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Unformatted text preview: i 2 derives from a succession of symmetric splittings of i R (reversing the procedure of reducing those six resistors to Fnd R in part (a)). We Fnd i 2 = 1 2 µ 1 2 i R ¶ = 0 . 50 A and is clearly downward. (c) Using our conclusions from part (a) in Eq. 2814, we obtain P = i 1 E 1 = (4)(16) = 64 W supplied. (d) Using results calculated in part (a) in Eq. 2814, we obtain P = i E 2 = (2)(8) = 16 W absorbed....
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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.
 Fall '08
 SPRUNGER
 Physics, Current

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