P28_018 - them in series. Thus, R 1 R 2 / ( R 1 + R 2 ) = 3...

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18. Let the resistances of the two resistors be R 1 and R 2 . Note that the smallest value of the possible R eq must be the result of connecting R 1 and R 2 in parallel, while the largest one must be that of connecting
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Unformatted text preview: them in series. Thus, R 1 R 2 / ( R 1 + R 2 ) = 3 . 0 and R 1 + R 2 = 16 . So R 1 and R 2 must be 4 . 0 and 12 , respectively....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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