Second Exam-solutions

Second Exam-solutions - Version 050 Second Exam...

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Version 050 – Second Exam – Hofmann – (56745) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. Many o± the answers do not show units. Read beginning o± each problem very care±ully ±or in±ormation about the missing units. Also, some answers equate ±orces to unit vectors. ²or these answers you should interpret the ”=” sign to mean ”IS IN THE DIRECTION.” 001 10.0 points Consider two cylindrical conductors made o± the same ohmic material. Conductor 1 has a radius r 1 and length 1 while conductor 2 has a radius r 2 and length 2 . Denote: The currents o± the two conductors as I 1 and I 2 , the potential diferences between the two ends o± the conductors as V 1 and V 2 , and the electric Felds within the conductors as E 1 and E 2 . V 1 v E I r 1 b 2 r 2 b ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 1 and V 2 = V 1 , Fnd the ratio R 2 R 1 o± 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 o± the resistances is R 2 R 1 = ρ ℓ 2 π r 2 2 π r 2 1 ρ ℓ 1 = 2 1 r 2 1 r 2 2 = 3 1 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). A±ter 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 em± o± 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 .
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Version 050 – Second Exam – Hofmann – (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 micro±arads. Consider the capacitor circuit 64 V 33 μ ± 48 51 30 What is the efective 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 ±. Explanation: Let : C 1 = 33 μ ± , C 2 = 48 μ ± , C 3 = 51 μ ± , C 4 = 30 μ ± , and E B = 64 V . E B C 1 C 2 C 3 C 4 ±or capacitors in series, 1 C series = s 1 C i V series = s V i , and the individual charges are the same. ±or parallel capacitors, C parallel = s C i Q parallel = s Q i , and the individual voltages are the same. C 1 and C 2 are in parallel, so C 12 = C 1 + C 2 = 33 μ ± + 48 μ ± = 81 μ ± . C 3 and C 4 are in parallel, so C 34 = C 3 + C 4 = 51 μ ± + 30 μ ± = 81 μ ± . E B C 12 C 34
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Version 050 – Second Exam – Hofmann – (56745) 3 C 12 and C 34 ( Note: C 12 = C 34 ) are in series with the battery, so C 1234 = 1 1 C 12 + 1 C 34 = C 12 C 34 C 12 + C 34 = (81 μ F) (81 μ F) 81 μ F + 81 μ F = 40 . 5 μ F .
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Second Exam-solutions - Version 050 Second Exam...

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