ID: B
4
UCSD Physics 2B
Unit Exam 3B
Current, Resistance, Circuits
Answer Section
MULTIPLE CHOICE
1. ANS: A
Two ways
to do this:
1.
This is a compound circuit, so we must find the current around the entire loop, and
therefore the
equivalent resistance of the
entire
circuit. First, reduce the parallel leg to its equivalent. Since
R
1
=
R
2
,
R
par
=
R
N
=
50
Ω
2
=
25
Ω
. Now this is in series with R
3,
so
R
Circuit
=
R
12
+
R
3
=
25.0
Ω+
50.0
Ω=
75.0
Ω
. The total circuit current is thus
I
=
V
R
Series
=
120
V
75.0
Ω
=
1.60
A
Since the entire circuit current must pass through
R
3
P
3
=
I
2
R
3
=
1.60
A
()
2
50.0
Ω
=
128
W
2.
Think “voltage divider”. Since
R
3
=
2
R
,
V
3
=
2
3
V
=
2
3
V
=
80
V
Now apply the power formula locally to
R
3
P
3
=
V
3
2
R
3
=
V
2
Ω
=
W
TOP:
NET + POWER
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View Full DocumentID: B
2
2. ANS: C
Let's think of this device as being made up of many coaxial shells (layers) of length
L
, "width"
w
=
2
π
r
and
thickness
dr
.
Since the current flows directly outward, each layer can be treated as a
parallel plate
resistor
whose resistance is
R
=
ρ
d
A
. The
differential
element for this layer is
dR
=
d
d
A
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
dr
Ar
()
=
Lw
=
2
rL
Since each layer acts as one element in a
series
network (why?), the total resistance is the sum of these layers,
that is, an integral:
R
=
∫
=
2
L
r
a
b
∫
=
2
L
ln
b
a
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
Plugging in the numers, we get
R
=
2
L
b
a
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
22
Ω⋅
m
2
1.36m
2.33
2.22
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
125
m
Ω
Note that the diameters
a
and
b
are so close that you can get an
approximate
answer by treating the whole
device as a rolledup plate resistor with
d
=
b
−
a
and
A
=
2
aL
. Then
R
≈
d
A
=
b
−
a
2
=
b
/
a
−
1
2
L
=
m
2.33
2.22
−
1
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
2
=
128
m
Ω
TOP:
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 Summer '09
 Lever
 Current, Resistance, Resistor, Electrical resistance, dr dr Á dR

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