PH203 Chapter 23 solutions
Tactics Box 23.1 Using Kirchhoff's Loop Law
Description: Knight/Jones/Field Tactics Box 23.1 Using Kirchhoff’s loop law is illustrated.
Learning Goal: To practice Tactics Box 23.1 Using Kirchhoff’'loop law.
Circuit analysis is based on Kirchhoff's laws, which can be summarized as follows:
• Kirchhoff's junction law says that the total current into a junction must equal the total current
leaving the junction.
• Kirchhoff's loop law says that if we add all of the potential differences around the loop formed
by a circuit, the sum of these potential differences must be zero.
While Kirchhoff's junction law is needed only when there are one or more junctions in a circuit,
Kirchhoff's loop law is used for analyzing any type of circuit, as explained in the following Tactics Box.
Using Kirchhoff’s loop law
Draw a circuit diagram. Label all known and unknown quantities.
Assign a direction to the current. Draw and label a current arrow to show your choice.
Choose the direction of the current based on how the batteries or sources of emf "want" the
current to go. If you choose the current direction opposite the actual direction, the final value
for the current that you calculate will have the correct magnitude but will be negative, letting
you know that the direction is opposite the direction you chose.
"Travel" around the loop. Start at any point in the circuit; then go all the way around the
loop in the direction you assigned to the current in step 2. As you go through each circuit
TACTICS BOX 23.1
1.
2.
3.
element,
is interpreted to mean
.

For a battery with current in the negative-to-positive direction:

For a battery in the positive-to-negative direction (i.e., the current is going into the
.
positive terminal of the battery):

