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:

S = V × I

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:

For single-phase systems, use S = V × I. For three-phase systems, use S = √3 × (V × I).

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.

Solve the equation S = √3 × (V × I) using the specified values of S and V:

100 kVA = √3 × (50 × I)

100 kVA ÷ ( √3 × 50) = I

I = 100 ÷ (1.732 × 50) = 1.155 kiloamperes

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