In electromagnetic physics, a *volt-ampere,* the notation for which is VA, is a measure of apparent power and uses watts for units. For some problems, you may be required to determine the current, I, flowing through a circuit and measured in amperes. If so, you may be given the value of this apparent power, often provided in kilo-volt-amperes, or kVA.

The equation you need for such problems is:

Where S is apparent power – sometimes the same as actual power, as in the case of a purely resistive circuit, but usually less – V is the potential difference in volts and I is the current in amperage. Since power, as mentioned, can be equivalently expressed in watts or volt-amperes, you can see that the units match up.

One wrinkle is that the equation needs to be modified in the case of three-phase systems. In these instances, a constant multiplying factor of √3 must be appended to the right-hand side.

To convert from kVA to amperes:

## Step 1: Determine the Phase of the System

For single-phase systems, use S = V × I. For three-phase systems, use:

Assume for this sample problem that you are addressing a three-phase system with an apparent power of 100 kVA and a potential difference of 50 V.

## Step 2: Determine the Current in Kiloamperes

Solve the equation using the specified values of S and V:

## Step 3: Convert From Kiloamperes to Amperes

Since 1 kA = 1,000 A, 1.155 kA = 1,155 A.

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.