Discussion :: RC Circuits - General Questions (Q.No.1)
In the complex plane, the number 14 – j5 is located in the
Answer: Option D
No answer description available for this question.
|Neeraj Patel said: (Dec 11, 2010)|
|Because the fourth quadrant is positive. The angle is 19.65 this is positive.|
|Chaitanya said: (Dec 25, 2010)|
|I think in 4th quadrant you will get -19.65.|
|Rupal said: (Jan 13, 2011)|
|Option D is wright.because r is on x-axis and xl on y-axis.r is (+)and xl is (-).so its possible on 4 th qud.|
|Naeem said: (Mar 30, 2012)|
|If we draw the coordinate as taking real value on x-axis and imaginary value on y-axis then the (-j5) will be in -y-axis and real value on +x-axis. Then it is in fourth quadrant.|
|Hitesh said: (Aug 18, 2012)|
|Here real value in X axis and imaginari value in Y axis so(+,-) ans is D|
|Srikanth said: (Aug 28, 2012)|
|This is in the form of x+jy so x=14 and y=-5 so this is in the fourth quadrant....|
|Sidhu said: (Dec 11, 2012)|
|1 quad= +ve real +ve img
2 quad= -ve real +ve img
3 quad= -ve real -ve img
4 quad= +ve real -ve img
This is the reason.
|Dhwani said: (Sep 20, 2013)|
|x axes contains real parts and y axes imaginary.
So if we write X+iY then 'X' is real part and 'Y' is imaginary.
Similarly the given equation 14-j5 , 14 will be Real part situated on +x axes & -j5 on the negative y axes(4th quadrant).
|Milan Patel said: (Feb 18, 2014)|
|In quadrant co-ordinate is write in form (cos@, sin@). So it also write as cos@+jsin@.
So on that the answer will be (cos@, -sin@) so it in fourth quadrant.
|Vamshi said: (Jul 29, 2015)|
|Do we need to convert to polar for answer.|
|Raj Kumar said: (Sep 21, 2015)|
|Can you explain please?|
|Kiran said: (Feb 3, 2016)|
|I don't know anything. Please anyone explain me?|
|Vikas said: (Aug 31, 2016)|
|Fiords said: (Nov 24, 2016)|
|Sameeksh said: (Sep 14, 2018)|
|The tan inverse(-5/14)=-ve answer.
That implies it's in 4th quadrant or 2nd quadrant. But the real coefficient is +ve. So obviously it's in 4th quadrant.
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