8.022pset9solution

# 8.022pset9solution - Massachusetts Institute of Technology...

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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 9: AC Currents . TA: Yi Mao, [email protected] 1. Purcell 8.3: Parallel RLC. The impedance of a parallelled RLC combination is Z = • 1 R + i ( ωC- 1 ωL ) ‚- 1 . (1) R=1000 ohm, C=500 × 10- 12 farad, L=2 × 10- 3 henry. For f = 10 kilohertz, ω = 2 π × 10 4 sec- 1 , Z ≈ 16 + 124 i ohms. For f = 10 megahertz, ω = 2 π × 10 7 sec- 1 , Z ≈ 1- 32 i ohms. The absolute value of impedance is | Z | = fl fl fl fl 1 R + i ( ωC- 1 ωL ) fl fl fl fl- 1 = µ 1 R 2 + ( ωC- 1 ωL ) 2 ¶- 1 / 2 . (2) To reach the maximum of | Z | , we should look for the minimum of the expression within the big parentheses. Both terms inside are positive or zero. Obviously the maximum of | Z | takes place at ωC = 1 ωL or ω = 1 √ LC = 10 6 sec- 1 f = 1 2 π × 10 6 ≈ 1 . 6 × 10 5 Hertz. (3) 2. Purcell 8.7: Resonant cavity. The resonant cavity is equivalent to a simple LC circuit. The inductor is a configuration of currents uniformly flowing up and down on the surface of the inner conducting cylinder and the outer one. The capacitor is a pair of parallel plates of seperation s and area A = πa 2 . Quote the result in Problem 5 of pset 8: the inductance per unit length is L/l = 2 c 2 ln ( b a ) . This gives the total inductance L = ( h- s ) 2 c 2 ln ( b a ) . [You could also use Purcell (58) of Chapter 7; you would get h in place of h- s , which is good enough for small s .] The capacitor is C = A/ 4 πs = a 2 / 4 s . So the resonant frequency is ω = 1 √ LC = ˆ a 2 ( h- s ) 2 sc 2 ln( b a ) !...
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8.022pset9solution - Massachusetts Institute of Technology...

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