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**Unformatted text preview: **fierro (jmf2547) – HW 16 – berk – (60290) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 10 . 7 g wire has a density of 6 . 9 g / cm 3 and a resistivity of 1 . 37 × 10 − 7 Ω m. The wire has a resistance of 80 Ω. How long is the wire? Correct answer: 30 . 0921 m. Explanation: Let : m = 10 . 7 g = 0 . 0107 kg , ρ = 6 . 9 g / cm 3 = 6900 kg / m 3 , r = 1 . 37 × 10 − 7 Ω m , and R = 80 Ω . ρ is density, and r is the resistivity of the wire, so m = ρ V V = m ρ = A ℓ A = m ρ ℓ Thus the resistance is R = r ℓ A = r ℓ m ρ ℓ = r ρ ℓ 2 m ℓ 2 = R m ρ r ℓ = radicalBigg R m ρ r = radicalBigg (80 Ω) (0 . 0107 kg) (6900 kg / m 3 ) (1 . 37 × 10 − 7 Ω m) = 30 . 0921 m . 002 (part 2 of 2) 10.0 points The wire is made up of atoms with valence 4 and mass 26 kg, where u = 1 . 6605 × 10 − 27 kg. What is the drift speed of the electrons when there is a voltage drop of 176 V across the wire? Correct answer: 0 . 000171615 m / s. Explanation: Let : n con = 4 , m μ = 26 kg = 26 kg , V = 176 V , N A = 6 . 02 × 10 23 , ρ = 6 . 9 g / cm 3 = 6900 kg / m 3 , ℓ = 30 . 0921 m , and q e = 1 . 602 × 10 − 19 C , where m μ is atomic mass, n con is number of conduction electron per an atom, and N A is Avogadro’s number. Since the current I is given by I = n q A v d , where n is the density of charge carriers, q the charge on an electron and v d the drift speed, we have v d = I n q e A = V n q e A R . (1) The density of charge carriers is found from knowing how many electrons there are per atom and then the total number of atoms in the wire. This density is n = n con ρ parenleftbigg m μ N A parenrightbigg = n con ρ N A m μ = (4) (6900 kg / m 3 ) (6 . 02 × 10 23 ) (26 kg) = 6 . 39046 × 10 26 m − 3 . Here, the mass of one atom is given by parenleftbigg Atomic mass Avogadro ′ s number parenrightbigg . fierro (jmf2547) – HW 16 – berk – (60290) 2 Since the volume is the cross-sectional area multiplied by the length this tells us that A ℓ = m ρ , so A = m ρ ℓ = (26 kg) (6900 kg / m 3 ) (30 . 0921 m) = 0 . 00012522 m 2 , thus v d = V n q e A R (1) = (176 V) (6 . 39046 × 10 26 m − 3 ) × 1 (1 . 602 × 10 − 19 C) × 1 (0 . 00012522 m 2 ) × 1 (80 Ω) = . 000171615 m / s . 003 10.0 points A length of wire is cut into 8 equal pieces. The 8 pieces are then connected parallel, with the resulting resistance being 3 Ω. What was the resistance r of the original length of wire? Correct answer: 192 Ω. Explanation: Let : n = 8 and R p = 3 Ω ....

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