# Polar forms

## Homework Statement

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

## 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?

## The Attempt at a Solution

Last edited:

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

why result is completely different compared with MatLAB

gneill
Mentor
why result is completely different compared with MatLAB
I couldn't say; I haven't seen your Matlab work.

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

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

• 1 person
That doesn't look right. (34.2460)/(22.1009) does not yield 1.1278. Maybe you copied the wrong value down?
sorry
IN=34.2460<-9.32954/22.1009<30.3658=1.5495<-39.6953

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

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

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

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

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

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 :)

gneill
Mentor
fx>> A=33.79310-5.51724i
fx>> magn=abs(A)
fx>> magn=34.2405
fx>>ANGLE=angle(A)*180/pi
fx>>ANGLE=-9.2726
Right. Compare these values with the ones you provided in post #5.

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

gneill
Mentor
Ohh, for calculation of Vth I used calculator
sqr(33.793102+5.517242)=34.2460
then tan-1(5.51724/33.79310)=-9.32954°
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°.

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

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