P28_074 - counterclockwise: 3 ( E − ir ) − (2 i ) R = 0...

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

View Full Document Right Arrow Icon
74. (a) From symmetry we see that the current through the top set of batteries ( i )i sthesamea sthe current through the second set. This implies that the current through the R =4 . 0 Ω resistor at the bottom is i R =2 i .T h u s ±w i
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: counterclockwise: 3 ( E − ir ) − (2 i ) R = 0 . This yields i = 3 . 0 A. Consequently, i R = 6 . 0 A. (b) The terminal voltage of each battery is E − ir = 8 . 0 V. (c) Using Eq. 28-14, we obtain P = i E = (3)(20) = 60 W. (d) Using Eq. 27-22, we have P = i 2 r = 36 W....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online