# M g d v m g d tan θ q 0 000148 kg 9 8 m s 2 0 059 m

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m g d V = m g d tan θ q = (0 . 000148 kg) (9 . 8 m / s 2 ) (0 . 059 m) 2 . 7 × 10 - 8 C × tan 18 = 1 . 0298 kV . 021 (part 1 of 2) 10 points Consider the circuit shown below. There are three resistors, r 1 , r 2 , and R ; and two emf’s, E 1 and E 2 . The directions of the cur- rents i 1 , i 2 , and i 3 are shown in the figure. E F A B D C A E 1 E 2 i 1 r 1 i 3 R i 2 r 2 Apply Kirchhoff’s rules. Neglect the differ- ence of an overall sign, if applicable. What equation does the loop ABCDA yield? 1. E 1 + E 2 - i 2 r 2 + i 1 r 1 = 0 2. E 1 - E 2 - i 1 r 2 + i 2 r 1 = 0 3. E 1 + E 2 + i 2 r 2 + i 1 r 1 = 0 4. E 1 - E 2 + i 2 r 2 + i 1 r 1 = 0 5. E 1 + E 2 + i 2 r 2 - i 1 r 1 = 0 6. E 1 - E 2 - i 1 r 2 - i 2 r 1 = 0 7. E 1 + E 2 - i 2 r 2 - i 1 r 1 = 0 8. E 1 - E 2 - i 2 r 2 + i 1 r 1 = 0 9. E 1 - E 2 - i 2 r 2 - i 1 r 1 = 0 10. E 1 - E 2 + i 2 r 2 - i 1 r 1 = 0 correct Explanation: Recall that Kirchhoff’s loop rule states that the sum of the potential differences across all the elements around a closed circuit loop is zero. If a resistor is traversed in the direction of the current, the change in potential is - i R . If an emf source is traversed from the - to + terminals, the change in potential is + E . Apply the opposite sign for traversing the elements in the opposite direction. Hence, by inspection up to an overall sign, following is the correct equation ABCDA : E 1 - E 2 + i 2 r 2 - i 1 r 1 = 0 . 022 (part 2 of 2) 10 points Let E 1 = E 2 = 5 V, and r 1 = r 2 = 4 . 5 Ω, and R = 1 . 9 Ω. E F A B D C A 5 V 5 V i 1 4 . 5 Ω i 3 1 . 9 Ω i 2 4 . 5 Ω Find the current i 3 . Hint: From symmetry, one expects i 1 = i 2 . Correct answer: 1 . 20482 A. Explanation: We are given that E 1 = E 2 and r 1 = r 2 . This implies that i 1 = i 2 . Hence the junction rule yields i 1 + i 2 = 2 i 2 = i 3 i 2 = i 3 2 .
Platt, David – Quiz 2 – Due: Oct 18 2005, 10:00 pm – Inst: Ken Shih 13 Substituting this into the loop equation DCFED gives E 2 - i 3 R - i 3 2 r 2 = 0 . Solving for i 3 yields i 3 = E 2 R + r 2 2 = (5 V) (1 . 9 Ω) + (4 . 5 Ω) 2 = 1 . 20482 A . 023 (part 1 of 2) 10 points Consider two conductors 1 and 2 made of the same ohmic material; i.e. , ρ 1 = ρ 2 . Denote the length by , 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 ~ E 1 I 1 1 r 1 V 2 ~ E 2 I 2 2 r 2 If A 2 = 2 A 1 , ‘ 2 = 2 1 and V 2 = V 1 , find the ratio E 2 E 1 of the electric fields. 1. E 2 E 1 = 1 8 2. E 2 E 1 = 1 3. E 2 E 1 = 2 4. E 2 E 1 = 1 16 5. E 2 E 1 = 1 3 6. E 2 E 1 = 4 7. E 2 E 1 = 1 2 correct 8. E 2 E 1 = 8 9. E 2 E 1 = 1 12 10. E 2 E 1 = 1 4 Explanation: E 2 E 1 = V 2 2 V 1 1 = 1 2 = 1 2 1 = 1 2 . 024 (part 2 of 2) 10 points Determine the ratio I 2 I 1 of the currents . 1. I 2 I 1 = 8 2. I 2 I 1 = 1 3 3. I 2 I 1 = 2 4. I 2 I 1 = 4 5. I 2 I 1 = 1 2 6. I 2 I 1 = 1 correct 7. I 2 I 1 = 1 8 8. I 2 I 1 = 1 4 9. I 2 I 1 = 1 16 10. I 2 I 1 = 1 12 Explanation:
Platt, David – Quiz 2 – Due: Oct 18 2005, 10:00 pm – Inst: Ken Shih 14 I 2 I 1 = V 2 R 2 V 1 R 1 = R 1 R 2 = ρ 1 1 A 1 ρ 2 2 A 2 = 1 A 1 2 1 2 A 1 = 1 .
Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih 1 This print-out should have 27 questions.
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