# 6 - karna (pk4534) – HW05 – li – (59050) 1 This...

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Unformatted text preview: karna (pk4534) – HW05 – li – (59050) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If a metal wire carries a current of 81.0 mA, how long does it take for 9.00 × 10 20 electrons to pass a given cross-sectional area anywhere along the wire? The magnitude of the charge on an electron is 1 . 6 × 10 − 19 C. Correct answer: 1777 . 78 s. Explanation: Let : q e = 1 . 6 × 10 − 19 C , I = 81 . × 10 − 2 A and N = 9 . 00 × 10 20 . I = Δ Q Δ t = N q e Δ t Δ t = N q e I = (9 × 10 20 )(1 . 6 × 10 − 19 C) . 081 A = 1777 . 78 s . 002 (part 1 of 2) 10.0 points The current in a conductor varies over time as shown in the figure. 1 2 3 4 5 6 7 1 2 3 4 5 6 0 1 2 3 4 5 6 7 1 2 3 4 5 6 Current(A) Time(s) How much charge passes through a cross section of the conductor in the time interval t = 0 s to t = 4 . 5 s? Correct answer: 18 C. Explanation: Let : Δ t = 4 . 5 s . The charge is the area under the I vs t graph, represented here by two rectangles and two triangles: Δ Q = (2 A)(4 . 5 s) + (6 A- 2 A)(1 s) + 1 2 (6 A- 2 A)(1 s) + 1 2 (6 A- 2 A)(1 s) = 18 C . 003 (part 2 of 2) 10.0 points What constant current would transport the same total charge during the 4 . 5 s interval as does the actual current? Correct answer: 4 A. Explanation: I = Δ Q Δ t = 18 C 4 . 5 s = 4 A . 004 10.0 points The current in a wire decreases with time according to the relationship I = (1 . 95 mA) × e − a t where a = 0 . 13328 s − 1 . Determine the total charge that passes through the wire from t = 0 to the time the current has diminished to zero. Correct answer: 0 . 0146309 C. Explanation: I = dq dt q = integraldisplay t t =0 I dt = integraldisplay ∞ t =0 (0 . 00195 A) e − . 13328 s- 1 t dt = (0 . 00195 A ) e − . 13328 s- 1 t- . 13328 s − 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ∞ = . 0146309 C . karna (pk4534) – HW05 – li – (59050) 2 005 (part 1 of 3) 10.0 points Consider two cylindrical conductors made of the same ohmic material. Conductor 1 has a radius r 1 and length ℓ 1 while conductor 2 has a radius r 2 and length ℓ 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential differences between the two ends of the conductors as V 1 and V 2 , and the electric fields within the conductors as E 1 and E 2 . V 1 vector E 1 I 1 ℓ 1 r 1 b V 2 vector E 2 I 2 ℓ 2 r 2 b If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 ℓ 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 3 4 correct 2. R 2 R 1 = 1 2 3. R 2 R 1 = 1 4 4. R 2 R 1 = 2 5. R 2 R 1 = 4 3 6. R 2 R 1 = 3 2 7. R 2 R 1 = 1 3 8....
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## This note was uploaded on 11/02/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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6 - karna (pk4534) – HW05 – li – (59050) 1 This...

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