Three Phase Current

The calculation of current in a three-phase system has been brought up on our site feedback, and it is a discussion I seem to get involved in from time to time. While some colleagues prefer to recall formulas or factors, I prefer to solve problems step by step using fundamental principles. I thought it would be useful to document how I perform these calculations. Hopefully, it will be useful to someone else.

Power and Current in Three Phases

The power consumed by a circuit (single or three phase) is measured in watts W. (or kW). The apparent power is the product of voltage and current and is measured in VA (or kVA). The power factor (pf) describes the relationship between kVA and kW:

Single phase systems are the easiest to work with. The kVA can be easily calculated given the kW and power factor. The current is calculated by dividing the kVA by the voltage. Consider the following load, which consumes 23 kW of power at 230 V and has a power factor of 0.86:

The voltage is the main difference between a three phase system and a single phase system. The line to line voltage (VLL) and phase voltage (VLN) in a three phase system are related by:

Converting three-phase problems to single-phase problems seems to me to be the simplest way to solve them. Consider a three-phase motor (with three identical windings) that consumes a given kW. The total must be divided by three to get the kW per winding (single phase). A transformer (with three identical windings) supplying a given kVA will have each winding supplying one-third of the total power. To convert a three-phase problem to a single-phase problem, multiply the total kW (or kVA) by three.

Consider the following balanced three-phase load with a power factor of 0.86 and a line-to-line voltage of 400 V (VLL):

Simple enough. To calculate the power given a current, multiply it by the voltage and then by the power factor. For a three-phase system, multiply the total power by three.

**Using Formulas**

The above method is based on remembering a few basic principles and manipulating the problem to find the solution.

To achieve the same result, more traditional formulas can be used. These are easily derived from the preceding, yielding, for example:

**Unbalanced Three Phase Systems**

The preceding discusses balanced three-phase systems. That is, each phase has the same current and delivers or consumes the same amount of power. This is common in power transmission systems, electric motors, and similar equipment.

When single phase loads are involved, such as in residential and commercial buildings, the system can become unbalanced, with each phase delivering or consuming a different amount of power.

Voltages that are balanced

Fortunately, in practice, voltages are usually fixed or vary only slightly. In this case, and with a little thought, the above calculation can be extended to unbalanced current three phase systems. The key to accomplishing this is to ensure that the sum of power in each phase equals the total power of the system.

Consider a 400 V (VLL) three-phase system with the following loads: Phase 1 equals 80 A, Phase 2 equals 70 A, and Phase 3 equals 82 A.

VLN = 400/3 = 230 V line to neutral (phase) voltage

80 x 230 = 18,400 VA = 18.4 kVA = phase 1 apparent power

apparent power in phase 2 = 70 x 230 = 16,100 VA = 16.1 kVA

82 x 230 = 18,860 VA = 18.86 kVA = phase 3 apparent power

Total three-phase power = 53.36 kVA = 18.4 + 16.1 + 18.86 kVA

Similarly, given the power in each phase, the phase currents could be easily calculated. You can convert between kVA and kW if you also know the power factor, as shown earlier.

Voltages that are unbalanced

If the voltages become unbalanced or there are other considerations (e.g., unbalanced phase shift), more traditional network analysis is required. System voltages and currents can be calculated by fully describing the circuit and applying Kirchhoff’s laws and other network theorems.

Reactive Power & Efficiency

Other factors to consider when performing calculations include the efficiency of the equipment. Knowing that the efficiency of power-consuming equipment is defined as output power divided by input power, this can be easily accounted for. for The article does not discuss reactive power, but more information can be found in other notes (just use the site search).

Summary

Any three phase problem can be simplified by remembering that a three phase power (kW or kVA) is simply three times the single phase power. To calculate kVA, divide kW by the power factor. Because VA is simply the current multiplied by the voltage, knowing this and the voltage can provide the current. Use the phase voltage, which is related to the line voltage by the square root of three, to calculate the current. Using these rules, you can solve any three-phase problem without having to remember or rely on formulas.