39. (a) The symmetry of the problem allows us to use i2as the current in bothof the R2resistors and i1for the R1resistors. We see from the junction rule that i3= i1– i2. There are only two independent loop rule equations: ()22 1111123020iR iRiRi i Rε−−=−−−=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.