# HW 3 - segovia(cs39966 – Homework 3 – Spurlock...

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Unformatted text preview: segovia (cs39966) – Homework 3 – Spurlock – (14441) 1 This print-out should have 51 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 The potential difference in a simple circuit is 5 V and the resistance is 32 Ω. What current I flows in the circuit? Correct answer: 0 . 15625 A. Explanation: Let : V = 5 V and R = 32 Ω . The current is I = V R = 5 V 32 Ω = . 15625 A . 002 (part 2 of 2) 10.0 points How many electrons pass a given point in the circuit in 7 min? Correct answer: 4 . 10156 × 10 20 . Explanation: Let : I = 0 . 15625 A , t = 7 min = 420 s , and q e = 1 . 6 × 10 − 19 C . I = q t , where q is in coulombs and t in seconds; that is, q = I t, and the total charge for n electrons is q = n q e , so n = q q e = I t q e = (0 . 15625 A) (420 s) 1 . 6 × 10 − 19 C = 4 . 10156 × 10 20 . 003 10.0 points A 0 . 51 V potential difference is maintained across a 1 . 7 m length of tungsten wire that has a cross-sectional area of 0 . 46 mm 2 and the resistivity of the tungsten is 5 . 6 × 10 − 8 Ω · m. What is the current in the wire? Correct answer: 2 . 46429 A. Explanation: Let : V = 0 . 51 V , ℓ = 1 . 7 m , A = 0 . 46 mm 2 = 4 . 6 × 10 − 7 m 2 , and ρ = 5 . 6 × 10 − 8 Ω · m . The resistance is R = V I = ρ ℓ A , so the current is I = V A ρ ℓ = (0 . 51 V) (4 . 6 × 10 − 7 m 2 ) (5 . 6 × 10 − 8 Ω · m) (1 . 7 m) = 2 . 46429 A . 004 10.0 points At 31 . 5 ◦ C, the resistance of a segment of gold wire is 142 . 2 Ω. When the wire is placed in a liquid bath, the resistance decreases to 95 . 1 Ω. What is the temperature of the bath? The temperature coefficient for gold is . 0034 ( ◦ C) − 1 at 20 ◦ C. Correct answer:- 69 . 7278 ◦ C. Explanation: Let : T = 31 . 5 ◦ C , R 1 = 142 . 2 Ω , R 2 = 95 . 1 Ω , α = 0 . 0034 ( ◦ C) − 1 , and T = 20 ◦ C . segovia (cs39966) – Homework 3 – Spurlock – (14441) 2 The relation between resistance and tem- perature is R 1 = R [1 + α (Δ T )] R = R 1 1 + α Δ T = 142 . 2 Ω 1 + bracketleftBig . 0034 ( ◦ C) − 1 bracketrightBig (11 . 5 ◦ C) = 136 . 849 Ω . At the unknown temperature, R 2 = R [1 + α ( T b- T )] R 2 R = 1 + α ( T b- T ) R 2 R- 1 α = T b- T T b = T + R 2 R- 1 α = 20 ◦ C + 95 . 1 Ω 136 . 849 Ω- 1 . 0034 ( ◦ C) − 1 =- 69 . 7278 ◦ C . 005 10.0 points A 13 . 3 V battery delivers 95 . 8 mA when con- nected to a 31 . 8 Ω load. Determine the internal resistance of the battery. Correct answer: 107 . 031 Ω. Explanation: Let : E = 13 . 3 V , I = 95 . 8 mA = 0 . 0958 A , and R = 31 . 8 Ω . R i = E I- R = 13 . 3 V . 0958 A- 31 . 8 Ω = 107 . 031 Ω ....
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## This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.

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HW 3 - segovia(cs39966 – Homework 3 – Spurlock...

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