(faa335) aljabr Hw09 Ross (89251) This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points The current I = a t2 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 = t1 ? 1. q = a t2 b t1 + c 1 a3 b2 t t + c t1 correct 31 21 b 3. q = a t3 t2 + c t1 1 21 2. q = 4. q = a t3 b t2 + c t1 1 a3 b2 t t +c 31 21 Explanation: The unit of current is Coulomb per second dq I= or dq = I dt. dt To nd the total charge in coulombs that passes through the conductor, one must integrate the current over the time interval. 5. q =
t1 t1
1
Let :
a = 3 C/s3 , b = 3 C/s2 , c = 8 C/ s , t1 = 2 s , and t2 = 4 s .
t2
Since a3 b2 q= t t + ct 3 2 then 3 C/s3 3 C/s2 3 q2 = (4 s) (4 s)2 3 2 + (8 C/s)(4 s) = 72 C , and q1 = 3 C/s3 3 C/s2 (2 s)3 (2 s)2 3 2 + (8 C/s)(2 s) = 18 C . q21 = q2 q1 = 54 C . 003 10.0 points Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 50 mm2 when carrying a current of 10 A (values similar to those for the electric wire to your study lamp). Assume one electron per atom of copper contributes to the current. The atomic mass of copper is 63.5 g/mol and its density is 8.93 g/cm3 . Avogadros number NA is 6.02 1023 . Correct answer: 1.47651 105 m/s. Explanation: Let : N = 1 , M = 63.5 g/mol , = 8.93 g/cm3 , A = 50 mm2 = 5 105 m2 , I = 10 A , and qe = 1.6 1019 C/electron . ,
t1
so that
q=
0 t1
dq =
0
I dt
=
0
(a t2 b t + c) dt
t1 0
=
a3 b2 t t + ct 31 21 b a = t 3 t 2 + c t1 . 1 3 21
002 (part 2 of 2) 10.0 points If I is in A, and a = 3 C/s3 , b = 3 C/s2 , and c = 8 C/s, what quantity of charge moves across the section of the conductor from t1 = 2 s to t2 = 4 s? Correct answer: 54 C. Explanation:
aljabr (faa335) Hw09 Ross (89251) We rst calculate n, the number of currentcarrying electrons per unit volume in copper. Assuming one free conduction electron per NA n atom, = , where NA is Avogodros M number and and M are the density and the atomic weight of copper, respectively n 1 electron atom NA . M
2
stretched to a length , the volume will remain the same. The new cross-sectional area is then, A = A A A A A = = = . 1.86 1.86 The resistance of a conductor of resistivity is given by R= , A so R A (1.86 ) A =A= = A R A A 1.86 = (1.86)2 . Thus R = (1.86)2 R = 3.4596 R . 005 10.0 points A 0.59 V potential dierence is maintained across a 1.8 m length of tungsten wire that has a cross-sectional area of 0.79 mm2 and the resistivity of the tungsten is 5.6 108 m. What is the current in the wire? Correct answer: 4.62401 A. Explanation: Let : V = 0.59 V , = 1. 8 m , A = 0.79 mm2 = 7.9 107 m2 , = 5.6 108 m . The resistance is R= V = , I A and
n=
6.02 1023 atoms 63.5 g/mol 106 cm3 8.93 g/cm3 1 m3 = 8.46592 1028 electrons/m3 . 1 electron atom
Drift Speed in Copper Wire vd = = I n qe A 10 A 8.46592 1028 electrons/m3 1 19 C/electron 1.6 10 1 5 105 m2
= 1.47651 105 m/s . 004 10.0 points A wire with a resistance R is lengthened to 1.86 times its original length by pulling it through a small hole. Find the resistance of the wire after it is stretched. Correct answer: 3.4596 R . Explanation: Let : = 1.86 . The volume V of the wire is given by V = A, where is the length of the wire and A is the cross-sectional area. When the wire is
so the current is VA I= (0.59 V) (7.9 107 m2 ) = (5.6 108 m) (1.8 m) = 4.62401 A .
aljabr (faa335) Hw09 Ross (89251) 006 (part 1 of 2) 10.0 points An electric heater operating at full power draws a current of 9.3 A from a 191 V circuit. What is the resistance of the heater? Correct answer: 20.5376 . Explanation: Let : I = 9.3 A and V = 191 V . From Ohms law, R= 191 V V = = 20.5376 . I 9. 3 A
3
007 (part 2 of 2) 10.0 points Assuming constant R, how much current should the heater draw in order to dissipate 950 W? Correct answer: 6.80122 A. Explanation: Let : P = 950 W . Since P = I 2 R, I1 = P = R 950 W = 6.80122 A . 20.5376