Electrical Engineering - RC Circuits - Discussion
Discussion Forum : RC Circuits - General Questions (Q.No. 1)
1.
In the complex plane, the number 14 – j5 is located in the
Discussion:
18 comments Page 1 of 2.
Neeraj patel said:
1 decade ago
Because the fourth quadrant is positive. The angle is 19.65 this is positive.
Chaitanya said:
1 decade ago
I think in 4th quadrant you will get -19.65.
Rupal said:
1 decade ago
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:
1 decade ago
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:
1 decade ago
Here real value in X axis and imaginari value in Y axis so(+,-) ans is D
Srikanth said:
1 decade ago
This is in the form of x+jy so x=14 and y=-5 so this is in the fourth quadrant....
Sidhu said:
1 decade ago
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.
2 quad= -ve real +ve img
3 quad= -ve real -ve img
4 quad= +ve real -ve img
This is the reason.
DHWANI said:
1 decade ago
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).
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:
1 decade ago
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.
So on that the answer will be (cos@, -sin@) so it in fourth quadrant.
Vamshi said:
1 decade ago
Do we need to convert to polar for answer.
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