ch20-h2-solutions - pan (zp695) ch20-h2 chiu (56565) 1 This...

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Unformatted text preview: pan (zp695) ch20-h2 chiu (56565) 1 This print-out should have 22 questions. Multiple-choice 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 . 6 W; in circuit 2, ohmic resistor R 2 dissipates 20 . 09 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: 15 . 1515 . 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 . 6 W = 15 . 1515 . 002 (part 2 of 7) 10.0 points What is the resistance of R 2 ? Correct answer: 4 . 9776 . Explanation: As in part 1, we can write R 2 = ( V R ) 2 P 2 = (10 V) 2 20 . 09 W = 4 . 9776 . 003 (part 3 of 7) 10.0 points Resistor R 1 is made of a very thin metal wire that is 3 . 7 mm long, with a diameter of 0 . 1 mm. What is the electric field inside this metal resistor? Correct answer: 2702 . 7 V / m. Explanation: The electric field is given by V = E L E = V L = 10 V 3 . 7 mm = 2702 . 7 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 pan (zp695) ch20-h2 chiu (56565) 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 5. Potential 10 V R 1 R 2 6. Potential 10 V R 1 R 2 correct 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 20 . 1291 = 0 . 496793 A V 1 = I R 1 = (0 . 496793 A)(15 . 1515 ) = 7 . 52716 V V 2 = I R 2 = (0 . 496793 A)(4 . 9776 ) = 2 . 47284 V If the negative terminal of the battery if ground ( V = 0), then the electric potential V drops from 10 V to 2 . 47284 V to zero. It is 2 . 47284 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 ++ ++ + + + + ++ ++ ++-------------- pan (zp695) ch20-h2 chiu (56565) 3 3. R 1 10 V R 2 Circuit 3 ++ ++ + + + + ++ ++ ++ ++ ++---------- 4....
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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ch20-h2-solutions - pan (zp695) ch20-h2 chiu (56565) 1 This...

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