Electrical Engineering - RC Circuits - Discussion

Discussion :: RC Circuits - General Questions (Q.No.15)


What is the angular difference between +j4 and –j4?

[A]. 30°
[B]. 90°
[C]. 180°
[D]. 270°

Answer: Option C


No answer description available for this question.

Faizan said: (Jan 26, 2012)  
+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: (May 18, 2012)  
How you have decide the quadrant?

Dilip Verma said: (Sep 6, 2013)  
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: (May 20, 2014)  
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.

The angular difference will be 90-(-90) = 180 degree.

Karan said: (May 4, 2015)  
As per quadrant 4 = (positive real negative image). So its will be 90 - (-90) = 180 degree.

Roja said: (Dec 12, 2016)  
If the quadrant of +j is 180 and -j is 270 then why we can't take the answer as 270?

Madhu Gorle said: (Dec 27, 2016)  
Only consider in Y axis because its imaginary part.

So, Y-axis +ve =90, -ve = 270. So, the differences is 180.

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