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°

answer: 500 wph.

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