Name: ________________________________________ Code # _________ Summer Session, 2008 ID: X
UCSD Physics 2B Practice Exam 3 Current & Resistance
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Do this exam a few times... Once without the answers,
Once after seeing the solutions & Once again with your formula sheet
This is a good time to add formulas/constants to your sheet.
Remember, this is a timed exam... work swiftly so you have time to check your answers.
A heating coil operates at 117 V. If it draws 10.5 Amps of current, find its resistance .
A.
128
Ω
B.
8.97
×
10
−
2 Ω
C.
1.23
×
10
3 Ω
D.
11.1
Ω
Two lamps each of resistance R = 10.0
Ω
are connected in series to voltage source V
=
120 V as shown in the diagram below. What is the current through each lamp?
A.
6.00
A B.
12.0 k A C.
130 A D.
83 m A
1

Name: ________________________
Find the equivalent resistance of the network shown below if R
1
=
25.0
Ω
and R
2
=
17.3
Ω
ID: A
A.
42.3
Ω
B.
9.71
m
Ω
C.
10.2
Ω
D.
7.70
Ω
Find the power dissipated by resistor R
1
in the circuit shown below if V
=
50 V , R
1
R
2
=
17.2
Ω
.
=
25.0
Ω
and
A.
145 W B.
59.2 W C.
100 W D.
2.91 W
It takes 6.75 seconds for a capacitor to be “charged” with 175 Coulombs. What average current flows into the capacitor during this time?
A.
25.9 A B.
38.6 mA C.
25.9 mA D.
38.6 A
2

Name: ________________________ ID: A
If 4.70
×
10
16 electrons pass a particular point in a wire every second, what current flows through the wire?
A.
7.52 mA B.
133 mA C.
7.52 A D.
133 A
How much electric energy (in Joules) is delivered in 5.00 minutes to an electric motor that draws
0.500 Amps from a 6.00 Volt battery?
A.
900 J B.
15 J C.
60 J D.
54 kJ
Find the current through resistor R
2 in the circuit diagram below if R
V
=
75.0
Volts
1
=
R
2
=
R
3
=
30.0
Ω
and
A.
833 mA B.
400 mA C.
2.50 A D.
1.67 A
3

Name: ________________________ ID: A
A bicycle wheel of radius 15 inches is given an electric charge Q
=
750 mC which is distributed evenly over the surface of the tire as shown below. If the wheel is set spinning at 120 rpm, what is the electric current that a stationary observer at the rim would measure?
A.
1.50 A B.
90.0 A C.
12.5 mA D.
667 mA
In the circuit below, R
1
=
11.5
Ω
, R
2
=
16.5
Ω
1.34 A, what is the voltage of the battery?
, and R
3
=
22.3
Ω
. If the current through the circuit is
A.
67.4 V B.
1.74 V C.
32.3 V D.
50.3 V
4

Name: ________________________ ID: A
If 4.0 x 10 18 electrons and 1.5 x 10 18 protons move in opposite directions past a given section of a hydrogen discharge tube every second, the average current in the tube is
A.
0.40 A B.
0.56 A C.
0.88 A D.
1.5 A
Wire A, which is of the same length and material as wire B, has twice the cross-sectional area of wire B. If the resistance of wire B is R
B
, what is the resistance R
A of wire A?
A.
R
B
/ 4 B.
2 R
B
C.
R
B
/ 2 D.
4 R
B
If electric energy costs 10 cents per kilowatt-hour (you wish!), how many cents does it cost to keep a 660-W toaster in steady operation for 30 minutes?
A.
15 cents B.
12 cents C.
6.9 cents D.
3.3 cents
A coil of wire of length L in a heating unit used to regulate the temperature in a chemical reaction vessel operates on a power line voltage of 240 V. If another unit must be constructed to operate at
120 V using the same diameter wire, how long must the wire be in order to produce the same amount of heat (dissipate the same power )?
A.
L / 4 B.
L / 2 C.
4 L D.
2 L
5

