CourseNotes.51 - Note that we have dropped the term inside...

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Note that we have dropped the φ term inside of the cosine because the problem says that the capacitor starts off fully charged. The cosine term will have completed N complete cycles when ω 0 t = 2 πN t = 2 πN ω 0 The book says the we can generally treat ω 0 as if it were ω , but keeping it as ω 0 does not complicate this problem much so we will do so. Hence, t N = 2 πN q ω 2 - ( R 2 L ) 2 = 2 πN q 1 LC - ( R 2 L ) 2 After a complete cycle the cosine term will return to 1 so we only need to worry about the exponential term when computing the magnitude of the new charge. q N = q 0 exp - R 2 L 2 πN q 1 LC - ( R 2 L ) 2 = q 0 exp - 2 πN q 4 L R 2 C - 1 If we plug in the numbers we find q 5 = 5 . 85 μC q 10 = 5 . 52 μC q 100 = 1 . 93 μC q 10 , 000 = 1 . 16 × 10 - 50 μC Problem 31.34 An ac generator has emf E = E m sin ω d t , with E m = 25 V and ω d = 377 rad s . It is connected to a 12 . 7 H inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is - 12 . 5 V and increasing in magnitude, what is the current? Although we derived it above, lets rederive the current in this circuit. The potential across the
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