P27_068 - V 2 − V 3 | = V D = iR D = 10 V . (b) See...

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

View Full Document Right Arrow Icon
68. (Second problem of Cluster ) (a) We use Eq. 27-16 to compute the resistances in SI units: R C = ρ C L C πr 2 C = ( 2 × 10 6 ) 1 π (0 . 0005) 2 =2 . 5Ω R D = ρ D L D πr 2 D = ( 1 × 10 6 ) 1 π (0 . 00025) 2 =5 . 1Ω . The voltages follow from Ohm’s law: | V 1 V 2 | = V C = iR C =5 . 1V |
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V 2 − V 3 | = V D = iR D = 10 V . (b) See solution for part (a). (c) and (d) The power is calculated from Eq. 27-22: P = i 2 R = ½ 10 W for R = R C 20 W for R = R D...
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