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Unformatted text preview: thwing (jrt2329) – Homework 05 – Yao – (59110) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The current in a wire decreases with time according to the relationship I = (1 . 54 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 . 0115546 C. Explanation: I = dq dt q = integraldisplay t t =0 I dt = integraldisplay ∞ t =0 (0 . 00154 A) e − . 13328 s − 1 t dt = (0 . 00154 A ) e − . 13328 s − 1 t . 13328 s − 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ∞ = . 0115546 C . 002 10.0 points If 5 × 10 21 electrons pass through a 20 Ω resis tor in 14 min, what is the potential difference across the resistor? Correct answer: 19 . 0714 V. Explanation: Let : n = 5 × 10 21 , Δ t = 14 min , R = 20 Ω , and q e = 1 . 602 × 10 − 19 C . The current is defined as I = Δ Q Δ t . Since each electron carries a q e , we have I = n q e Δ t = ( 5 × 10 21 ) (1 . 602 × 10 − 19 C) 14 min parenleftbigg 1 min 60 s parenrightbigg = 0 . 953571 A . Then from Ohm’s Law, V = I R = (0 . 953571 A) (20 Ω) = 19 . 0714 V . 003 10.0 points A conductor with crosssectional area 4 cm 2 carries a current of 9 A. If the concentration of free electrons in the conductor is 8 × 10 28 electrons / m 3 , what is the drift velocity of the electrons? Correct answer: 0 . 00175542 mm / s. Explanation: Let : I = 9 A , n = 8 × 10 28 electrons / m 3 , q e = 1 . 60218 × 10 − 19 C , and A = 4 cm 2 = 0 . 0004 m 2 . The current in a conductor is given by I = n q v d A , where n is the number of charge carriers per unit volume, q is the charge per carrier, v d is the drift velocity of the carriers and A is the cross section of the conduction. Solving for v d , we have v d = I n q A = 9 A 8 × 10 28 electrons / m 3 × 1 1 . 60218 × 10 − 19 C × 1 . 0004 m 2 × 1000 mm 1 m = . 00175542 mm / s . thwing (jrt2329) – Homework 05 – Yao – (59110) 2 004 (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....
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This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas.
 Spring '09
 YAO
 Physics, Current, Work

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