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!(adsbygoogle = window.adsbygoogle || []).push({});

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([tex]\theta[/tex])=v2/v1 =

tan([tex]\theta[/tex])=5/5

([tex]\theta[/tex])= acrtan(1)

([tex]\theta[/tex])= (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([tex]\theta[/tex])=v2/v1 =

tan([tex]\theta[/tex])=5/5

([tex]\theta[/tex])= acrtan(1)

([tex]\theta[/tex])= (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<([tex]\theta[/tex])1 * V2<([tex]\theta[/tex]) = V1V2 < (([tex]\theta[/tex])1 + ([tex]\theta[/tex])2)

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

25<90 in polar?

d) A/B ->

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

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

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

not sure how to get started here...

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Adding/Sub. Phasors

**Physics Forums | Science Articles, Homework Help, Discussion**