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Unformatted text preview: This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 7) 10.0 points Consider the two circuits depicted in the figure below. In circuit 1, ohmic resistor R 1 dissipates 6 . 9 W; in circuit 2, ohmic resistor R 2 dissipates 20 . 2 W. The wires and bat teries have negligible resistance. The circuits contain 10 V batteries. R 1 10 V Circuit 1 R 2 10 V Circuit 2 What is the resistance of R 1 ? Correct answer: 14 . 4928 . Explanation: We can see that the voltage di ff erences in the circuits are the same. We use the expres sion P = I V. Since V R = emf , (from the loop rule) and V R = I R, we can write P = V R R V R = ( V R ) 2 R R 1 = ( V R ) 2 P 1 = (10 V) 2 6 . 9 W = 14 . 4928 . 002 (part 2 of 7) 10.0 points What is the resistance of R 2 ? Correct answer: 4 . 9505 . Explanation: As in part 1, we can write R 2 = ( V R ) 2 P 2 = (10 V) 2 20 . 2 W = 4 . 9505 . 003 (part 3 of 7) 10.0 points Resistor R 1 is made of a very thin metal wire that is 4 . 1 mm long, with a diameter of 0 . 1 mm. What is the electric field inside this metal resistor? Correct answer: 2439 . 02 V / m. Explanation: The electric field is given by V = E L E = V L = 10 V 4 . 1 mm = 2439 . 02 V / m . 004 (part 4 of 7) 10.0 points The same resistors are used to construct cir cuit 3 as in the figure below, using the same 10 V battery as before. R 1 10 V R 2 Circuit 3 Which of the following graphs of potential versus location is correct? 1. Potential 10 V R 1 R 2 2. Potential 10 V R 1 R 2 3. Potential 10 V R 1 R 2 4. Potential 10 V R 1 R 2 correct 5. Potential 10 V R 1 R 2 6. Potential 10 V R 1 R 2 Explanation: From the loop rule, which tells us that emf I R 1 I R 2 = 0, we can write I = emf R 1 + R 2 = 10 V 19 . 4432 = 0 . 514317 A V 1 = I R 1 = (0 . 514317 A)(14 . 4928 ) = 7 . 45387 V V 2 = I R 2 = (0 . 514317 A)(4 . 9505 ) = 2 . 54613 V If the negative terminal of the battery if ground ( V = 0), then the electric potential V drops from 10 V to 2 . 54613 V to zero. It is 2 . 54613 V between R 1 and R 2 , 10 V at x = 0, and zero after R 2 . 005 (part 5 of 7) 10.0 points Of the following diagrams, which indicates the correct surface charge distribution of +s and s on the circuit? 1. R 1 10 V R 2 Circuit 3 ++ ++ + + + + ++ cor rect 2. R 1 10 V R 2 Circuit 3 ++ ++ + + + + ++ ++ ++ ++ ++ 3. R 1 10 V R 2 Circuit 3 ++ ++ + + + + ++ ++ ++ 4. R 1 10 V R 2 Circuit 3 ++ ++ + + + + ++ ++ ++ 5. R 1 10 V R 2 Circuit 3 ++ ++ ++ ++ ++ ++ + + + + ++ Explanation: Since the circuit as a whole is neutral, there should be an equal number of positive and negative surface charges. Since the posi tive terminal of the battery should be pos itively charged and negative terminal nega tively charged, we can say that all the points in the circuit with potential between 05V (i...
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 Spring '08
 Turner
 Physics

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