Ttempt all questions q1 in a 3 phase 4 wire system the...

Q#1: In a 3-phase, 4-wire system, the currents in R, Y, and B lines under abnormal conditions of loading are as follows: IR = 100 ∠30° A, IY = 50 ∠300° A, IB = 30 ∠180° A. Calculate the positive, negative, and zero sequence currents in the R-line and the return current in the neutral wire. Q#2: A balanced star-connected load takes 90 A from a balanced 3-phase, 4-wire supply. If the fuses in the Y and B phases are removed, find the symmetrical components of the line currents: (i) before the fuses are removed, (ii) after the fuses are removed.

Transcript

00:01 So here it is given that for three phase system the current flowing throughout the red wire let's say ir this is given to us 100 angle of 30 degree amperes.

00:12 Similarly throughout the blue wire ib is equals to 30 angle of 180 degrees amperes means these are in polar form and the current the yellow wire is 50 angle 30 degree ampere sorry 300 degree amperes.

00:37 So the formula which we are going to use here is the matrix in which first of all let us say at initial the current flowing throughout the red wire is ir naught then ir 1 then ir 2 this should be equals to 1 by 3 times of the matrix 1 1 1 1 alpha alpha square 1 alpha square alpha times of the new terminals that is ir iy and i of b where if you see the value of alpha would be equals to 1 angle of 120 degree and hence alpha square will also be equals to 1 angle of 230 degree which would be equals to 1 angle of minus of 120 degree.

01:43 So let us calculate the current flowing throughout the red wire.

01:49 Okay because we are converting three phase four wire system to ryb system right so for that we need to calculate this currents.

02:00 So let us calculate ir at initial that is zero so it would be equals to 1 by 3 times of into the bracket ir plus iy plus ib this will be equals to 1 by 3.

02:15 Now let us substitute the values so this is 100 angle of 30 plus 50 angle of 300 plus 30 angle of 180 degrees so this will comes out to be equals to 27 .3 angle of 4 .7 degree ampere.

02:40 Now similarly we will try to calculate the other currents that is ir of 1 so this will be equals to 1 by 3 times of ir plus alpha times of iy plus alpha square times of ib...

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