PHYS1200
PHYSICS II
SPRING 2005
HOMEWORK SOLUTIONS
CLASS 6.
27.18, 27.36, 27.68, 27.73;
Q29.01.2, Q29.08
27.18.
a)
From the diagram, it is clear that there are three parallel paths from
F
to
H
. The
upper and the lower both consist of two 5.0
Ω
resistors in series, so the resistance of
each is 10
Ω
. The middle path has a resistance of 5.0
Ω
. Therefore, the equivalent
resistance of the three paths is given by:
Ω
=
Ω
+
Ω
+
Ω
=
Ω
+
Ω
+
Ω
=
10
4
10
2
10
1
10
1
5
1
10
1
10
1
1
FH
R
, and
4
10
Ω
=
FH
R
R
FH
= 2.5
Ω
b)
This part is easier if the diagram is redrawn as shown below.
From this sketch, it is apparent that there are two parallel paths from
F
to
G
. The
lower path has a resistance of 5.0
Ω
, but it will take some work to compute the
resistance of the upper path. First, calculate the equivalent resistance of the three
5.0
Ω
resistors in the upper left of the diagram, and replace them with their
equivalent,
R’
. To get
R’
, note that there are two parallel paths, the upper has two
5.0
Ω
resistors in series, so its resistance is 10
Ω
. The lower path has a resistance of
5.0
Ω
. Therefore, the equivalent resistance of the two paths is given by:
Ω
=
Ω
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 Spring '06
 Stoler
 Work, Resistor, Series and parallel circuits, equivalent resistance

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