This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: kazi (mik88) – HW03 – tsoi – (57210) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The potential difference in a simple circuit is 6 V and the resistance is 47 Ω. What current I flows in the circuit? Correct answer: 0 . 12766 A. Explanation: Let : V = 6 V and R = 47 Ω . The current is I = V R = 6 V 47 Ω = . 12766 A . 002 (part 2 of 2) 10.0 points How many electrons pass a given point in the circuit in 4 min? Correct answer: 1 . 91489 × 10 20 . Explanation: Let : I = 0 . 12766 A , t = 4 min = 240 s , and q e = 1 . 6 × 10 19 C . I = q t , where q is in coulombs and t in seconds; that is, q = I t, and the total charge for n electrons is q = n q e , so n = q q e = I t q e = (0 . 12766 A) (240 s) 1 . 6 × 10 19 C = 1 . 91489 × 10 20 . 003 (part 1 of 2) 10.0 points Consider two conductors 1 and 2 made of the same ohmic material; i.e. , ρ 1 = ρ 2 . Denote the length by ℓ , the cross sectional area by A . The same voltage V is applied across the ends of both conductors. The field E is inside of the conductor. 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 A 2 = 2 A 1 , ℓ 2 = 2 ℓ 1 and V 2 = V 1 , find the ratio E 2 E 1 of the electric fields. 1. E 2 E 1 = 2 2. E 2 E 1 = 1 16 3. E 2 E 1 = 8 4. E 2 E 1 = 1 3 5. E 2 E 1 = 1 4 6. E 2 E 1 = 1 12 7. E 2 E 1 = 1 8. E 2 E 1 = 1 2 correct 9. E 2 E 1 = 1 8 10. E 2 E 1 = 4 Explanation: E 2 E 1 = V 2 ℓ 2 V 1 ℓ 1 = ℓ 1 ℓ 2 = ℓ 1 2 ℓ 1 = 1 2 . 004 (part 2 of 2) 10.0 points Determine the ratio I 2 I 1 of the currents . kazi (mik88) – HW03 – tsoi – (57210) 2 1. I 2 I 1 = 2 2. I 2 I 1 = 1 3 3. I 2 I 1 = 1 8 4. I 2 I 1 = 1 4 5. I 2 I 1 = 1 16 6. I 2 I 1 = 4 7. I 2 I 1 = 8 8. I 2 I 1 = 1 correct 9. I 2 I 1 = 1 2 10. I 2 I 1 = 1 12 Explanation: I 2 I 1 = V 2 R 2 V 1 R 1 = R 1 R 2 = ρ 1 parenleftbigg ℓ 1 A 1 parenrightbigg ρ 2 parenleftbigg ℓ 2 A 2 parenrightbigg = ℓ 1 A 1 2 ℓ 1 2 A 1 = 1 . 005 10.0 points Which of the following copper conductor con ditions has the least resistance? 1. thick, short, and cool correct 2. thin, long, and hot 3. thick, long, and hot 4. thin, long, and cool 5. thick, long, and cool 6. thin, short, and cool 7. thick, short, and hot 8. thin, short, and hot Explanation: We know that the resistance R of a conduc tor is determined by the resistivity ρ , cross sectional area A and length ℓ : R = ρ ℓ A ....
View
Full
Document
This note was uploaded on 02/21/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.
 Spring '08
 morrison
 Physics

Click to edit the document details