Name: ________________________ ID: A
Find the equivalent resistance of the network shown below if each resistor has value R
=
15
Ω
A.
45
Ω
B.
5.0
Ω
C.
0.20
Ω
D.
0.02
Ω a slab 14.0 cm long, 2.00 cm wide and 3.00 cm tall. A wire have been attached to each end-face of the long dimension one of which is shown below. The resistance of the slab is measured and found to be 320 Ohms. What is the resistivity
ρ
of the unknown material?
A.
2.29
Ω
• m B.
1.37
Ω
• m C.
26.9 k
Ω
• m D.
74.7 k
Ω
• m
A beam of Alpha particles (Helium nuclei = 2 protons + 2 neutrons) in an accelerator has a particle density n
=
41
×
10 current I
=
1.1
×
10
12 m
−
4
A
−
3
. If the beam has cross-sectional area
, what is the drift velocity
A
=
0.23 c m
2
and the measured electric
of the particles in the beam?
A.
1.8 x 10 5 m/s B.
7.3 x 10 5 m/s C.
1.9 x 10 -4 m/s D.
3.6 x 10 5 m/s
6

Name: ________________________ ID: A
In the circuit below the switch has been connected to point B for a long time and the capacitor is fully discharged . At time t = 0, the switch is moved to point A and the capacitor begins charging through the resistor as shown. What will be the voltage across the capacitor at time t
=
4.00
s if
V
=
225 V , C
=
20.0
μ
F and R
=
250 k
Ω
?
A.
501 V B.
124 V C.
10.1 V D.
64.5 V internal resistor. If a voltmeter across the battery reads 12.0 V with no load attached and the same meter reads 10.0 V when a 60-Ohm load resistor is connected, what is the internal resistance?
A.
20
Ω
B.
10
Ω
C.
12
Ω
D.
50
Ω
7

Name: ________________________ ID: A special resistor is made in the following way. A resistive material is sandwiched in between two coaxial metal shells, and current is made to flow radially outward through the material as shown in the cross-section below. The radius of the inner shell is a the radius of the outer one is b and the overall length of the gadget is L . If the resistivity of the material is
ρ
, what is the resistance R of the device?
A.
4
ρ
π
L
1 a
−
1 b
B.
2
ρ
π
L ln
Ê
ÁÁ
ÁÁÁÁ
Ë b a
C.
2
ρ
π
L
( b
− a ) D.
ρ
L
( b
− a )
8

Name: ________________________ ID: A
A cylindrical conductor is carrying an electric current along its axis which varies with distance from its axis. The current density as a funtion of r is measured and found to be J r
=
K r , where
K
=
12.2 Amperes / meter
3
. If the radius of the wire R
=
16.6
cm what total current is it carrying?
Hint : The total current I
=
∫
J
• dA integrated over the entire circular cross-section of the cylinder and the differential area elements dA are just rings at distance r and thickness dr .
A.
117 A B.
10.6 kA C.
117 mA D.
10.6 A
9

ID: A
UCSD Physics 2B Practice Exam 3 Current & Resistance
Answer Section
MULTIPLE CHOICE
When I see current, resistance & voltage (V), I immediately think of Ohm’s Law:
V
R
=
I
−
117 V
10.5
A
=
11.1
Ω
Series equivalent resistance is the sum of the resistances.
R
=
R series 1
+
R
2
=
10.0
Ω +
10.0
Ω =
20.0
Ω
In series, the same current flows through all devices. So applying Ohm's Law,
I
=
V
R
=
120 V
20.0
Ω =
6.00 A
TOP: SERIES + OHM'S LAW
R
Series
A
=
R
1
+
R
2
=
25.0
Ω +
17.3
Ω =
42.3
Ω
TOP: SERIES
The resistors are in parallel, so they’re at the same voltage and independent of each other.
P
1
=
V
2
R
1
=
( 50.0
V )
2
25.0
Ω =
100 W
TOP: POWER
By definition of average current: I
=
Q
= t
175 C
6.75 s
=
25.9 A
TOP: CURRENT
Current is equal to charge per unit time, but we must convert electron charge into Coulombs
I
=
Q t
=
4.7
×
10
16
1 s e
−
1.6
×
10
−
19
1 e
−
C
=
7.52
×
10
−
3
A
=
7.52 mA
TOP: CURRENT
1

