Calculation of Load Current (I, Ampere)

Finding out the current consumed by a load is very essential in designing the Breakers, Cables etc in an electrical system.

The best thing to explain this will be through practical examples of 3 Phase and Single phase power.

Before Starting some basic formulas should be kept in mind

                          V_LL   =  V_LN  X √3             V_LL    = LINE TO LINE VOLATGE 

                                                                   V_LN     = LINE TO NEUTRAL VOLATGE 

                                                                   Ex:-  240V x √3  = 415V

                            KW = KVA X CosØ             

                        Power =  V x I x CosØ             ( For 1 Phase )

                                      √3 x V x I x CosØ       ( For 3 Phase )

                                  (√3 is only applicable for 3 Phase Power; for 1Phase omit the √3 )

 (Where V depends on the Voltage; 1 phase = 240V, 3 Phase = 415V)

                                  (I = Current;  CosØ = Power Factor  )

Single Phase Calculation

Consider with a n example Load = 25KW ; V = 240V  ; CosØ = 0.8

P = V x I x CosØ         

   

25 x 10³  = 240 x I x 0.8

            I  =  25000 / (240x0.8)

               =  130.20 Amps

So For this above Load of 130.20 Amps, We will be using a Single Pole Breaker which is 25% more current capacity than the rated current

ie ;      130.20 x 125%

        = 162.75 Amps

If same size breaker is not available, the breaker adjacent to the value above the calculated current is been chosen.

Three Phase Calculation

Method 1 : Using Direct Formula

Consider with a n example Load = 7.5KW ; V = 415V  ; CosØ = 0.8

Power =  √3 x V x I x CosØ

              7.5x1000 = √3 x 415 x I x 0.8

                            I = 7500/ 1.73x415x0.8

                              =  13.05 Amps

Method 2 : Converting 3Phase problem to 1Phase (Only on Balanced Loads)

Balanced Loads : The loads which are balanced with the three phases

                            Ex : Total Load 7.5 Kw  (R = 2.5 Kw ; Y= 2.5 Kw; B= 2.5 Kw)

Total Load = 7.5Kw

Load per Phase = 7.5Kw / 3 = 2.5Kw ( Only on Balanced Load)

Single phase power Equation P      =V x I x CosØ

                                      Given V_LL  = 415V

                                                 V_LN = 415 / √3

                                                         = 239.88V

Equating in 1 Phase Power Eqn, P = V x I x CosØ

                                                        =  239.88 x I x 0.8

                                          2.5x10³  =  239.88 x I x 0.8

                                              I_Amps  =  2.5x10³ /  (239.88 x 0.8)

                                                        =  13.02 Amps

So For this above Load of 13.02 Amps, We will be using a Tripple Pole or Four Pole Breaker which is 25% more current capacity than the rated current

ie ;      13.02 x 125%

        = 16.275 Amps

If same size breaker is not available, the breaker adjacent to the value above the calculated current is been chosen.