# AC Circuits: Phasors/Polar to rectangular transformation

1. Jun 27, 2010

### sgonzalez90

40<50degrees + 20<-30 degrees,
I get how to convert to rectangular,

I got 43.03 + j20.64, but converting it back to polar.... how exactly do you do so? The answer is 47.72<25.63degrees..... my book doesn't explain it.

Also with
(2+j4)(3-j5).... how exactly do you tackle this sort of problem? I tried the FOIL method and it didn't exactly work.

Thank you!

2. Jun 27, 2010

### sgonzalez90

1. The problem statement, all variables and given/known data

40<50degrees + 20<-30 degrees,
I get how to convert to rectangular,

I got 43.03 + j20.64, but converting it back to polar.... how exactly do you do so? The answer is 47.72<25.63degrees..... my book doesn't explain it.

Also with
(2+j4)(3-j5).... how exactly do you tackle this sort of problem? I tried the FOIL method and it didn't exactly work.

Thank you!

3. Jun 27, 2010

Length: |I|= sqrt(Re2+Im2)

sqrt(432+212)=48

Angle: < = arctan (Im/Re)

arctan (21/43) = 26 (deg)

Draw a phasor diagram and you see it easily. Simple geo/trig.

Part 2:
Either convert to polar form and multiply the length and add the angles, or just multiply out the two parenthesis inn a normal fashion. Remember j*j=-1

4. Jun 27, 2010

### Zryn

Rectangular $$\rightarrow$$ Polar

x + jy $$\rightarrow$$ $$\sqrt{x^{2} + y^{2}}$$ $$\angle$$ $$tan^{-1}(\frac{y}{x})$$

There are alot of calculators which can do this for you too!

(2+j4)(3-j5) ... FOIL is the way, expand the brackets first

2*3 + 2*-j5 + j4*3 + j4*-j5 ... simplify

6 - 10j + 12j - 20j*j ... keep in mind j = $$\sqrt{-1}$$, so $$j^{2}$$ = -1

6 +2j + 20

26 + 2j