Second Exam-solutions

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

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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
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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