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37. (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
ε
2
=
3
and
R
2
= 2
R
1
, from symmetry we know
that the currents through
2
and
3
are the same:
i
2
=
i
3
=
i
. Therefore, the current through
1
is
i
1
= 2
i
. Then from
V
b
– V
a
=
2
–
iR
2
=
1
+ (2
R
1
)(2
i
) we get
()
21
12
4.0V
2.0V
0.33A.
44
1
.
0
2
.
0
i
RR
−−
==
=
+Ω
+
Ω
Therefore, the current through
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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, Resistance

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