Advanced Placement
PHYSICS B
Circuits & Resistance
Student
2013-2014
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
ΣI int o a node = ΣI out of
a node
the sum of the inverse of each of the individual resistors. (Ohms Ω)
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 "lost energy"
is converted into heat or light.
Energy is power times time. (Joules J)
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
Important Concepts
• The charge on one electron (or one proton) is called the elementary charge (e) and is equal to 1.6×10-19C.
Hence, it takes 6.25×1018 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,
IRequivalent = IR1 + IR2 + IR3
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Circuits and Resistance
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:
===
VVVV
Total
123
By Ohm’s Law , V = IR, this is equivalent to
V
Requivalent
=
V1 V2 V3
+ +
R1 R2 R3
Factor out the potential difference, V.
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Circuits and Resistance
⎛ 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
Req = R1 + R2 + R3...
•
1
1
1
1
= +
+ ...
Req R1 R2 R3
VT = V1 + V2 + V3...
VT = V1 = V2 = V3...
IT = I1 = I2 = I3...
IT = I1 + I2 + I3...
Req>R1, R2, R3…
Req<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
V = IR so I =
R
36V
I=
= 1.26A for each resistor in series, but now we need to split this 1.26 A across
28.67Ω
each of the resistors in the parallel branch. This means we first need to know the voltage
across each resistor in our new drawing.
V = IR
V = 1.26 A × 1.67Ω = 2.1V
V = 1.26 A × 7Ω = 8.8V across the 7Ω resistor
V = 1.26 A × 10Ω = 12.6V across EACH of the 10Ω 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:
I=
V 2.1V
=
= 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 × 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 whose current you wish to measure. Ammeters have very low resistance
and thus little affect on the current.
Voltmeters must be hooked up in parallel. This allows them to touch a circuit at two different points with very little
current passing through them because they have very high resistance and hence, little affect on 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 VA − 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.
b.
c.
d.
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.
• Capacitors
1. An empty capacitor does not resist the flow of current, and thus acts like a wire.
2. A capacitor which is fully charged 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. Thus, the net charge in a capacitor is
always zero; however, the process of charging creates an electric field between the plates.
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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 can be expressed as C =
•
•
•
•
•
•
Capacitance is the ratio of charge to voltage. (Farads F or
C
)
V
Energy stored in a capacitor. (Joules J)
kε A
Q
or C = 0 where:
V
d
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.
ε 0 is the permittivity constant = 8.85 x 10-12 C2 / Nm2
kε 0 A
, if you pull the two plates of a capacitor apart you decrease its
d
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.
As you can see from looking at the formula C =
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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.
B. Determine the value of the unknown resistor R.
C. Determine the heat dissipated by resistor R in one minute.
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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.
E. Determine the total amount of energy stored in the capacitor.
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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.
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.
C. From your graph, determine the value of the resistor.
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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.
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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
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
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
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Circuits and Resistance
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
5. Which of the following statements are true concerning the currents in resistors R1, R2,
and R3?
I. I1 > I2
II. I1 > I3
III. I1< I2 + I3
A)
B)
C)
D)
E)
I only
II only
III only
I and II only
I, II, & III
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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
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
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
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Circuits and Resistance
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
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
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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
A
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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
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
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
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Circuits and Resistance
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
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
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Circuits and Resistance
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)
½ Co
3/2 Co
Co
3 Co
6 Co
E
C=
Q ε0A
3
so C = C 0 = 6C 0
=
V
d
.5
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2007 AP PHYSICS B FREE-RESPONSE QUESTIONS (Form B)
3. (15 points)
In the circuit above, a 12.0 V battery is connected to two resistors, one of resistance 1000 W and the other of
resistance 500 W . A capacitor with a capacitance of 30 ¥ 10 -6 F is connected in parallel with the 500 W
resistor. The circuit has been connected for a long time, and all currents have reached their steady states.
(a) Calculate the current in the 500 W resistor.
(b)
i. Draw an ammeter in the circuit above in a location such that it could measure the current in the 500 W
resistor. Use the symbol
to indicate the ammeter.
ii. Draw a voltmeter in the circuit above in a location such that it could measure the voltage across the
1000 W resistor. Use the symbol
to indicate the voltmeter.
(c) Calculate the charge stored on the capacitor.
(d) Calculate the power dissipated in the 1000 W resistor.
(e) The capacitor is now discharged, and the 500 W resistor is removed and replaced by a resistor of greater
resistance. The circuit is reconnected, and currents are again allowed to come to their steady-state values. Is
the charge now stored on the capacitor larger, smaller, or the same as it was in part (c)?
____ Larger
____ Smaller
____ The same as
Justify your answer.
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2003 AP® PHYSICS B FREE-RESPONSE QUESTIONS
2. (15 points)
A circuit contains two resistors (10 Ω and 20 Ω) and two capacitors (12 mF and 6 mF) connected to a 6 V battery,
as shown in the diagram above. The circuit has been connected for a long time.
(a) Calculate the total capacitance of the circuit.
(b) Calculate the current in the 10 W resistor.
(c) Calculate the potential difference between points A and B.
(d) Calculate the charge stored on one plate of the 6 mF capacitor.
(e) The wire is cut at point P . Will the potential difference between points A and B increase, decrease, or
remain the same?
____ increase
____ decrease
____ remain the same
Justify your answer.
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2002 AP® PHYSICS B FREE-RESPONSE QUESTIONS (Form B)
3. (15 points)
Lightbulbs of fixed resistance 3.0 W and 6.0 W, a 9.0 V battery, and a switch S are connected as shown in the
schematic diagram above. The switch S is closed.
(a) Calculate the current in bulb A.
(b) Which lightbulb is brightest? Justify your answer.
(c) Switch S is then opened. By checking the appropriate spaces below, indicate whether the brightness of each
lightbulb increases, decreases, or remains the same. Explain your reasoning for each lightbulb.
i. Bulb A: The brightness
___ increases
___ decreases
___ remains the same
___ increases
___ decreases
___ remains the same
___ increases
___ decreases
___ remains the same
Explanation:
ii. Bulb B: The brightness
Explanation:
iii. Bulb C: The brightness
Explanation:
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2002 AP® PHYSICS B FREE-RESPONSE QUESTIONS
3. (15 points)
Two lightbulbs, one rated 30 W at 120 V and another rated 40 W at 120 V, are arranged in two
different circuits.
(a) The two bulbs are first connected in parallel to a 120 V source.
i. Determine the resistance of the bulb rated 30 W and the current in it when it is connected
in this circuit.
ii. Determine the resistance of the bulb rated 40 W and the current in it when it is connected in
this circuit.
(b) The bulbs are now connected in series with each other and a 120 V source.
i. Determine the resistance of the bulb rated 30 W and the current in it when it is connected in
this circuit.
ii. Determine the resistance of the bulb rated 40 W and the current in it when it is connected in
this circuit.
(c) In the spaces below, number the bulbs in each situation described, in order of their brightness.
(1 = brightest, 4 = dimmest)
____30 W bulb in the parallel circuit
____40 W bulb in the parallel circuit
____30 W bulb in the series circuit
____40 W bulb in the series circuit
(d) Calculate the total power dissipated by the two bulbs in each of the following cases.
i. The parallel circuit
ii. The series circuit
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