This preview shows page 1. Sign up to view the full content.
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
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details