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Unformatted text preview: Version 046 – TEST01 – tsoi – (57210) 1 This printout should have 16 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 following diagram shows part of a closed electrical circuit. I E X Y 8 Ω 37 Ω 36 Ω Find the electric resistance R XY of the part of the circuit shown between point X and Y . 1. R XY = 6 Ω . 2. R XY = 10 Ω . 3. R XY = 8 Ω . 4. R XY = 18 Ω . 5. R XY = 15 Ω . 6. R XY = 42 Ω . 7. R XY = 20 Ω . correct 8. R XY = 9 Ω . 9. R XY = 3 Ω . 10. R XY = 24 Ω . Explanation: I E X Y R 1 R 2 R 3 Let : R 1 = 8 Ω , R 2 = 37 Ω , and R 3 = 36 Ω . Since R 1 and R 2 are in series, their equiva lent resistance R 12 is R 12 = R 1 + R 2 = 8 Ω + 37 Ω = 45 Ω . I E X Y R 12 45 Ω R 3 36 Ω Since R 12 and R 3 are connected parallel, their equivalent resistance R XY is 1 R XY = 1 R 12 + 1 R 3 = R 3 + R 12 R 12 R 3 R XY = R 12 R 3 R 12 + R 3 = (45 Ω) (36 Ω) 45 Ω + 36 Ω = 1620 Ω 2 81 Ω = 20 Ω . 002 (part 2 of 2) 10.0 points When there is a steady current in the circuit, the amount of charge passing a point per unit time is 1. greater at point X than at point Y . 2. greater in the 37 Ω resistor than in the 36 Ω resistor. 3. the same everywhere in the circuit. 4. greater in the 36 Ω resistor than in the 37 Ω resistor. correct 5. greater in the 8 Ω resistor than in the 37 Ω resistor. Version 046 – TEST01 – tsoi – (57210) 2 Explanation: The amount of charge passing a point per unit of time (the same as the current at a point) is not the same everywhere, but it is the same at point X as at point Y ; i.e. , it is the same in the 8 Ω resistor as in the 37 Ω resistor. It is greater in the 36 Ω resistor than in the 8 Ω or 37 Ω resistor. From Ohm’s Law E = I R , we have I 12 = E R 12 = E 45 Ω I 3 = E R 3 = E 36 Ω I 3 > I 12 . When there is a steady current in the cir cuit, the amount of charge passing a point per unit time is greater in the smaller 36 Ω resistor than in the larger 37 Ω resistor. 003 (part 1 of 2) 10.0 points Consider the RC circuit shown. The emf of the battery is V , the resistance R , and the capacitance C . The capacitor is uncharged. R C V S Consider the following statements. A1: I = 0 A2: I = V R A3: I = V R e 1 /RC B1: V C = 0 B2: V C = V B3: V C = V e 1 /RC Immediately after the switch is closed, the current, I , and the potential across the capac itor, V C , are respectively given by 1. A3, B1 2. A3, B3 3. A2, B1 correct 4. A2, B2 5. A3, B2 6. A1, B2 7. A1, B3 8. A2, B3 9. A1, B1 Explanation: It takes time to charge the capacitor. Im mediately after the switch is closed, the capac itor has zero charge, so V C = 0. Based on the loop equation V IR V C = 0, V IR = 0 or I = V R ....
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This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas.
 Spring '10
 TSOI
 Physics

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