{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

problem26_05

# problem26_05 - 26.5 a Req 1 2.4 1 1.6 1 4.8 1 0.8 b I 2.4...

This preview shows page 1. Sign up to view the full content.

26.5: a) . 8 . 0 8 . 4 1 6 . 1 1 4 . 2 1 1 eq = + + = - R b) ; A 5 . 17 ) 6 . 1 ( ) V 28 ( ; A 67 . 11 ) 4 . 2 ( ) V 28 ( 6 . 1 6 . 1 4 . 2 4 . 2 = = = = = = R ε I R ε I . A 83 . 5 ) 8 . 4 ( ) V 28 ( 8 . 4 8 . 4 = = = R ε I c) . A 35 ) 8 . 0 ( ) V 28 ( = = = total total R ε I d) When in parallel, all resistors have the same potential difference over them, so here all have V = 28 V. e) = = = = = = ) 6 . 1 ( ) A 5 . 17 ( ; W 327 ) 4 . 2 ( ) A 67 . 11 ( 2 6 . 1 2 6 . 1 2 4 . 2 2 4 . 2 R I P R I P W. 163 ) 8 . 4 ( ) A 83 . 5 ( ; W 490 2 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern