ch27-p039 - 39. (a) The symmetry of the problem allows us...

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39. (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: () 22 11 11 1 2 3 0 20 iR iR iR i i R ε −= −− = where in the latter equation, a zigzag path through the bridge has been taken. Solving, we
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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