# R - Piazza ```*
AP PHYSICS B
Circuits and Resistance
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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Circuits and Resistance
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions on circuits and resistance.
Standards
Circuits is addressed in the topic outline of the College Board AP* Physics Course Description
Guide as described below.
III. Electricity and Magnetism
C. Electric Circuits
1. Current, Resistance, Power
2. Steady State Direct Current Circuits with batteries and resistors only
3. Capacitors in Circuits
AP Physics Exam Connections
Topics relating to circuits and resistance are tested every year on the multiple choice and in most
years on the free response portion of the exam. The list below identifies free response questions
that have been previously asked over circuits. These questions are available from the College
Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com.
2007
2003
2002
2001
Free Response Questions
Question 3
2007 Form B Question 3
Question 2
2003 Form B Question 2
Question 3
Question 5
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Circuits and Resistance
What I Absolutely Have to Know to Survive the AP* Exam
•
•
•
•
•
•
Draw diagrams for series and parallel circuits using accepted symbols, including proper placement of
ammeters and voltmeters.
Calculate total resistance and total potential difference for series, parallel, and combination circuits.
Determine power in a circuit.
Find the charge stored in a capacitor (in steady state).
Determine the short term and long term (steady state) behavior of a capacitor.
Apply Kirchhoff’s Rules (especially in multi-loop circuits).
Key Formulas and Relationships
Δq
Δt
V = IR
I=
Current is the rate at which charge passes a point in an electrical circuit. (Amperes A)
The potential difference across a resistor is equal to the current through the resistor
times the size of the resistor. (Volts V)
R=
ρL
Resistance of a resistor is proportional to the length L and inversely proportional to the
A
Req = R1 + R2 + R3 + ...
cross sectional area A. The resistance is also proportional to the resistivity ρ (Ωm)
which is a characteristic of the material.
The equivalent resistance of a series combination of resistors equals the sum of the
individual resistors in the combination. (Ohms Ω)
1
1
1
1
= +
+ + ... The inverse of the equivalent resistance of a parallel combination of resistors equals
Req R1 R2 R3
the sum of the inverse of each of the individual resistors. (Ohms Ω)
ΣI int o a node = ΣI out of
a node
Kirchoff's First law (conservation of charge) the current that flows into a node must
equal the current that flows out of a node.
Kirchoff's Second Law (conservation of energy) the sum of the voltage changes around
any closed path within a circuit must equal zero.
Power provided or dissipated to a circuit by a circuit element is equal to the current
ΣΔV = 0
P = IV
through the element times the voltage across the element. (Watts W or
P = I 2R =
E = Pt
V2
R
J
)
S
Power dissipated by a resistor. When power is dissipated by a resistor, the &quot;lost energy&quot;
is converted into heat or light.
Energy is power times time. (Joules J)
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Circuits and Resistance
Important Concepts
• The charge on one electron (or one proton) is called the elementary charge (e) and is equal to 1.6&times;10-19C.
Hence, it takes 6.25&times;1018 electrons to generate one coulomb of charge. 1 Ampere is the amount of current that
moves one Coulomb of charge through the circuit each second.
• A node is a junction where two wires meet within a circuit. A branch is a section of circuitry between two
nodes. The current in a particular branch will be the same everywhere within the branch.
• Conventional current—tracks the flow of positive charge.
Electron current tracks the path of the electron flow
and it is opposite to conventional current flow.
•
Ohm’s Law: V = IR
V is the potential difference or voltage, measured in volts.
I is current, measured in amps which is equal to coulombs/second.
R is resistance, measured in ohms and symbolized by the Greek letter omega, Ω
•
RESISTORS IN SERIES
Note that there is only one path for the current to travel; therefore, the current through each of the resistors is the same.
