P28_033.PDF
(
62 KB
)
Pobierz
Chapter 28 - 28.33
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
.Then
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
=
2
.
0V
4(1
.
0Ω)+2
.
0Ω
=0
.
33A
.
4
.
0V
−
Therefore, the current through
E
1
is
i
1
=2
i
=0
.
67A, flowing downward. The current through
E
2
is 0
.
33A, flowing upward; the same holds for
E
3
.
(b)
V
a
−
V
b
=
iR
2
+
E
2
=
−
(0
.
333A)(2
.
0Ω)+4
.
0V=3
.
3V
.
−
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