# Three Phase Current - Simple Calculation

**Steven McFadyen**on March 1st, 2009

The calculation of current in a three phase system has been brought up on our site feedback and is a discussion I seem to get involved in every now and again. While some colleagues prefer to remember formulas or factors, I prefer to resolve the problem step by step using basic principles. I thought it would be good to write how I do these calculations and hopefully it may prove useful to someone else.

The calculation of current in a three phase system has been brought up on our site feedback and is a discussion I seem to get involved in every now and again. While some colleagues prefer to remember formulas or factors, I prefer to resolve the problem step by step using basic principles. I thought it would be good to write how I do these calculations. Hopefully it may prove useful to someone else.

**Three Phase Power and Current**

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

* Single phase system* - this is the easiest to deal with. Given the kW and power factor the kVA can be easily worked out. The current is simply the kVA divided by the voltage. As an example, consider a load consuming 23 kW of power at 230 V and a power factor of 0.86:

Note:you can do these equations in either VA, V and A or kVA, kV and kA depending on the magnitude of the parameters you are dealing with. To convert from VA to kVA just divide by 1000.

* Three phase system* - The main difference between a three phase system and a single phase system is the voltage. In a three phase system we have the line to line voltage (V

_{LL}) and the phase voltage (V

_{LN}), related by:

or alternatively as:

to better understand this or gain more insight, you can read the Introduction to Three Phase Electric Power post

To me the easiest way to solve three phase problems is to convert them to a single phase problem. Take a three phase motor (with three windings, each identical) consuming a given kW. The kW per winding (single phase) has to be the total divided by 3. Similarly a transformer (with three windings, each identical) supplying a given kVA will have each winding supplying a third of the total power. To convert a three phase problem to a single phase problem take the total kW (or kVA) and divide by three.

As an example, consider a balanced three phase load consuming 36 kW at a power factor of 0.86 and line to line voltage of 400 V (V_{LL}) :

the line to neutral (phase) voltage V

_{LN}= 400/√3 = 230 V

three phase power is 36 kW, single phase power = 36/3 = 12 kW

now simply follow the above single phase method

Easy enough. To find the power given current, multiply by the voltage and then the power factor to convert to W. For a three phase system multiply by three to get the total power.

Personal note on the methodAs a rule I remember the method (not formulae) and rework it every time I do the calculation. When I try to remember formulae I always forget them soon or become unsure if I am remembering them correctly. My advice would be to always try remember the method and not simply memorize formula. Of course if you do have some super ability at remembering formula, you can always keep to this approach.

**Using Formulas**

**Derivation of Formula - Example**

Balanced three phase system with total power P (kW), power factor pf and line to line voltage V_{LL}

Convert to single phase problem:

Single phase apparent power Q_{1ph} (kVA):

Phase current I (kA) is the single phase apparent power divided by the phase to neutral voltage (and given V_{LN}=V_{LL} / √3):

Simplifying (and with 3 = √3 x √3):

The above method relies on remembering a few simple principals and manipulating the problem to give the answer.

More traditionally formulas may be used to give the same result. These can be easily derived from the above, giving for example:

**Unbalanced Three Phase Systems**

The above deals with balanced three phase systems. That is the current in each phase is the same and each phase delivers or consumes the same amount of power. This is typical of power transmission systems, electrical motors and similar types of equipment.

Often where single phase loads are involved, residential and commercial premises for example, the system can be unbalanced with each phase have a different current and delivering or consuming a differing amount of power.

**Balanced Voltages**

Luckily in practice voltages tend to be fixed or very by only small amounts. In this situation and with a little thought it is possible to extend the above type of calculation to unbalanced current three phase systems. The key to doing this is that the sum of power in each phase is equal to the total power of the system.

For example, take a 400 V (V_{LL}) three phase system with the following loads: phase 1 = 80 A, phase 2 = 70 A, phase 3 = 82 A

the line to neutral (phase) voltage V

_{LN}= 400/√3 = 230 V

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

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

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

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

Similarly given the power in each phase you could easily find the phase currents. If you also know the power factor you can convert between kVA and kW as shown earlier.

**Unbalanced Voltages**

If the voltages become unbalanced or there are other considerations (i.e. unbalanced phase shift), then it is necessary to revert to more traditional network analysis. System voltages and currents can be found by drawing out the circuit in full detail and using Kirchhoff's laws and other network theorems.

Network analysis is not the intent of this note. If you interested in an introduction you can view our post: Network Theory – Introduction and Review

**Efficiency & Reactive Power**

Other things to consider while carrying out calculations may include the efficiency of equipment. Knowing that efficiency of power consuming equipment is the output power divided by the input power, again this can easily accounted.for Reactive power is not discussed in the article and more details can be found in other notes (just use the site search).

**Summary**

By remembering that a three phase power (kW or kVA) is simply three times the single phase power, any three phase problem can be simplified. Divide kW by the power factor to get the kVA. VA is simply the current times the voltage, so knowing this and the voltage can give the current. When calculating the current use the phase voltage which is related to the line voltage by the square root of three. Using these rules it is possible to work out any three phase problem without the need to remember and/or resort to formulas.

Hi All, This is wrong calaculation. Please check it .if u divide kw by p.f u will get kVA not KW. three phase power is 36 kW, single phase power = 36/3 = 12 kW now follow the above single phase method kVA = kW / power factor = 12/0.86 = 13.9 kW (13900 W) Current = W / voltage = 13900/230 = 60 A