V
Requivalent
=
V1 V2 V3
+
+
R1 R2 R3
Also, the voltage “drops” across the resistors must add up to the total voltage supplied by the battery:
VTotal = V1 + V2 + V3
Ohm’s Law states that V=IR, then VTotal = IR1 + IR2 + IR3
BUT, Ohm’s Law must also be satisfied for the complete circuit: VTotal = IRequivalent
Set these two equations equal to each other since they each equal Vtotal,
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Circuits and Resistance
IRequivalent = IR1 + IR2 + IR3
Factor out the current I. IRequivalent = I ( R1 + R2 + R3 )
Since the current is the same through each resistor in the circuit, it divides out, and we see that Requivalent is simply the
sum of the resistors connected in series: Req = R1 + R2 + R3
•
RESISTORS IN PARALLEL
Note that there are now three paths for the current to travel between points A and B. The current is split along these
three paths as it travels from point A to point B. Furthermore the current is the same at both points A and B so, the
sum of the currents through the three branches is the same as the current at both point A [where the current splits] and
at point B [where the current reunites].This means that
ITotal = I1 + I 2 + I 3
At both points A and B, the potential difference must be the same, so between points A and B the potential must be the
same. That is, each of the three resistors in the parallel circuit must have the same voltage across it:
VTotal = V1 = V2 = V3
By Ohm’s Law , V = IR, this is equivalent to
V
Requivalent
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=
V1 V2 V3
+ +
R1 R2 R3
Circuits and Resistance
Factor out the potential difference, V.
⎛
1
V⎜
⎜ Requivalent
⎝
⎞
⎛ 1
1
1 ⎞
+ ⎟
⎟⎟ = V ⎜ +
⎝ R1 R2 R3 ⎠
⎠
Since the potential difference, V, is the same across each resistor in the circuit, it divides out, and we see that
1
1
1
1
= +
+
Requivalent can be found by
Req R1 R2 R3
•
SUMMARY
Series
Parallel
1
1
1
1
=
+
+ ...
Req R1 R2 R3
Req = R1 + R2 + R3...
•
VT = V1 + V2 + V3...
VT = V1 = V2 = V3...
IT = I1 = I2 = I3...
IT = I1 + I2 + I3...
Req&gt;R1, R2, R3…
Req&lt;R1, R2, R3…
COMPLEX CIRCUIT DIAGRAMS
Complex circuits involve resistors in series in combination with resistors in parallel.
Steps in simplifying complex circuits:
1.
Calculate the equivalent resistance of each set of resistors in parallel.
2.
Calculate the total resistance of the circuit.
3.
Calculate the total current using Ohm’s Law.
4.
Apply Ohm’s Law to each individual series resistor to determine the individual resistor’s voltage drop.
5.
Apply Ohm’s Law to each resistor in the parallel group to find the current in that branch.
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Circuits and Resistance
•
EXAMPLE
Three 5-Ω resistors are connected in parallel. This parallel combination is in series with a 7-Ω, and two 10-Ω
resistors as pictured below. Calculate the total resistance in the circuit. Use Ohm’s Law to calculate the total
current in the circuit. Calculate the voltage drop across each resistor.
o First, calculate the equivalent resistance of the parallel network of resistors.
1
Req ( parallel )
=
1 1 1 3
5
+ + = ∴ Req = = 1.67
5 5 5 5
3
RT = Req ( parallel ) + R2 + R3 + R4
RT = 1.67Ω + 7Ω + 10Ω + 10Ω = 28.67Ω
o Second, redraw the circuit so that the equivalent resistance is in series with the remaining series resistors.
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Circuits and Resistance
Use Ohm's Law to solve for the current:
V = IR so I =
V
R
36V
= 1.26A for each resistor in series, but now we need to split this 1.26 A across
28.67 Ω
each of the resistors in the parallel branch. This means we first need to know the voltage
I=
across each resistor in our new drawing.
V = IR
V = 1.26 A &times; 1.67 Ω = 2.1V
V = 1.26 A &times; 7 Ω = 8.8V across the 7Ω resistor
V = 1.26 A &times; 10 Ω = 12.6V across EACH of the 10 Ω resistors
This gives a total voltage of [2.1 +8.8+2(12.6)]=36.1 V (due to rounding) as expected since
we have a 36 V battery.
Now we can determine the current through EACH resistor in the parallel branch:
V 2.1V
I= =
= 0.42A across each of the 5Ω resistors
R
5Ω
The grand total of all of the current through EACH resistor in the parallel branch should match our 1.26 A total current:
(3 &times; 0.42A) = 1.26A
• Meters
You must know how to hook up and use ammeters and voltmeters for the exam.
