Single-Phase Calculations

Basic electrical theory tells us that for a single-phase system,

kW = (V × I × PF) ÷ 1,000.

For the sake of simplicity, let's assume the power factor (PF) is unity. Therefore, the above equation becomes

kW = (V × I) ÷ 1,000.

Solving for I, the equation becomes

I = 1,000kW ÷ V (Equation 1)

Now, if we look at the 1,000 ÷ V portion of this equation, you can see that by inserting the respective single-phase voltage for V and dividing it into the 1,000, you get a specific number (or constant) you can use to multiply kW to get the current draw of that load at the respective voltage.

For example, the constant for the 120V calculation is 8.33 (1,000 ÷ 120). Using this constant, Equation 1 becomes

I = 8.33kW.

So, if you have a 10kW load, you can calculate the current draw to be 83.3A (10 × 8.33). If you have a piece of equipment that draws 80A, then you can calculate the relative size of the required power source, which is 10kW (80 ÷ 8.33).

Table 1. Constants used on single-phase systems

By using this same procedure but inserting the respective single-phase voltage, you get the following single-phase constants, as shown in Table 1.

3-Phase Calculations

For 3-phase systems, we use the following equation:

kW = (V × I × PF × 1.732) ÷ 1,000.

Again, assuming unity PF and solving this equation for I, you get:

I = 1,000kW ÷ 1.732V.

Table 2. Constants used on 3-phase systems

Now, if you look at the 1,000 4 1.732V portion of this equation, you can see that by inserting the respective 3-phase voltage for V and multiplying it by 1.732, you can then divide that quantity into the 1,000 to get a specific number (or constant) you can use to multiply kW to get the current draw of that 3-phase load at the respective 3-phase voltage. Table 2 lists each 3-phase constant for the respective 3-phase voltage obtained from the above calculation.