Question

Purely resistive loads of 24kW, 18kW and 12kW are connected between the neutral and the red, yellow and the blue respectively of 3 phase, four-wire system. The line voltage is 415. What is the current in each line conductor ?

Answer 1

The phase current in each line conductor are;

`I_(R) = 100.17 < 0A`

`I_(Y) = 75.13< - 120A`

`I_(B) = 50.08 <120A`

Step-by-step explanation:

Given the following data;

Red phase = 24kW,

Yellow phase = 18kW

Blue phase = 12kW

Line voltage = 415V

For a star connected system, we have;

`Phase voltage (V_(p) ) = (Line voltage)/(√(3))`

`Phase voltage (V_(p) ) = (415)/(√(3))`

`Phase voltage (V_(p) ) = 239.6V`

The phase sequence for RYB is given by;

`V_(R) = 239.6<0\\V_(Y) = 239.6<120\\V_(B) = 239.6<-120`

`Phase current (I) = (Phase power)/(Phase voltage)`

`Hence, I = (P)/(V)`

For the Red phase;

`I_(R) = (24000)/(239.6<0)`

`I_(R) = 100.17 < 0A`

For the Yellow phase;

`I_(Y) = (18000)/(239.6<120)`

`I_(Y) = 75.13< - 120A`

For the Blue phase;

`I_(B) = (12000)/(239.6<-120)`

`I_(B) = 50.08 <120A`

For the line neutral;

`I_(N) =\sqrt{ (I_(R)^(2) +I_(Y)^(2)+I_(B)^(2)-I_(R)I_(Y)-I_(Y)I_(B)-I_(R)I_(B)`

Substituting we have,
`I_(N) = 43.29A`