Ammeters must be placed in series with the component you are measuring the current through. In theory anywhere in a
circuit that is in series with a component should have identical current, due to approximations like ideal wires. Proper
lab procedure, however, is to place the ammeter just before the component you wish to measure the current flowing
through it.
Voltmeters must be hooked up in parallel. This allows them to touch a circuit at two different points without any
current passing through them so that they do not affect the circuit they are measuring.
• Resistance of a wire
You must understand the relationship between resistance of a material, its length and cross sectional area. Fat wires
have less resistance to electric current in a manner similar to wide highways with many lanes having less resistance to
the flow of traffic. Shorter wires have less resistance as well. This is typically tested in a multiple choice question of
ρL
variable manipulation. R=
A
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Circuits and Resistance
• Batteries
An ideal battery can maintain a constant voltage when the current flowing through it is very small or very large. The
exam often assumes batteries to be ideal. Real batteries, however, do not maintain a constant voltage when you begin
to draw large currents from them. They act as if there is a small resistor inside them impeding the current. You can
tell when a real battery is going bad by looking at the voltage between the terminals when it is not under load and then
again when it is under load. If the numbers change a lot then the battery has a large internal resistance. The idea of
using a voltmeter on a battery gives us an equation V A − VB = ε − Ir where the left hand side of the equation is the
voltage measured under load, ε is the voltage when it is not under load and r is the internal resistance of the battery. A
very good battery like a car battery can get ΔV very close to ε so the approximation of an ideal battery is not a bad
one.
• Kirchhoff’s Rules for Multi-loop Circuits
1. The net (total) current entering a node (junction) must equal the net current out of a node (junction). This is
sometimes called the junction rule, and is a statement of conservation of charge. I in = I out
2. The sum of the potential rises and drops (voltage differences) around a closed loop of a circuit must equal zero.
This is sometimes called the loop rule, and is a statement of conservation of energy. ΣV = 0
a.
If we pass a battery from negative to positive, we say that there is a rise in
potential, +ε.
If we pass a battery from positive to negative, we say that there is a drop in
potential, - ε.
If we pass a resistor against the direction of our arbitrarily chosen current, we
say there is a rise in potential across the resistor, + IR.
If we pass a resistor in the direction of our arbitrarily chosen current, we say
there is a drop in potential across the resistor, - IR.
b.
c.
d.
• Capacitors
1. An empty capacitor does not resist the flow of current, and thus acts like a wire.
2. A capacitor which is full of charge will not allow current to flow, and thus acts like a broken wire.
Capacitors in DC circuits are used to store energy. They do this when a certain amount of charge is placed on one
plate while an equal amount of opposite charge is placed on the other plate. This creates an electric field between
the plates. The energy is stored in the electric field.
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Circuits and Resistance
CP = C1 + C2 + C3 + ...
Capacitors in parallel add like resistors in series.
1
1
1
1
= +
+
+ ...
Cs C1 C2 C3
Capacitors in series add like resistors in parallel.
q
V
1
1
1 Q2
2
U E = CV = QV =
2
2
2 C
q = CV or C =
Capacitance is the ratio of charge to voltage. (Farads F or
C
)
V
Energy stored in a capacitor. (Joules J)
Capacitance can be expressed as C = Q/V or C = KA/d where:
•
•
•
•
•
Q is the charge held on the capacitor.
V is the voltage across the conductive plates.
A is the area of the conducting plate.
d is the distance between the plates.
K represents the dielectric constant (measures how well the insulator does its job). On the AP Exam k=1 since it
is normally air or vacuum.
As you can see from looking at the formula C = κA d , if you pull the two plates of a capacitor apart you decrease its
capacitance. It takes work to accomplish this if there is charge on the capacitor since you are moving the plates against
the electric field. If you decrease the area of the plates it has a similar effect of decreasing the capacitance.
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Circuits and Resistance
Free Response
Question 1 (15 pts)
S
_
2Ω
R
+
12 V
4Ω
The circuit shown above includes a 12-volt battery, a resistor of unknown resistance R, an
open switch, a 2-Ω resistor, and a 4-Ω resistor. The switch is then closed so that a total
current of 3.0 A flows in the circuit.
