# Polar forms

1. Nov 27, 2013

### shaltera

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

A=1.5495<21.0363°x(22.1009<30.3658°/69.9667<9.1884°)

2. Relevant equations

3. The attempt at a solution
A=1.5495<21.0363x(22.1009/69.9667(30.3658-9.1884)=1.5495<21.0363(0.3159<21.1774)=(1.5495x0.3159)(21.0363+21.1774)=0.4895<42.2137°

Solution above is it correct or I have to convert polar forms to complex number, and final result to convert to polar form?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Nov 27, 2013
2. Nov 27, 2013

### Staff: Mentor

You can do it either way. Your solution looks fine.

3. Nov 27, 2013

### shaltera

why result is completely different compared with MatLAB

4. Nov 27, 2013

### Staff: Mentor

I couldn't say; I haven't seen your Matlab work.

5. Nov 27, 2013

### shaltera

IL=IN(ZTH/ZTH+RL)
IN=VTH/ZTH

FOR:
VTH=33.79310-j5.51724
ZTH=19.06896+j11.172413
RL=50

Here comes the tricky part:
I convert VTH and ZTH in polar forms I got
VTH=34.2460<-9.32954
ZTH=22.1009<30.3658
and
ZTH+RL=19.06896+j11.172413+50=69.06896+j11.172413=69.9667<9.1884

Then

IN=34.2460<-9.32954/22.1009<30.3658=1.1278<-39.6953

IL=1.1278<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.1278<-39.6953(0.3159<21.1774)=0.3563<-18.5169

6. Nov 27, 2013

### Staff: Mentor

That doesn't look right. (34.2460)/(22.1009) does not yield 1.1278. Maybe you copied the wrong value down?

7. Nov 27, 2013

### shaltera

sorry
IN=34.2460<-9.32954/22.1009<30.3658=1.5495<-39.6953

8. Nov 27, 2013

### Staff: Mentor

Okay, that looks better.

I note that in your various conversions you are specifying several decimal places but the values after the first decimal place or so aren't always accurate... it looks like some truncation or rounding was done on some previous intermediate values, so the extra digits aren't helping (or useful). Make sure to keep extra digits in all intermediate values.

For example, you converted Vth from rectangular to polar as

33.79310-j5.51724 → 34.2460 ∠-9.32954

If I do the same conversion, being sure to retain all digits for intermediate steps, I obtain:

33.79310-j5.51724 → 34.24052 ∠-9.2726°

Angles are notorious sources of rounding/truncation errors if you skimp on digits, because the functions involved are not linear; in some regions a tiny difference can blow up into a big difference.

9. Nov 27, 2013

### shaltera

therefor

IL=1.5495<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.5495<-39.6953(0.3159<21.1774)=0.4895<-18.5179

10. Nov 27, 2013

### Staff: Mentor

Yep. Again, rounding/truncation errors are suspect: I get 0.489 ∠-18.461° .

11. Nov 27, 2013

### shaltera

interesting I calculate using a MatLAB, and that's what MatLAB gives me?

12. Nov 27, 2013

### shaltera

fx>> A=33.79310-5.51724i
fx>> magn=abs(A)
fx>> magn=34.2405
fx>>ANGLE=angle(A)*180/pi
fx>>ANGLE=-9.2726

13. Nov 27, 2013

### shaltera

Ohh, for calculation of Vth I used calculator
sqr(33.793102+5.517242)=34.2460
then tan-1(5.51724/33.79310)=-9.2726°

I don't know what I have done to get -9.32954 but obviously is not right :)

14. Nov 27, 2013

### Staff: Mentor

Right. Compare these values with the ones you provided in post #5.

15. Nov 27, 2013

### shaltera

yes you are right angle is a bit off.And it has been corrected

16. Nov 27, 2013

### Staff: Mentor

The inverse tan value looks suspect. Did you truncate a radian value before converting to degrees? My "calculator" gives 0.161837 radians for the arctan. Converting that to degrees gives 9.2726°.

17. Nov 27, 2013

### shaltera

you are correct, I made a mistake.I copied the wrong value on the forum.I have 9.2726, and MatLAB confirmed it.Thanks

18. Nov 27, 2013

Excellent