Solve Complex Math Problem: Adding/Subtracting Phasors

[Quelsita]

Ok, I'm trying to wrap my head around phasor notation. I think part a is correct, not sure of the difference for subtraction and converting phasor to polar notation. Any help is appreciated!

Problem:
For A=5<36.9 and B=5<53.1

a)Calculate the sum A+B
b)Calculate the difference A-B
c)Calculate the product AB in both polar and rectangular forms.
d)Calculate the quotient A/B in both polar and rectangular forms.
e)If A = 3 +j4, compute 1/A in both polar and rectangular forms.


a) A+B->
tan(\theta)=v2/v1 =
tan(\theta)=5/5
(\theta)= acrtan(1)
(\theta)= (pi/4)

Vr=V1^2 +V2^2 = 5^2 + 5^2 = 50V^2
Vr=sqrt(50V^2)
Vr= 5sqrt(2)V

thus, A+B= 5sqrt(2)< (pi/4)

b) A-B -> not certain of the difference between adding and subtracting phasors...?
tan(\theta)=v2/v1 =
tan(\theta)=5/5
(\theta)= acrtan(1)
(\theta)= (pi/4)

Vr=V1^2 - V2^2 = 5^2 - 5^2 = 0
Vr=0

thus, A+B= 0(pi/4)?? This means 0 voltage?


c) A*B
V1<(\theta)1 * V2<(\theta) = V1V2 < ((\theta)1 + (\theta)2)

5*5< (36.9+53.1) = 25<90

25<90 in polar?


d) A/B ->

[V1<(\theta)1]/[V2<(\theta)2] = (V1/V2) <(\theta)1 -(\theta)2

5/5 < (39.6 - 53.1) = 1< -16.2



e) A = 3 +j4, compute 1/A ->

not sure how to get started here...

 

[collinsmark]

Ok, I'm trying to wrap my head around phasor notation. I think part a is correct, not sure of the difference for subtraction and converting phasor to polar notation. Any help is appreciated!

'Just to start, phasor notation and polar notation are the same thing. The conversions you need to do involve converting to and fro polar (i.e phasor) notation and rectangular (i.e. Cartesian) notation.

Problem:
For A=5<36.9 and B=5<53.1

a)Calculate the sum A+B
b)Calculate the difference A-B
c)Calculate the product AB in both polar and rectangular forms.
d)Calculate the quotient A/B in both polar and rectangular forms.
e)If A = 3 +j4, compute 1/A in both polar and rectangular forms.a) A+B->
tan(\theta)=v2/v1 =
tan(\theta)=5/5
(\theta)= acrtan(1)
(\theta)= (pi/4)

Vr=V1^2 +V2^2 = 5^2 + 5^2 = 50V^2
Vr=sqrt(50V^2)
Vr= 5sqrt(2)V

thus, A+B= 5sqrt(2)< (pi/4)

Start over from scratch with parts a) and b). The general outline for both is to
(1) Convert to rectangular coordinates.
(2) Add or subtract components appropriately.
(3) Convert back to polar coordinates (if you are instructed to do so).

b) A-B -> not certain of the difference between adding and subtracting phasors...?
tan(\theta)=v2/v1 =
tan(\theta)=5/5
(\theta)= acrtan(1)
(\theta)= (pi/4)

Vr=V1^2 - V2^2 = 5^2 - 5^2 = 0
Vr=0

thus, A+B= 0(pi/4)?? This means 0 voltage?c) A*B
V1<(\theta)1 * V2<(\theta) = V1V2 < ((\theta)1 + (\theta)2)

5*5< (36.9+53.1) = 25<90

25<90 in polar?d) A/B ->

[V1<(\theta)1]/[V2<(\theta)2] = (V1/V2) <(\theta)1 -(\theta)2

5/5 < (39.6 - 53.1) = 1< -16.2

c) and d) are correct for the calculations done in polar coordinates, but I'm guessing you also need to do the calculations in rectangular coordinates too. See below for a hint on how to approach the calculations in terms of rectangular coordinates.

e) A = 3 +j4, compute 1/A ->

not sure how to get started here...

Here is a hint, for calculations in rectangular (Cartesian) form (you should already be able to do this in polar form):

\frac{1}{\vec A} = \frac{1}{3 + j4} = \frac{1}{(3 + j4)} \frac{(3-j4)}{(3 - j4)}

You can take it from there.

===========================================
General formulas

A \angle \theta_A = A \cos \theta_A + j A \sin \theta_A

|A| = \sqrt{(\Re \{ \vec A \})^2 + (\Im \{ \vec A \})^2}

\theta_A = \arctan \left( \frac{\Im \{ \vec A \} } {\Re \{ \vec A \}} \right)