ID: A
This is basically a unit conversion problem if you think about the units:
Power
= energy time
⇒ energy
= power
× time
E
=
P
× t
=
( V
×
I )
× t
=
ÁÁÁÁ
Ë
6.00
J
C
0.500
C s
( 5.00 min )
60 s min
=
900 J
TOP: ENERGY
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
1
R
Parallel
=
1
R
1
+
1
R
2
=
1
30.0
Ω +
1
30.0
Ω =
1
15.0
Ω
so that R
Parallel
=
15.0
Ω
This is in series with R
3, so
R
Series
=
R
1
+
2
+
R
3
=
15.0
Ω +
30.0
Ω =
45.0
Ω
The total circuit current is thus I
=
V
R
Series
=
75.0
V
45.0
Ω =
1.67
A
Since R
1
= R
2
, the circuit current is equally divided between them, so I
2
=
I
2
=
0.833
A
*** There is a faster way. If we recognize this network as a voltage divider R
12
in series with R
3
, then the ratio of the voltages equals the ratio of the resistances. R
12
is easy to calculate - it's half
30
Ω =
15
Ω
. Hence V
12
/ V
3
=
15 / 30
⇒
V
12
=
25 V . Now we have
I
2
=
V
2
R
2
=
25 V
30
Ω =
0.833
A
Every time the wheel makes one complete turn, all the charge must pass a fixed point. Hence
I
=
Q t
=
0.750
C rev
120 rev min
1 min
60 s
=
1.50
A
R
A
=
R
(
+
R
Series 1
V
=
IR
=
1.34
A
2
)
+
R
(
3
50.3
=
1.5
+
16.5
+
22.3
=
50.3
Ω
Ω
)
=
67.4
V
TOP: NET + OHMS LAW
2

ID: A
TOP: CURRENT
Resistance varies inversely with cross section area. Hence, if R
B
= ρ
L
A
, then R
A
= ρ
L
2 A
=
R
B
/ 2
TOP: RESISTIVITY
This is a conversion problem -
( 660 W ) × k W
1000 W
× ( 30 min ) × hr
60 min
×
10 cents kW
• hr
=
3.3 cents
TOP: POWER & ENERGY
P
=
V
2
R
=
Ê
ËÁÁ
V 2
ˆ
¯˜˜
2
R '
=
V
2
4 R '
⇒
R '
=
R / 4
R
= ρ
L
A
⇒
R '
= ρ
L / 4
A
TOP: RESISTIVITY + POWER when all resistors are identical R
PAR
=
R
N
=
15
Ω
3
=
5.0
Ω
TOP: PARALLEL
R
= ρ
L
A
⇒ ρ =
RA
L
=
Rwh
L
=
( 320
Ω )
Ê
2.00
Ë
×
10
−
2 m
14.0
×
10
ˆ
¯
3.00
×
10
−
2
Ë m
ˆ
−
2 m
¯
=
1.37
Ω • m
TOP: RESISITIVITY
ANS:
I
=
JA
= nqv d
A v d
=
I nqA
=
Ë
41
×
10
12 m
−
3
¯ Ë
1.1
×
10
−
4
A
ˆ
2
×
1.60
×
10
−
19
C
ˆ
¯ Ë
0.23
cm
2
¯
100 cm m
2
=
3.65
×
10
5 m / s
3

ID: A
Calculate the "time constant" and plug it into the "charging" formula -
τ =
RC
=
Ë
250
×
10
3 Ω
ˆ
¯ Ë
20.0
×
10
−
6
F
ˆ
¯
=
5.00
s
V t
=
V
0 exp
ÁÁÁÁ
Ë
−
τ t
=
( 225 V )
ÔÔÔÔÔ
Ó
1
− exp
ÁÁÁÁ
Ë
−
4.00
s
5.00
s
ÔÔÔÔ
ÔÔÔÔÔ
˝
˛
=
124 V
If no load is connected, we have an open circuit in which no current flows . Hence the voltage drop across R
INT
must be zero V
INT
=
IR
INT
, so the meter reads the "ideal" battery voltage:
V
=
12 V
V
INT
=
V
−
V
LOAD
=
12.0
V
−
10.0
V
=
2.0
V
I
=
V
INT
R
INT
=
V
LOAD
R
LOAD
⇒
R
INT
=
( 60
Ω
)
2.0
V
10.0
V
=
12
Ω
ANS:
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 with resistance dR
= ρ d
A
= ρ dr
Lw
= ρ
2 dr
π 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
=
∫ dR
=
ρ
2
π
L
∫ dr r
= a b
ρ
2
π
L ln
Ê
ÁÁ
ÁÁÁÁ
Ë b a
TOP: RESISTIVITY + INTEGRATION
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
= ∫
0
R
J r 2
π r dr
= ∫
0
R
2
π
Kr
2 dr
=
2
π
K
R
3
3
=
2
3
( 3.14
)
Ë
12.2
A / m
3
ˆ
¯ Ë
16.6
×
10
−
2 m
ˆ 3
=
117 mA
TOP: CURRENT DENSITY + INTEGRATION
4