# 47132_HW5 - ragsdale(zdr82 – HW5 – ditmire –(58335 1...

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Unformatted text preview: ragsdale (zdr82) – HW5 – ditmire – (58335) 1 This print-out should have 20 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 quantity of charge passing through a sur- face of area 2 . 6 cm 2 varies with time as q = q 1 t 3 + q 2 t + q 3 , where q 1 = 5 . 8 C / s 3 , q 2 = 6 . 7 C / s, q 3 = 6 . 2 C, and t is in seconds. What is the instantaneous current through the surface at t = 1 . 1 s? Correct answer: 27 . 754 A. Explanation: Let : q 1 = 5 . 8 C / s 3 , q 2 = 6 . 7 C / s , q 3 = 6 . 2 C , and t = 1 . 1 s . I ≡ d q dt = 3 q 1 t 2 + q 2 = 3 ( 5 . 8 C / s 3 ) (1 . 1 s) 2 + 6 . 7 C / s = 27 . 754 A . 002 (part 2 of 2) 10.0 points What is the value of the current density at t = 1 . 1 s? Correct answer: 1 . 06746 × 10 5 A / m 2 . Explanation: Let : a = 2 . 6 cm 2 = 0 . 00026 m 2 and t = 1 . 1 s . J ≡ I A = 27 . 754 A . 00026 m 2 = 1 . 06746 × 10 5 A / m 2 . 003 10.0 points An incandescent bulb has a resistance of 27 Ω when it is at room temperature (25 ◦ C) and 432 Ω when it is hot and delivering light to the room. The temperature coefficient of re- sistivity of the filament is 0 . 006 ( ◦ C) − 1 , where the base resistance R is determined at 0 ◦ C. What is the temperature of the bulb when in use? Correct answer: 2900 ◦ C. Explanation: Let : R 1 = 27 Ω , R 2 = 432 Ω , and α = 0 . 006 ( ◦ C) − 1 . Temperature dependence of resistance is R T = R [1 + α ( T- ◦ C)] = R [1 + α Δ T ] . (1) Using Eq. (1), at room temperature; e.g. , R 1 = 27 Ω and Δ T 1 = 25 ◦ C R = R 1 [1 + α Δ T 1 ] (2) = 27 Ω 1 + [0 . 006 ( ◦ C) − 1 ] (25 ◦ C) = 23 . 4783 Ω . Using Eq. (1), with resistance R 2 , Δ T 2 = R 2 R- 1 α (3) = 432 Ω 23 . 4783 Ω- 1 . 006 ( ◦ C) − 1 = 2900 ◦ C . Or directly substituting Eq. (1) into (2), we have Δ T 2 = R 2 R- 1 α ragsdale (zdr82) – HW5 – ditmire – (58335) 2 = R 2 R 1 [1 + α Δ T 1 ]- 1 α = R 2 + α R 2 Δ T 1- R 1 α = R 2- R 1 α R 1 + R 2 R 1 Δ T 1 (4) = 432 Ω- 27 Ω [0 . 006 ( ◦ C) − 1 ] (27 Ω) + 432 Ω 27 Ω (25 ◦ C) = 2900 ◦ C . 004 (part 1 of 4) 10.0 points If the current carried by a metallic conductor is doubled, (a) What happens to the charge carrier density? Assume that the temperature of the metallic conductor remains constant. 1. It is unchanged. correct 2. It quadruples. 3. It doubles. 4. It is quartered. 5. It is halved. Explanation: J = σ E = n q v d v d = q E m τ . The charge carrier density n is unaffected. This is simply a property of the material, and is related to the mass density and the number of valence electrons available per atom. 005 (part 2 of 4) 10.0 points (b) What happens to the current density?...
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47132_HW5 - ragsdale(zdr82 – HW5 – ditmire –(58335 1...

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