# Hw15_pdf - mcconnell(kam2342 – homework15 – Turner...

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Unformatted text preview: mcconnell (kam2342) – homework15 – Turner – (60230) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (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 2. R 2 R 1 = 2 3. R 2 R 1 = 3 2 4. R 2 R 1 = 2 3 5. R 2 R 1 = 4 3 6. R 2 R 1 = 1 3 7. R 2 R 1 = 4 8. R 2 R 1 = 3 4 correct 9. R 2 R 1 = 1 2 10. R 2 R 1 = 1 4 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 ℓ 2 = 3 ℓ 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 = ℓ 2 r 2 1 ℓ 1 r 2 2 = (3 ℓ 1 ) r 2 1 ℓ 1 (2 r 1 ) 2 = 3 4 . 002 (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 fields. 1. E 2 E 1 = 3 4 2. E 2 E 1 = 1 2 3. E 2 E 1 = 3 2 4. E 2 E 1 = 1 4 5. E 2 E 1 = 4 3 6. E 2 E 1 = 4 7. E 2 E 1 = 2 3 8. E 2 E 1 = 1 3 correct 9. E 2 E 1 = 3 mcconnell (kam2342) – homework15 – Turner – (60230) 2 10....
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## This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.

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Hw15_pdf - mcconnell(kam2342 – homework15 – Turner...

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