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

Answer:

The phase current in each line conductor are;

$I_{R} = 100.17 < 0A$

$I_{Y} = 75.13< - 120A$

$I_{B} = 50.08 <120A$

Explanation:

Given the following data;

Red phase = 24kW,

Yellow phase = 18kW

Blue phase = 12kW

Line voltage = 415V

For a star connected system, we have;

$Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}$

$Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}$

$Phase voltage (V_{p} ) = 239.6V$

The phase sequence for RYB is given by;

$V_{R} = 239.6<0\\V_{Y} = 239.6<120\\V_{B} = 239.6<-120$

$Phase current (I) = \frac{Phase power}{Phase voltage}$

$Hence, I = \frac{P}{V}$

For the Red phase;

$I_{R} = \frac{24000}{239.6<0}$

$I_{R} = 100.17 < 0A$

For the Yellow phase;

$I_{Y} = \frac{18000}{239.6<120}$

$I_{Y} = 75.13< - 120A$

For the Blue phase;

$I_{B} = \frac{12000}{239.6<-120}$

$I_{B} = 50.08 <120A$

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, $I_{N} = 43.29A$