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# HW10Sol - Platt David Homework 10 Due 4:00 am Inst Ken Shih...

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Platt, David – Homework 10 – Due: Sep 30 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 1) For 002, v is the average thermal speed of the free electron, not the average drift speed. For the same carriers the average thermal speed only depends on Temperature . 2) For 003, Think about metallic material only. Semiconductor has different story. 001 (part 1 of 1) 10 points A length of copper wire has a resistance 49 Ω. The wire is cut into three pieces of equal length. The pieces are then connected as parallel lengths between points A and B. What resistance will this new “wire” of length L 0 3 have between points A and B? Correct answer: 5 . 44444 Ω. Explanation: Let : R 0 = 49 Ω . The new wire has length L = L 0 3 and cross- section A = 3 A 0 , so its resistance is R = ρ L A = ρ L 0 3 3 A 0 = 1 9 ρ L 0 A 0 = R 0 9 = 49 Ω 9 = 5 . 44444 Ω . 002 (part 1 of 3) 10 points Consider the application of two potential dif- ferences across the same metallic ohmic con- ductor where the conductor is maintained at a constant temperature of 300 K. V 1 J 1 L R V 2 J 2 L R Case 1: The potential difference is V 1 , the current density is J 1 , and the average thermal speed of the free electron within the conductor is v 1 . Case 2: The corresponding quantities are V 2 , J 2 , and v 2 . If V 2 = 2 V 1 , the corresponding ratio J 2 J 1 of current densities is given by 1. J 2 J 1 = 1 8 . 2. J 2 J 1 = 1 . 3. J 2 J 1 = 2 . correct 4. J 2 J 1 = 1 2 . 5. J 2 J 1 = 16 . 6. J 2 J 1 = 1 16 . 7. J 2 J 1 = 8 . 8. J 2 J 1 = 1 4 . 9. J 2 J 1 = 4 . Explanation: Using the relation between current density for an ohmic material E = ρ J , we find J 2 J 1 = E 2 ρ E 1 ρ = E 2 E 1

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Platt, David – Homework 10 – Due: Sep 30 2005, 4:00 am – Inst: Ken Shih 2 = V 2 L V 1 L = V 2 V 1 = 2 .
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