Previous in Forum: CT Magnetisation Current Ie   Next in Forum: Brushes
Close
Close
Close
5 comments
Rate Comments: Nested
Active Contributor
United Kingdom - Member - New Member Engineering Fields - Power Engineering - New Member

Join Date: Dec 2011
Location: Manchester, UK
Posts: 12

Electrical Help

04/15/2013 8:50 AM

Hello i need help understanding some information i read in a text book,

'This information is from J O Bird - Electrical circuit theory and technology 2nd rev.'

i dont know how to get from step 1 (in red) to step 2 (in green) and to step 3 (in blue)

i understand the determinants ie (ad-cb) but i dont under stand how to add the voltages with the phase angle

Any help would be great

__________________
There Is No Spoon
Register to Reply
Pathfinder Tags: JOBird
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Guru

Join Date: Feb 2011
Posts: 7025
Good Answers: 207
#1

Re: Electrical Help

04/15/2013 9:18 AM

homework gets tougher at this level

Register to Reply
Guru

Join Date: Oct 2008
Posts: 42355
Good Answers: 1693
#2

Re: Electrical Help

04/15/2013 10:42 AM

Why not solve the equations? That should get you there.

Register to Reply
Guru
Popular Science - Cosmology - New Member Technical Fields - Technical Writing - New Member Engineering Fields - Energy Engineering - New Member Engineering Fields - Electrical Engineering - New Member Engineering Fields - Control Engineering - New Member Engineering Fields - Electromechanical Engineering - Old Member, New Association

Join Date: Apr 2008
Location: Lexington, KY
Posts: 1639
Good Answers: 73
#3

Re: Electrical Help

04/15/2013 1:09 PM

Well, obviously, you can't add magnitudes of different phase angles. You must first convert from Polar to Rectangular, add the Rectangular components, and then convert back to Polar. Most engineering calculators have this function.

__________________
A great troubleshooting tip...."When you eliminate the impossible, whatever remains, however improbable, must be the truth." Sir Arthur Conan Doyle
Register to Reply
Guru

Join Date: Oct 2012
Location: 100 miles North from the World Center
Posts: 879
Good Answers: 42
#4
In reply to #3

Re: Electrical Help

04/16/2013 4:56 AM

It is a very good idea. However, one may transform 3+4i and 6+8i in 5<53.13

and respectively 10<53.13 and then the product will be 415<120*10<53.13=4150<173.13 and 415<0*5<53.13=2075<53.13 as was written in this book.

__________________
Julius
Register to Reply
Guru

Join Date: Oct 2012
Location: 100 miles North from the World Center
Posts: 879
Good Answers: 42
#5
In reply to #4

Re: Electrical Help

04/17/2013 12:20 AM

Generally-for any real or complex number-the product of two numbers is:

a*x^n*b*x^m=a*b*x^(n+m)

Now if we should transform from Cartesian into polar it could be written:

re+j*im=abs*e^j*fi where: abs=sqrt(re^2+im^2) ; fi=arctan(im/re)

Then :

abs1*e^j*f1*abs2*e^j*fi2=abs1*abs2*e^j*(fi1+fi2)

Or written symbolic :

abs1*abs2*e^j*(fi1+fi2)=abs1*abs2<(fi1+fi2)

__________________
Julius
Register to Reply
Register to Reply 5 comments
Copy to Clipboard

Users who posted comments:

7anoter4 (2); Fredski (1); lyn (1); NotUrOrdinaryJoe (1)

Previous in Forum: CT Magnetisation Current Ie   Next in Forum: Brushes

Advertisement