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 ω0 t = 2 πN ⇒ t = 2 πN ω0 The book says the we can generally treat ω0 as if it were ω , but keeping it as ω0 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 = q0 exp -R 2 L 2 πN q 1 LC-( R 2 L ) 2 = q0 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
This is the end of the preview.
access the rest of the document.