This preview shows page 1. Sign up to view the full content.
Unformatted text preview: must be i = i ''’ + i 1 = 11.25 A upward (which is the "forward" direction for that battery). Thus, battery 3 is supplying energy. (h) Eq. 2717 leads to P 3 = 56.3 W. 84. (a) We reduce the parallel pair of resistors (at the bottom of the figure) to a single R’ =1.00 Ω resistor and then reduce it with its series ‘partner’ (at the lower left of the figure) to obtain an equivalence of R” = 2.00 Ω +1.00 Ω =3.00 Ω . It is clear that the current through R” is the i 1 we are solving for. Now, we employ the loop rule, choose a path that includes R” and all the batteries (proceeding clockwise). Thus, assuming i 1 goes leftward through R” , we have 5.00 V + 20.0 V − 10.0 V − i 1 R” = 0 which yields i 1 = 5.00 A. (b) Since i 1 is positive, our assumption regarding its direction (leftward) was correct....
View
Full
Document
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

Click to edit the document details