Plotting Points on a Graph (4, 0°) and (3, 27°)

[maali5]

Homework Statement

I was wondering if you could help me on how to draw these on a graph

(4, 0°) and (3, 27°)

The Attempt at a Solution

Just a hint, Please

 

[Ibix]

These numbers are in polar form. Have a look at the Wikipedia article on complex numbers and see if that helps.

 

[maali5]

they gave sin and cos answers but my questions is saying

These can be written as (4, 0°) and (3, 27°)
Find the result of these voltages as a single combined Polar phasor.

On a graph, plot the 4V phasor in blue.

On the same graph, plot the 3V phasor in red

On the same graph, plot the combined phasor in purple (all between 0° and 360°)

 

[Ibix]

A phasor is a line from the origin to the point. All it's asking you to do is add together the two complex numbers to get a third, then draw lines from the origin to each of the three points.

 

[lazyaditya]

One way is to convert them into complex number !

 

[maali5]

A phasor is a line from the origin to the point. All it's asking you to do is add together the two complex numbers to get a third, then draw lines from the origin to each of the three points.

Thanks for your help. Do you have an example ? So i can follow

 

[lazyaditya]

Details are given to you ! (4^2 = P^2 + B^2) and tan (0) = P/B solve for both phasors convert in complex form and mark on graph !

 

[maali5]

Details are given to you ! (4^2 = P^2 + B^2) and tan (0) = P/B solve for both phasors convert in complex form and mark on graph !

Thanks for explain but I have no idea. Sorry

"(4^2 = P^2 + B^2)" May you explain where that came from

 

[lazyaditya]

actually the phasor is like a line with certain magnitude drawn in anticlockwise direction with certain angle.

 

[lazyaditya]

the magnitude of the phasor is the hypotenuse which is given as the square root of the sum of the square of the perpendicular and the base. and the angle of the phasor from the base is given by the tan inverse of the P/B.

P= perpendicular
B= Base
thus the angle can give relation between P and B and you can calculate the values of P and B from the magnitude equation and thus you ave two points available which will show your complex number.

 

[maali5]

the magnitude of the phasor is the hypotenuse which is given as the square root of the sum of the square of the perpendicular and the base. and the angle of the phasor from the base is given by the tan inverse of the P/B.

P= perpendicular
B= Base
thus the angle can give relation between P and B and you can calculate the values of P and B from the magnitude equation and thus you ave two points available which will show your complex number.

Thanks. I will try it and attached after my work

 

[Ibix]

Thanks for your help. Do you have an example ? So i can follow

If you look under the 'polar form' heading on the Wikipedia page on complex numbers, the blue arrow is a phasor. The diagram shows how to construct it from an angle and a magnitude, which you've got.

All you have to do apart from that is add the two together to get their sum. As has been pointed out, this is easier to do if you convert to x+iy form (as per the wiki). Alternatively, just make a copy of the second phasor coming from the tip of the first (as per the diagram under addition and subtraction in Wikipedia). The sum is the straight line completing the triangle.

 

[lazyaditya]

One more method would be to convert them in rectangular coordinates ..

by Y= rsin (theta)
X= rcos(theta)
theta the angle and r is the magnitude given to you !

 

[maali5]

One more method would be to convert them in rectangular coordinates ..

by Y= rsin (theta)
X= rcos(theta)
theta the angle and r is the magnitude given to you !

I have done my calculation and received r = 6.81v



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