problem26_11

problem26_11 - 26.11: Using the same circuit as in Problem...

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

View Full Document Right Arrow Icon
26.11: Using the same circuit as in Problem 27.10, with all resistances the same: = + = + + + = + = - - 00 . 6 50 . 4 3 50 . 4 1 1 1 1 1 4 3 2 1 234 1 eq R R R R R R R . a) . A 500 . 0 3 1 A, 50 . 1 00 . 6 V 00 . 9 1 4 3 2 eq 1 = = = = = = = I I I I R ε I b) . W 125 . 1 9 1 , W 13 . 10 ) 50 . 4 ( ) A 50 . 1 ( 1 4 3 2 2 1 2 1 1 = = = = = = = P P P P R I P c) If there is a break at , 4 R then the equivalent resistance increases: . 75 . 6 50 . 4 2 50 . 4 1 1 1 1 3 2 1 23 1 eq = + = + + = + = - - R R R R R R And so: . A 667 . 0 2 1 A, 33 . 1 75 . 6 V 00 . 9 1 3 2 eq 1 = = = = = = I I I R ε I d) . W 99 .
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online