A. Determine the current passing through the unknown resistor.
(4 points max)
V
12 volts
I 2Ω and 4Ω = =
= 2.0 A
R 2Ω + 4Ω
I R = I total − I 2 Ω and 4Ω = 3.0 A − 2.0 A =1.0 A
1 point for the application of Ohm’s law
1 point for finding the current through
the 2Ω and 4Ω resistors by dividing by
6Ω
1 point for determining the difference
between the total current and the current
through the two resistors
1 point for the correct answer with
correct units
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Circuits and Resistance
B. Determine the value of the unknown resistor R.
(1 point max)
V 12 volts
R= =
= 12 Ω
1.0 A
I
1 point for the correct answer with
correct units consistent with the
C. Determine the heat dissipated by resistor R in one minute.
1 point for a correct equation for
power
(4 points max)
V 2 144 volts
=
= 12 W
P = IV = I R =
12Ω
R
2
P = IV = (1.0 A )(12 volts ) = 12 watts = 12
J⎞
⎛
Heat energy = Pt = ⎜ 12 ⎟ ( 60 s ) = 720 J
s⎠
⎝
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J
s
1 point for the correct value for
power
1 point for the correct relationship
between power and heat energy
1 point for the correct answer for
heat energy with correct units and
reasonable number of significant
digits
Circuits and Resistance
The resistor R is removed and replaced with a 6μF parallel-plate capacitor, as shown on
the figure below. The switch is then closed.
S
_
– – – – – – –
2Ω
+
12 V
++ + + + + +
4Ω
D. Determine the total current in the 2-Ω and 4-Ω resistors after the switch has been
closed for a long time.
(3 points max)
After a long time, the capacitor is full, so no
current flows through it, and the current
flows through the 4Ω and 2Ω resistors.
V
12V
Io =
=
=2A
Rtot 6Ω
1 point for recognition that no current
flows through a full capacitor
1 point for the correct application of
Ohm’s law
1 point for the correct answer
including correct units
E. Determine the total amount of energy stored in the capacitor.
(3 points max)
1
1
2
U = CV 2 = ( 6 μ F )(12 V ) = 432 μ J
2
2
1 point for recognition that the voltage
drop across the capacitor is 12 V
1 point for the correct application of
the equation for the energy stored in a
capacitor
1 point for the correct answer
including correct units
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Circuits and Resistance
Question 2 (15 pts)
A student is given three identical resistors, five 1.5 Volt “D” cell batteries, a voltmeter,
an ammeter, and some wires. She is instructed to complete a circuit using one resistor
such that batteries can be added in series one at a time and in which both the current and
voltage can be properly measured.
A. In the box below, draw a schematic diagram of the circuit when 2 batteries are
connected in series with one resistor, and with both the ammeter and voltmeter connected
correctly.
1 point For two batteries connected
correctly in series
(3 points max)
1 point For a complete series circuit with
resistor
1 point For proper placement of both the
voltmeter and ammeter
After completing the circuit with one battery, the student measures the voltage and
current for the resistor, then continues adding batteries one at a time and makes the
following measurements:
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Circuits and Resistance
B. On the grid below, plot the necessary data that illustrates the mathematical relationship
between voltage and current. Use appropriate graphing techniques.
1 point For both axes properly labeled
with units and using an appropriate
scale
(3 point max)
see graph
1 point For plotting the points correctly
1 point For drawing an appropriate
best-fit line through the points
C. From your graph, determine the value of the resistor.
1 point For using the slope to find
the resistance
(3 points max)
m=
y2 − y1
6.5 V − 2.1 V
4.4 V
=
=
= 10 Ω
x2 − x1 0.66 A − 0.22 A 0.44 A
1 point For using points on the best
fit line
1 point For the correct answer with
units
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Circuits and Resistance
a
R1
R2
ε1= 4.5 V
R3
b
ε2= 3 V
D. The student arranges the three identical resistors and the five batteries in the
arrangement shown above such that there are neither series nor parallel combinations of
resistors. She then measures the current through each resistor. Determine I1, I2, and I3.
