*This preview shows
pages
1–3. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **fierro (jmf2547) – HW14 – berk – (60290) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electric field of 1824 V / m is applied to a section of silver of uniform cross section. Assume: The resistivity ρ of silver is 1 . 59 × 10 − 8 Ω · m at 20 ◦ C. The tempera- ture coefficient of silver is 0 . 0038( ◦ C) − 1 . Calculate the resulting current density if the specimen is at a temperature of 6 ◦ C. Correct answer: 1 . 21163 × 10 11 A / m 2 . Explanation: Let : ρ 20 = 1 . 59 × 10 − 8 Ω · m , T = 20 ◦ C , α = 0 . 0038( ◦ C) − 1 , and E = 1824 V / m . By Ohm’s Law, J = σE ρ = ρ 20 [1 + α ( T − T 20 )] = (1 . 59 × 10 − 8 Ω · m) × { 1 + bracketleftbig . 0038( ◦ C) − 1 bracketrightbig (6 ◦ C − 20 ◦ C) } = 1 . 50541 × 10 − 8 Ω · m . Thus, J = σE = E ρ = 1824 V / m 1 . 50541 × 10 − 8 Ω · m = 1 . 21163 × 10 11 A / m 2 . 002 10.0 points The resistance to the right of A ′ B ′ is the same as the resistance to the right of AB; that is, R AB = R → A ′ B ′ . r r r 2 r 2 r 2 r A A ′ B B ′ What is the resistance R AB between the terminals A and B of this infinitely repeating chain of resistors. 1. R AB = r + 2 r R AB r + R AB , therefore R AB = 5 2 r 2. R AB = r + 2 r − R AB , therefore R AB = 3 2 r 3. R AB = r + 2 r R AB 2 r + R AB , therefore R AB = 2 r correct Explanation: R series = R 1 + R 2 + R 3 + · · · 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + · · · The infinite chain can be redrawn as follows r 2 r R AB A A ′ B B ′ R AB = r + 1 1 2 r + 1 R AB = r + 2 r R AB R AB + 2 r ( R AB − r ) ( R AB + 2 r ) = 2 r R AB R 2 AB − r R AB − 2 r 2 = 0 R AB = r ± radicalbig r 2 + 4 (2 r 2 ) 2 = r + 3 r 2 = 2 r . 003 10.0 points Calculate the average drift speed of electrons traveling through a copper wire with a cross- sectional area of 80 mm 2 when carrying a current of 60 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 / cm 3 . Avogadro’s number N A is 6 . 02 × 10 23 . Correct answer: 5 . 5369 × 10 − 5 m / s. fierro (jmf2547) – HW14 – berk – (60290) 2 Explanation: Let : N = 1 , M = 63 . 5 g / mol , ρ = 8 . 93 g / cm 3 , A = 80 mm 2 = 8 × 10 − 5 m 2 , I = 60 A , and q e = 1 . 6 × 10 − 19 C / electron . We first calculate n , the number of current- carrying electrons per unit volume in copper....

View
Full
Document