Electrical Engineering - RC Circuits - Discussion
Discussion Forum : RC Circuits - General Questions (Q.No. 15)
15.
What is the angular difference between +j4 and –j4?
Discussion:
7 comments Page 1 of 1.
Faizan said:
1 decade ago
+j4 can be in 1st and 2nd quadrant while -j4 can be 3rd and 4th quadrant. So I think there can be 2 possibilties 180 and 270 both can be the right answer. If someone has another logic. Please help me out.
Hardeep said:
1 decade ago
How you have decide the quadrant?
Dilip verma said:
1 decade ago
Because of when we have consider x axis. At the left side all will be negative (-j4) and at the right all will be positive (+j4) and it is makes a x axis as a simple line so the angle is 180.
Yahya Rehman said:
1 decade ago
Here the question is clear cut and the values are given properly. There is no supposition. Therefore for +j4 phase angle is 90 degree and for -j4 phase angle is -90 degree.
So,
The angular difference will be 90-(-90) = 180 degree.
So,
The angular difference will be 90-(-90) = 180 degree.
Karan said:
1 decade ago
As per quadrant 4 = (positive real negative image). So its will be 90 - (-90) = 180 degree.
Roja said:
9 years ago
If the quadrant of +j is 180 and -j is 270 then why we can't take the answer as 270?
Madhu gorle said:
9 years ago
Only consider in Y axis because its imaginary part.
So, Y-axis +ve =90, -ve = 270. So, the differences is 180.
So, Y-axis +ve =90, -ve = 270. So, the differences is 180.
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