Purely resistive loads of 24 kW, 18 kW, and 12 kW are connected between the neutral

and the red, yellow and blue phases respectively of a 3-0, four-wire system. The line

voltage is 415 V. Calculate:

i. the current in each line conductor (i. e., IR, Iy and IB); and

ii. the current in the neutral conductor.

(i) IR = 100.167 A Iy = 75.125∠-120 IB = 50.083 ∠+120 (ii) IN = 43.374∠ - 30°

Explanation:

Solution

Given that:

Three loads 24 kW, 18 kW, and 12 kW are connected between the neutral.

Voltage = 415V

Now,

(1) The current in each line conductor

Thus,

The Voltage Vpn = vL√3

Gives us, 415/√3 = 239.6 V

Then,

IR = 24 K / Vpn ∠0°

24K/239.6 ∠0° = 100.167 A

For Iy

Iy = 18k/239. 6

= 75.125A

Thus,

Iy = 75.125∠-120 this is as a result of the 3 - 0 system

Now,

IB = 12K / 239.6

= 50.083 A

Thus,

IB is = 50.083 ∠+120

(ii) We find the current in the neutral conductor

which is,

IN = Iy + IB + IR

= 75.125∠-120 + 50.083∠+120 + 100.167

This will give us the following summation below:

-37.563 - j65.06 - 25.0415 + j 43.373 + 100.167

Thus,

IN = 37.563 - j 21.687

Therefore,

IN = 43.374∠ - 30°