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33. (a) We note that the
R
1
resistors occur in series pairs, contributing net resistance 2
R
1
in each branch
where they appear. Since
E
2
=
E
3
and
R
2
=2
R
1
, from symmetry we know that the currents
through
E
2
and
E
3
are the same:
i
2
=
i
3
=
i
. Therefore, the current through
E
1
is
i
1
=2
i
.Th
en
from
V
b
−
V
a
=
E
2
−
iR
2
=
E
1
+ (2
R
1
)(2
i
)weget
i
=
E
2
−E
1
4
R
1
+
R
2
=
4
.
0V
−
2
.
0V
4(1
.
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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.
 Fall '08
 SPRUNGER
 Physics, Current, Resistance

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