AC cct troubles

hi !
wayne-ok to post here? i thought it would be ok,and you would know to answer it too.

i was doing some revision,and used an online site for a textbook because we dont get decent ones that tell us what we need to know...

ok i was reading my notes,and i have one method that i understand,but cant tell if its a correct example,and the 2nd one,is an 'answer' to a question i did in an assignment,problem is i completely have no idea what hes doing in it.

thanks alot!

i know its simple,yet my tutor hasnt explained it well in a way i can use.

i have looked at a site but it uses different methods...(listed at bottom)

is this correct ??


(AC Parallel circuit.)
Obviously Convert mH or F into Xl/Xc Reactance

eg
5Ω

work out angle if any resistance included =
eg
ta-1= Xl/r
=-53°

Z = 5 Ω <-53°

I=V/Z

=20<0° / 5<-53°

=4A <53°
notice sign is opposite.(correct?)(did i have it correct in the first place!)

So now i have the impedence of that branch ,then i just use sin and cos to find the active and reactive parts

eg =
Active
I1=10 cos 0=10
I2=4 Cos-53=2.4
=12.4A
Reactive
I1=10 sin 0 =0
I2=4 cos -53=-2.4
=-3.2A

(easy)
______________
Is=/12.4^2 + 3.2 ^2

12.8A

ø=Ta-1 Is/Ir
Ta-1 -3.2/12.4
ø=-14.97°


Is=12.8A <-14.43°(correct sign/?)

OK that method i UNDERSTAND but im just not sure if my negatives are right....thats all!


Another example


(by a different tutor)
inductor + resistance in series

in parallel=Cap + resistance in series

calculate=total current,total impedence,phase angle of I and Vs

ok so im fine getting up to finding the I in each path..
Il=2.11A
Ic=3.14A
Ir=8.33A

Then, Tan ø =Xl/R=82°
(why is the angle of the Xl worked out ONLY ? )

Horizontal components=IL cos ø = 2.11 amps * cos 82 = 0.26A
Vertical components=Il sin ø = 2.11 amps * sin 82 = 2.09 A

Reactance=Il-Ic= +-1.05A
_________________
Current=It =/(horiz Il)^2-Ix^2

=1.085A

i really dont see his method! what the hell is he doing???

i think il abandon this last way,and use the first one!

Cheers!




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