hw5 - homework 05 LIN KEVIN Due 4:00 am Question 1 chap 27...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
homework 05 – LIN, KEVIN – Due: Feb 17 2008, 4:00 am 1 Question 1, chap 27, sect 3. part 1 of 2 10 points The quantity of charge passing through a surface of area 1 . 84 cm 2 varies with time as q = q 1 t 3 + q 2 t + q 3 , where q 1 = 3 . 9 C / s 3 , q 2 = 4 C / s, q 3 = 8 . 7 C, and t is in seconds. What is the instantaneous current through the surface at t = 1 . 5 s? Correct answer: 30 . 325 A (tolerance ± 1 %). Explanation: Let : q 1 = 3 . 9 C / s 3 , q 2 = 4 C / s , q 3 = 8 . 7 C , and t = 1 . 5 s . I d q dt = 3 q 1 t 2 + q 2 = 3 ( 3 . 9 C / s 3 ) (1 . 5 s) 2 + 4 C / s = 30 . 325 A . Question 2, chap 27, sect 3. part 2 of 2 10 points What is the value of the current density at t = 1 . 5 s? Correct answer: 164810 A / m 2 (tolerance ± 1 %). Explanation: Let : A = 1 . 84 cm 2 = 0 . 000184 m 2 and t = 1 . 5 s . J I A = 30 . 325 A 0 . 000184 m 2 = 164810 A / m 2 . Question 3, chap 27, sect 2. part 1 of 3 10 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 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 1 r 1 V 2 vector E 2 I 2 2 r 2 If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 1 4 2. R 2 R 1 = 3 2 3. R 2 R 1 = 3 4. R 2 R 1 = 1 2 5. R 2 R 1 = 4 6. R 2 R 1 = 1 3 7. R 2 R 1 = 2 3 8. R 2 R 1 = 4 3 9. R 2 R 1 = 3 4 correct 10. R 2 R 1 = 2
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
homework 05 – LIN, KEVIN – Due: Feb 17 2008, 4:00 am 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 . Question 4, chap 27, sect 2. part 2 of 3 10 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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern