This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Kapoor (mk9499) – homework15 – Turner – (60230) 1 This printout should have 12 questions. Multiplechoice 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 = 1 3 2. R 2 R 1 = 1 4 3. R 2 R 1 = 4 3 4. R 2 R 1 = 3 4 correct 5. R 2 R 1 = 1 2 6. R 2 R 1 = 4 7. R 2 R 1 = 3 2 8. R 2 R 1 = 2 3 9. R 2 R 1 = 2 10. R 2 R 1 = 3 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 2. E 2 E 1 = 1 4 3. E 2 E 1 = 2 4. E 2 E 1 = 1 2 5. E 2 E 1 = 4 3 6. E 2 E 1 = 2 3 7. E 2 E 1 = 1 3 correct 8. E 2 E 1 = 4 9. E 2 E 1 = 3 2 Kapoor (mk9499) – homework15 – Turner – (60230) 2 10....
View
Full Document
 Fall '08
 Turner
 Physics, Electric Potential, Work, Resistor, Electrical resistance

Click to edit the document details