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Unformatted text preview: Version 050 – Second Exam – Hoffmann – (56745) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Many of the answers do not show units. Read beginning of each problem very carefully for information about the missing units. Also, some answers equate forces to unit vectors. For these answers you should interpret the ”=” sign to mean ”IS IN THE DIRECTION.” 001 10.0 points Consider two cylindrical conductors made of the same ohmic material. Conductor 1 has a radius r 1 and length ℓ 1 while conductor 2 has a radius r 2 and length ℓ 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential differences between the two ends of the conductors as V 1 and V 2 , and the electric fields within the conductors as E 1 and E 2 . V 1 vector E 1 I 1 ℓ 1 r 1 b V 2 vector E 2 I 2 ℓ 2 r 2 b If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 ℓ 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 4 2. R 2 R 1 = 3 4 correct 3. R 2 R 1 = 1 4 4. R 2 R 1 = 2 5. R 2 R 1 = 1 2 6. R 2 R 1 = 3 7. R 2 R 1 = 4 3 8. R 2 R 1 = 2 3 9. R 2 R 1 = 1 3 10. R 2 R 1 = 3 2 Explanation: The relation between resistance and resis tivity is given by R = ρ ℓ A = ρ ℓ π r 2 . Thus the ratio of the resistances is R 2 R 1 = ρ ℓ 2 π r 2 2 π r 2 1 ρ ℓ 1 = ℓ 2 ℓ 1 r 2 1 r 2 2 = 3 ℓ 1 ℓ 1 r 2 1 (2 r 1 ) 2 = 3 4 . 002 10.0 points The answer choices do not show units. The units are Volts (V). After a 2 . 73 Ω resistor is connected across a battery with a 0 . 26 Ω internal resistance, the electric potential between the physical battery terminals is 15 V. What is the rated emf of the battery? 1. 10.3101 2. 11.5063 3. 4.19459 4. 9.43299 5. 15.4167 6. 16.4286 7. 8.52352 8. 8.24528 9. 4.2603 10. 13.2686 Correct answer: 16 . 4286 V. Explanation: Let : R = 2 . 73 Ω , r = 0 . 26 Ω , and V = 15 V . Version 050 – Second Exam – Hoffmann – (56745) 2 The current drawn by the external resistor is given by I = V R = 15 V 2 . 73 Ω = 5 . 49451 A . The output voltage is reduced by the inter nal resistance of the battery by V = E − I r , so the electromotive force is E = V + I r = 15 V + (5 . 49451 A) (0 . 26 Ω) = 16 . 4286 V . 003 10.0 points The answer choices do not show units. The units are microFarads. Consider the capacitor circuit 64 V 33 μ F 48 μ F 51 μ F 30 μ F What is the effective capacitance of the circuit? 1. 35.0 2. 32.0 3. 37.5 4. 38.0 5. 35.5 6. 40.5 7. 42.0 8. 31.5 9. 39.5 10. 29.5 Correct answer: 40 . 5 μ F. Explanation: Let : C 1 = 33 μ F , C 2 = 48 μ F , C 3 = 51 μ F , C 4 = 30 μ F , and E B = 64 V . E B C 1 C 2 C 3 C 4 For capacitors in series, 1 C series = summationdisplay 1 C i V series = summationdisplay V i , and the individual charges are the same....
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 Spring '08
 Turner
 Magnetic Field, Electric charge, answer choices

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