Step 4: Solve for Volts (or Amps)

You now have everything you need to determine the voltage of this system.

22,380 ÷ (1.728)(0.45)(800) = V

V = 35.98 volts

The equation needed to solve problems of this type is

P = (E)(Ph)(V)(A) ÷ 746,

Where P = power in HP, E = efficiency, Ph is a phase correction factor (1 for single-phase systems, 1.728 for three-phase systems), V is the voltage and I is the amperage.

References

About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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