(6 points max)
Junction Equation for A:
ΣI in junction a = ΣI out junction a
1 point For stating that the
net current into a junction
must equal the net current
coming out of the junction
I1 + I 3 = I 2
Loop Equations
Loop I: traversing clockwise from
4.5V battery through R1 , junction A, and R 2
4.5 − 10 I1 − 10 I 2 = 0 ⇒ 10 I1 + 10 I 2 = 4.5
I1 + I 2 = 0.45
Loop II: traversing counterclockwise from
3V battery through R 3 , junction A, and R 2
3 − 10 I 3 − 10 I 2 = 0 ⇒ 10 I 3 + 10 I 2 = 3
I 2 + I 3 = 0.3
1 point For stating that the
sum of the voltage
differences around a
closed loop must equal
zero
1 point For writing the
potential rises and drops
around a closed loop
either CW or CCW
1 point For determining
three equations with 3
unknowns
1 point For attempting to
simultaneously solve the
three equations for the
values of the three
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Circuits and Resistance
Solve simulataneously
I1 + I 3 = I 2
Equation One
I1 + I 2 = 0.45
Equation Two
I 2 + I 3 = 0.3
Equation Three
Substituting Equation one into Equation two
I1 + ( I1 + I 3 ) = 0.45 ⇒ 2 I1 + I 3 = 0.45 ⇒ I1 + 0.5 I 3 = 0.225
currents by substituting
the values of the resistors
and the emfs of the
batteries in the equations
1 point For the correct
with part C
Substituting Equation one into Equation three
( I1 + I3 ) + I 3 = 0.3 ⇒ I1 + 2 I 3 = 0.3
I1 + 0.5 I 3 = 0.225
− ( I1 + 2 I 3 = 0.3)
−1.5 I 3 = −0.075 ⇒ I 3 = 0.5 A
I 2 + I 3 = 0.3 ⇒ I 2 = 0.25 A
I1 + I 2 = 0.45 ⇒ I1 = 0.20 A
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Circuits and Resistance
Multiple Choice
Questions 1-3
1. What is the ε of the battery?
A) 3.0 V
B) 4.0 V
C) 12.0 V
D) 24.0 V
E) 48.0 V
ε = IR and I = 4 A
Rs = R1 + R2 = 4Ω + 2Ω = 6Ω
1
1
1
1
1
2
= +
=
+
=
RP R1 R2 6Ω 6Ω 6Ω
C
6Ω
= 3Ω
2
ε = IR = ( 4 A )( 3Ω ) = 12 V
Req =
2. What is the power dissipated by the 4.0 Ω resistor ?
A)
B)
C)
D)
E)
6.0 W
16 W
24 W
48 W
64 W
At junction X, the current splits in half so that 2 A flows through the 4Ω
B
P = IV =
V2
= I 2R
R
P = ( 2 A ) ( 4Ω ) = 16 W
2
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Circuits and Resistance
3. What is the voltage drop across the 2.0 Ω resistor?
A)
B)
C)
D)
E)
2.0 V
4.0 V
6.0 V
8.0 V
12.0 V
B
V = IR = ( 2 A)( 2Ω ) = 4 V
4. Which of the ammeters shown in the diagram above will correctly indicate the current
through the 5Ω resistor?
A)
B)
C)
D)
E)
1 and 2 only
2 and 3 only
2 and 4 only
2 only
3 only
C
Ammeters must be placed in series with the 5 Ω resistor, hence, 2 and 4 only
will measure the current through the resistor.
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Circuits and Resistance
5. Which of the following statements are true concerning the currents in resistors R1, R2,
and R3?
I. I1 &gt; I2
II. I1 &gt; I3
III. I1&lt; I2 + I3
A)
B)
C)
D)
E)
I only
II only
III only
I and II only
I, II, &amp; III
D
All the current flows through resistor R1 and portions of the total
current flow through resistors R2 and R3, such that I1 =I2+ I3.
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Circuits and Resistance
Questions 6-8 refer to the following circuits. Assume all the light bulbs and all the sources are
identical.
I
II
III
6. Which of the circuits would draw the most total current? If two or more circuits draw the same
total current, and this current is the largest, choose all the circuits in which this is the case.
A)
B)
C)
D)
E)
I only
II only
III only
I and III only
I, II, and III
C
Identical bulbs connected in parallel will have less total resistance than in
series, causing more current to be drawn in the circuit.
7. In which of the circuits would the light bulbs be the least bright? If two or more circuits are
the same brightness, and are also the least bright, choose all the circuits in which this is the case.
