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solution_pdf - dang(rtd356 – HW05 – li –(59050 This...

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dang (rtd356) – HW05 – li – (59050) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points IF a metal wire carries a current oF 100 mA, how long does it take For 13.0 × 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: 2080 s. Explanation: Let : q e = 1 . 6 × 10 19 C , I = 100 × 10 2 A and N = 13 . 0 × 10 20 . I = Δ Q Δ t = N q e Δ t Δ t = N q e I = (1 . 3 × 10 21 )(1 . 6 × 10 19 C) 0 . 1 A = 2080 s . 002 (part 1 oF 2) 10.0 points The current in a conductor varies over time as shown in the fgure. 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 0 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 = 2 s? Correct answer: 8 C. Explanation: Let : Δ t = 2 s . The charge is the area under the I vs t graph, represented here by two rectangles and two triangles: Δ Q = (2 A)(2 s) + (6 A - 2 A)(0 . 5 s) + 1 2 (6 A - 2 A)(0 . 5 s) + 1 2 (6 A - 2 A)(0 . 5 s) = 8 C . 003 (part 2 oF 2) 10.0 points What constant current would transport the same total charge during the 2 s interval as does the actual current? Correct answer: 4 A. Explanation: I = Δ Q Δ t = 8 C 2 s = 4 A . 004 10.0 points The current in a wire decreases with time according to the relationship I = (3 . 99 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 . 029937 C. Explanation: I = dq dt q = i t t =0 I dt = i t =0 (0 . 00399 A) e 0 . 13328 s - 1 t dt = (0 . 00399 A ) e 0 . 13328 s - 1 t - 0 . 13328 s 1 v v v v v 0 = 0 . 029937 C .
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dang (rtd356) – 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 1 while conductor 2 has a radius r 2 and length 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential diFerences between the two ends of the conductors as V 1 and V 2 , and the electric ±elds within the conductors as E 1 and E 2 . V 1 v E I r 1 b 2 r 2 b If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 1 and V 2 = V 1 , ±nd the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 1 3 2. R 2 R 1 = 3 4 correct 3. R 2 R 1 = 3 4. R 2 R 1 = 4 5. R 2 R 1 = 2 6. R 2 R 1 = 2 3 7. R 2 R 1 = 4 3 8. R 2 R 1 = 1 2 9. R 2 R 1 = 1 4 10. R 2 R 1 = 3 2 Explanation: The relation between resistance and resis- tivity is given by R = ρ ℓ A = ρ ℓ π r 2 . Then since r 2 = 2 r 1 and 2 = 3 1 , the ratio of the resistances is R 2 R 1 = ρ ℓ 2 π r 2 2 π r 2 1 ρ ℓ 1 = 2 r 2 1 1 r 2 2 = (3 1 ) r 2 1 1 (2 r 1 ) 2 = 3 4 . 006 (part 2 of 3) 10.0 points When the two conductors are attached to a battery of voltage V, determine the ratio E 2 E 1 of the electric ±elds. 1. E 2 E 1 = 3 2. E 2 E 1 = 1 2 3.
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