{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hwk5Solns - Assignment 5 Solutions Chpt 26 9 We use vd =...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment 5 Solutions Chpt 26 9. We use v d = J / ne = i / Ane . Thus, () ( ) ( )( ) 14 2 28 3 19 2 0.85m 0.21 10 m 8.47 10 / m 1.60 10 C / 300A 8.1 10 s 13min. d LLL A n e t vi A n e i −− ××× == = = = Chpt 27 12. (a) We solve i = ( ε 2 1 )/( r 1 + r 2 + R ) for R : R i rr = −− = × −−= × 21 12 3 2 30 20 99 10 .. . VV 1.0 10 A ΩΩ . Ω (b) P = i 2 R = (1.0 × 10 –3 A) 2 (9.9 × 10 2 Ω ) = 9.9 × 10 –4 W. 26. (a) R eq ( FH ) = (10.0 Ω )(10.0 Ω )(5.00 Ω )/[(10.0 Ω )(10.0 Ω ) + 2(10.0 Ω )(5.00 Ω )] = 2.50 Ω . (b) R eq ( FG ) = (5.00 Ω ) R /( R + 5.00 Ω ), where R = 5.00 Ω + (5.00 Ω )(10.0 Ω )/(5.00 Ω + 10.0 Ω ) = 8.33 Ω . So R eq ( FG ) = (5.00 Ω )(8.33 Ω )/(5.00 Ω + 8.33 Ω ) = 3.13 Ω . 32. Using the junction rule ( i 3 = i 1 + i 2 ) we write two loop rule equations: 10.0 V – i 1 R 1 – ( i 1 + i 2 ) R 3 = 0 5.00 V – i 2 R 2 – ( i 1 + i 2 ) R 3 = 0. (a) Solving, we find i 2 = 0, and (b) i 3 = i 1 + i 2 = 1.25 A (downward, as was assumed in writing the equations as we did). 41. (a) The batteries are identical and, because they are connected in parallel, the potential differences across them are the same. This means the currents in them are the 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
same. Let i be the current in either battery and take it to be positive to the left. According
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

Hwk5Solns - Assignment 5 Solutions Chpt 26 9 We use vd =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online