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Unformatted text preview: 1 + 2 R 2 k (c) We seek the conditions under which V ( ) = 2 V (0) . Using the result of Part (b), we find 2 R 5 k m 5 1 + 2 R 2 k = 4 R 5 k m = 1 + 2 R 2 k = 4 Solving for , we conclude that = 3 k 2 R 2...
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.
 Spring '06
 Shiflett

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