ID: A
5
UCSD Physics 2B
Unit Exam 3A
Current, Resistance, Circuits
Answer Section
MULTIPLE CHOICE
1. ANS: B
Each differential area element
dA
of a circle (cross section) is a ring of thickness
dr
and length
(circumference)
2
π
r
. The area of this ring is the product,
dA
=
2
π
r dr
I
=
J
•
dA
∫
=
J r
( )
2
π
r dr
0
R
∫
=
Kr
2
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
2
π
r dr
0
R
∫
=
2
π
K
r
3
dr
0
R
∫
=
2
π
K
R
4
4
=
1
2
3.14
(
)
122
A
m
4
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
166
×
10
−
2
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
4
=
1.45 k
A
TOP: CURRENT DENSITY + INTEGRATION
2. ANS: D
This is a series combination. When all resistors are identical
R
SER
=
NR
=
3
×
33
Ω =
99
Ω
TOP: SERIES
3. ANS: D
This is basically a
unit conversion
problem if you think about it:
Power
=
energy
time
⇒
energy
=
power
×
time
E
=
P
×
t
=
V
×
I
(
)
×
t
=
12.0
J
C
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
5.00
C
s
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
3.00 min
(
)
60 s
min
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
1.08
×
10
4
J
=
10.8k
J
TOP: ENERGY
4. ANS: B
Important to keep track of direction here. Since the protons and electrons are moving in the
same
direction, you must subtract the electron contribution to the flux since the charge is negative.
Calculate the number of particles times the charge/particle and then divide by time
I
=
Q
t
=
n
p
e
+
+
n
e
e
−
t
=
n
p
e
+
−
n
e
e
+
t
=
n
p
−
n
e
Ê
Ë
Á
Á
ˆ
¯
˜
˜
e
+
t
=
6.5
×
10
17
−
3.5
×
10
17
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
1.6
×
10
−
19
C
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
1
s
=
0.048
A
=
48m
A
TOP: CURRENT

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