P28_096 - 96. (Fourth problem of Cluster) (a) The symmetry...

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96. (Fourth problem of Cluster ) (a) The symmetry of the problem allows us to use i 2 as the current in both of the R 2 resistors and i 1 for the R 1 resistors. We see from the junction rule that i 3 = i 1 i 2 . There are only two independent loop rule equations: E− i 2 R 2 i 1 R 1 =0 2 i 1 R 1 ( i 1 i 2 ) R 3 . where in the latter equation, a zigzag path through the bridge has been taken. Solving, we ±nd
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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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