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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.
 Spring '08
 Any
 Physics, Current

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