ch20-h2-solutions

# ch20-h2-solutions - This print-out should have 22 questions...

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Unformatted text preview: 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 . 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 Ω . Explanation: We can see that the voltage di ff erences in the circuits are the same. We use the expres- sion P = I Δ V. Since Δ V R = emf , (from the loop rule) and Δ V R = I R, we can write P = Δ V R R Δ V R = ( Δ V R ) 2 R ⇒ R 1 = ( Δ V R ) 2 P 1 = (10 V) 2 6 . 9 W = 14 . 4928 Ω . 002 (part 2 of 7) 10.0 points What is the resistance of R 2 ? Correct answer: 4 . 9505 Ω . Explanation: As in part 1, we can write R 2 = ( Δ V R ) 2 P 2 = (10 V) 2 20 . 2 W = 4 . 9505 Ω . 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 Δ V = E L ⇒ E = Δ 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 Ω = 0 . 514317 A ⇒ Δ V 1 = I R 1 = (0 . 514317 A)(14 . 4928 Ω ) = 7 . 45387 V ⇒ Δ V 2 = I R 2 = (0 . 514317 A)(4 . 9505 Ω ) = 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 0-5V (i...
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ch20-h2-solutions - This print-out should have 22 questions...

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