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Unformatted text preview: homework 05 – FAUSAK, TAYLOR – Due: Feb 18 2008, 4:00 am 1 Question 1, chap 27, sect 1. part 1 of 2 10 points The current I = a t 2 b t + c in a section of a conductor depends on time. What quantity of charge moves across the section of the conductor from t = 0 to t = t 1 ? 1. q = a 3 t 3 1 b 2 t 2 1 + c 2. q = a 3 t 3 1 b 2 t 2 1 + c t 1 correct 3. q = a t 3 1 b t 2 + c t 1 4. q = a t 2 1 b t 1 + c 5. q = a t 3 1 b 2 t 2 1 + c t 1 Explanation: The unit of current is Coulomb per second I = d q dt or dq = I dt . To find the total charge in coulombs that passes through the conductor, one must inte grate the current over the time interval. q = integraldisplay t 1 dq = integraldisplay t 1 I dt = integraldisplay t 1 ( a t 2 b t + c ) dt = bracketleftbigg a 3 t 3 1 b 2 t 2 1 + c t bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle t 1 = a 3 t 3 1 b 2 t 2 1 + c t 1 . Question 2, chap 27, sect 1. part 2 of 2 10 points If I is in A, and a = 4 C / s 3 , b = 3 C / s 2 , and c = 8 C / s, what quantity of charge moves across the section of the conductor from t 1 = 2 s to t 2 = 4 s? Correct answer: 72 . 6667 C (tolerance ± 1 %). Explanation: Let : a = 4 C / s 3 , b = 3 C / s 2 , c = 8 C / s , t 1 = 2 s , and t 2 = 4 s . Since q = bracketleftbigg a 3 t 3 b 2 t 2 + c t bracketrightbigg vextendsingle vextendsingle vextendsingle vextendsingle t 2 t 1 , then q 2 = ( 4 C / s 3 ) 3 (4 s) 3 ( 3 C / s 2 ) 2 (4 s) 2 + (8 C / s)(4 s) = 93 . 3333 C , and q 1 = ( 4 C / s 3 ) 3 (2 s) 3 ( 3 C / s 2 ) 2 (2 s) 2 + (8 C / s)(2 s) = 20 . 6667 C . so that q 21 = q 2 q 1 = 72 . 6667 C . Question 3, chap 27, sect 3. part 1 of 1 10 points A current of 5 A flows in a copper wire 2 mm in diameter. The density of valence electrons in copper is roughly 9 × 10 28 m 3 ....
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This note was uploaded on 03/30/2010 for the course CH 369 taught by Professor Kbrowning during the Spring '07 term at University of Texas.
 Spring '07
 KBrowning

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