could someone please solve this problem for me,i don't know where to start, any help would be great. thank you

 

I wouldn't open a word document posted on a forum if my life depended on it. Sorry.

 

Maybe it would be easier for someone to help you if you uploaded the image and posted the link. Good luck, I'm struggling on Thevenin a bit myself.

 

I converted paulmc's word document to PDF.

hgmjr

 

The voltage source equivalent of 4A across 3 ohms is 12 Volts in series with 3 ohms. Now we can add 12 Volts plus 5 Volts to get 17 volts, and we can add 8 ohms and 3 ohms to get 11 ohms. So the Thevenin equivalent for Q6A is:

17 Volts in series with 11 ohms.

For Q6B the norton equivalent is 1.54... Amps across 11 Ohms

 

greetings paulmc,

You can get a feel for the technique that papabravo has used to arrive at the answers to your post by reviewing the information at the following links to the tutorials on www.allaboutcircuits.com.

Thevenin's Theorem link

http://www.allaboutcircuits.com/vol_1/chpt_10/7.html/thevenin


Norton's Theorem link

http://www.allaboutcircuits.com/vol_1/chpt_10/8.html/norton

hgmjr