A)
B)
C)
D)
E)
I only
II only
III only
I and III only
I, II, and III
B
Each bulb would get the least amount of current in circuit II since they are in
series, and the total resistance is the greatest of the three circuits.
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Circuits and Resistance
8. In which of the circuits would the light bulbs be the brightest? If two or more circuits are the
same brightness, and are also the brightest, choose all the circuits in which this is the case.
A)
B)
C)
D)
E)
I only
II only
III only
I and III only
I, II, and III
D
In circuits I and III, each bulb has the same voltage (from the battery) across
it, and is the same resistance. Thus, each bulb will have the same current
passing through it, giving equal brightness, as well as the greatest brightness.
9. A 1.5-volt battery is connected to a circuit in such a way that 4 Coulombs of charge pass a
point in the circuit in a time of 2 seconds. The current flowing in the circuit is
A)
B)
C)
D)
E)
2A
3A
4A
6A
8A
Current is defined as the amount of charge that passes a given point per unit
time:
A
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Circuits and Resistance
10. The internal resistance of the battery in the circuit shown below is most nearly
2A
6Ω
A)
B)
C)
D)
E)
20V
4Ω
7Ω
8Ω
10 Ω
12 Ω
A
By Ohm’s law, 2 A =
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20V
, so the internal resistance R = 4 Ω.
(6Ω + R )
Circuits and Resistance
11. In which of the circuits below are the ammeter and voltmeter correctly connected in order to
measure current through the light bulb and voltage across the light bulb?
A)
V
A
B)
V
A
C)
A
V
D)
A
V
E)
V
C
A
The ammeter should be connected in series with the light bulb, and the
voltmeter should be connected in parallel with the light bulb.
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Circuits and Resistance
Questions 12-14 refer to the circuit diagram shown below. The switch is initially open and the
ammeter labeled A2 indicates the current I2 through the 3-Ω light bulb is 3 A.
A1
A3
3.0 Ω
R2
9.0 V
A2
12. The power dissipated in the 3-Ω light bulb is
A)
B)
C)
D)
E)
1W
3W
9W
18 W
27 W
E
13. The switch is now closed. The first ammeter indicates a current of 9.0 A. Calculate the
current I3 through the third ammeter, A3:
A)
B)
C)
D)
E)
1.0 A
3.0 A
6.0 A
8.0 A
9.0 A
C
The current through the 3-Ω bulb must still be 3.0 A, so the remainder of the
9.0-A total current must pass through the other bulb and the ammeter A3.
Thus, the current through A3 is 9.0 A - 3.0 A = 6.0 A.
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Circuits and Resistance
14. The value of the resistance R is
A)
B)
C)
D)
E)
1.0 Ω
1.5 Ω
3.0 Ω
6.0 Ω
9.0 Ω
B
15. The equivalent resistance in the figure below is
A)
B)
C)
D)
E)
less than 2.0 Ω
between 2.0 and 4.0 Ω
between 4.0 and 6.0 Ω
between 6.0 and 10.0 Ω
greater than 10.0 Ω
C
The 6.0-Ω and 10.0-Ω resistors are in series, so their equivalent resistance is
16.0-Ω. This resistance is in parallel with the 4.0-Ω resistor, giving 16/5 Ω
for the parallel combination. Adding this value to the 2.0-Ω resistor gives
between 4.0-Ω and 6.0-Ω for the total equivalent resistance.
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Circuits and Resistance
16. A copper wire has a cross-sectional area, A and a length, L. Which of the following
would reduce the resistance of the wire by a factor of four?
ρ
A
L
A)
B)
C)
D)
E)
double both the length and the diameter
double only the length
double only the diameter
reduce the length by half
reduce both the length and the diameter by half
C
R=
ρL
A
Doubling the diameter will reduce the resistance of the wire by a
factor of four.
17. A parallel-plate capacitor has a capacitance Co. A second parallel-plate capacitor has
plates with three times the area and half the separation. The capacitance of the second
capacitor is most nearly
F)
G)
H)
I)
J)
&frac12; Co
3/2 Co
Co
3 Co
6 Co
E
C=
Q ε0 A
3
=
so C = C 0 = 6C 0
V
d
.5
&reg;