This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: pan (zp695) ch20h2 chiu (56565) 1 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 . 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) ch20h2 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) ch20h2 chiu (56565) 3 3. R 1 10 V R 2 Circuit 3 ++ ++ + + + + ++ ++ ++ ++ ++ 4....
View
Full
Document
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.
 Spring '08
 Turner

Click to edit the document details