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.
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.