4.
For a resistor:
.
.
Apply the loop law:
.
Part A
The current in the circuit shown in the figure is 0.20
. What is the potential difference
the battery traveling in the direction shown in the figure?
Hint
A.1
across
Find the potential difference across the resistor
What is the potential difference
across the resistor in the direction assigned to the current in
the figure?
Express your answer in volts.
ANSWER:
=
Now apply Kirchhoff's loop law to find the potential difference across the battery.
Express your answer in volts.
Express your answer in volts.
ANSWER:
=
Now apply Kirchhoff's loop law to find the potential difference across the battery.
Express your answer in volts.
ANSWER:
=
Part B
Find the current
Hint
B.1
in the circuit shown in the figure.
Find the potential difference across the battery
If we assign the counterclockwise direction to the current, what is the potential difference across
the battery,
?
Express your answer in volts.
ANSWER:
Hint
B.2
Find the potential difference across the 40-ohm resistor
If we assign the counterclockwise direction to the current
difference
across the 40-ohm resistor?
in the circuit, what is the potential
Hint
B.2
Find the potential difference across the 40-ohm resistor
If we assign the counterclockwise direction to the current
difference
in the circuit, what is the potential
across the 40-ohm resistor?
Express your answer in terms of the variable
.
ANSWER:
=
Hint
B.3
Find the potential difference across the 50-ohm resistor
If we assign the counterclockwise direction to the current
difference
in the circuit, what is the potential
across the 50-ohm resistor?
Express your answer in terms of the variable
.
ANSWER:
=
Now apply Kirchhoff's loop law and solve for
.
Express your answer in amperes.
ANSWER:
=
Part C
What is the potential difference
across the unknown element in the circuit shown in the figure?
Express your answer as if traveling across the element in the direction shown in the figure.
Hint
C.1
Apply Kirchhoff's loop law
Complete the expression below and write an equation for Kirchhoff's loop law applied to the circuit
in the figure.
Express your answer in terms of the variable
.
ANSWER:
=
Express your answer in volts.
ANSWER:
=
Series And Parallel Connections
Description: Several calculations of increasing complexity that help the students practice finding the
equivalent resistance of the circuits combining series and parallel connections.
Learning Goal: To learn to calculate the equivalent resistance of the circuits combining series and
parallel connections.
Resistors are often connected to each other in electric circuits. Finding the equivalent resistance of
combinations of resistors is a common and important task. Equivalent resistance is defined as the single
resistance that can replace the given combination of resistors in such a manner that the currents in the
rest of the circuit do not change.
Finding the equivalent resistance is relatively straighforward if the circuit contains only series and
parallel connections of resistors.
An example of a series connection is shown in the diagram:
For such a connection, the current is the same for all individual resistors and the total voltage is the sum
of the voltages across the individual resistors.
Using Ohm's law (
), one can show that, for a series connection, the equivalent resistance is
the sum of the individual resistances.
Mathematically, these relationships can be written as:
An example of a parallel connection is shown in the diagram:
For resistors connected in parallel the voltage is the same for all individual resistors because they are
all connected to the same two points (A and B on the diagram). The total current is the sum of the
currents through the individual resistors. This should makes sense as the total current "splits" at points A
and B.
Using Ohm's law, one can show that, for a parallel connection, the reciprocal of the equivalent
resistance is the sum of the reciprocals of the individual resistances.
Mathematically, these relationships can be written as:
NOTE: If you have already studied capacitors and the rules for finding the equivalent capacitance, you
should notice that the rules for the capacitors are similar - but not quite the same as the ones discussed
here.
In this problem, you will use the the equivalent resistance formulas to determine
combinations of resistors.
for various
Part A
For the combination of resistors shown, find the equivalent resistance between points A and B.
Express your answer in Ohms.
ANSWER:
=
These resistors are connected in series; the current through each is the same.
Part B
For the set-up shown, find the equivalent resistance between points A and B.
Express your answer in Ohms.
ANSWER:
=
Express your answer in Ohms.
ANSWER:
=
This is a parallel connection since the voltage across each resistor is the same.
Part C
For the combination of resistors shown, find the equivalent resistance between points A and B.
Hint
C.1
How to approach the question
You cannot say that all three resistors are connected either in series or in parallel: this circuit has to
be viewed as a combination of different connections.
Hint
C.1
How to approach the question
You cannot say that all three resistors are connected either in series or in parallel: this circuit has to
be viewed as a combination of different connections.
Find the equivalent resistance of the "4-Ohm-12 Ohm" combination first.
Hint
C.2
What kind of connection is this?
The 2-Ohm resistor is connected:
ANSWER:
in series with the 4-Ohm resistor
in series with the 12-Ohm resistor
in series with the combination of the 4-Ohm and the 12-Ohm resistors
in parallel with the 4-Ohm resistor
in parallel with the 12-Ohm resistor
in parallel with the combination of the 4-Ohm and the 12-Ohm resistors
Express your answer in Ohms.
ANSWER:
=
In this case, you cannot say that all three resistors are connected either in series or in parallel. You
have a combination of a series and a parallel connection.
Some circuits may contain a large number of resistors connected in various ways. To determine the
equivalent resistance of such circuits, you have to take several steps, carefully selecting the "subcombinations" of resistors connected in relatively obvious ways. Good record-keeping is essential
here.
The next question helps you practice this skill.
Part D
For the combination of resistors shown, find the equivalent resistance between points A and B.
Hint
D.1
How to approach the question
Find separately the equivalent resistances of the top and the bottom branches of the circuit; then
combine them.
Hint
D.2
Find
for the "4-6-12" combination
What is the equivalent resistance for the "4 ohm - 6 ohm - 12 Ohm" combination?
Express your answer in ohms.
ANSWER:
=
Hint
D.3
Find
for the top branch
What is the equivalent resistance for the top branch of the circuit (between C and D)?
Express your answer in ohms.
ANSWER:
=
Hint
D.3
Find
for the top branch
What is the equivalent resistance for the top branch of the circuit (between C and D)?
Express your answer in ohms.
ANSWER:
=
Hint
D.4
Find
for the bottom branch
What is the equivalent resistance for the bottom branch of the circuit (between E and F)?
Express your answer in ohms.
ANSWER:
=
Express your answer in Ohms.
ANSWER:
=
The next level of analyzing a circuit is to determine the voltages across and the currents through the
various branches of the circuit. You will practice that skill in the future.
Of course, there are circuits that cannot possibly be represented as combinations of series and parallel
connections. However, there are ways to analyze those, too.
Brightness of Light Bulbs Ranking Task
Description: Asks students to rank brightness of light bulbs in a mixed series and parallel circuit.
Part A
Consider a circuit containing five identical light bulbs and an ideal battery. Assume that the resistance
of each light bulb remains constant. Rank the bulbs (A through E) based on their brightness.
Hint
A.1
How to approach the problem
The greater the current in a bulb, the brighter the bulb is. Compare the currents in the light bulbs.
Hint
A.2
Comparing bulb A to bulb B
When a wire "splits," the current splits according to the resistance in each pathway. How does the
resistance in the pathway through bulb A compare to the resistance in the pathway through bulb B?
Hint
A.3
Comparing bulb D to bulb E
Bulb D and bulb E are connected in series. How much of the charge that goes through D also goes
through E?
Hint
A.4
Comparing bulb C to bulb D or E
When a wire "splits," the current splits according to the resistance in each pathway. How does the
resistance in the pathway of bulb C compare to the resistance in the pathway of bulbs D and E?
Bulb D and bulb E are connected in series. How much of the charge that goes through D also goes
through E?
Hint
A.4
Comparing bulb C to bulb D or E
When a wire "splits," the current splits according to the resistance in each pathway. How does the
resistance in the pathway of bulb C compare to the resistance in the pathway of bulbs D and E?
Recall that two bulbs connected in series have more resistance than a single bulb.
Hint
A.5
Comparing bulb C to bulb A or B
What fraction of the total current (through the battery) goes through the pathway containing bulb
A? Recall that the pathways through bulbs A and B have the same resistance, since the bulbs are
identical. Compare that number to the fraction of the current that goes through the pathway
containing bulb C.
Rank from brightest to dimmest. To rank items as equivalent, overlap them.
ANSWER:
View
Now consider what happens when a switch in the circuit is opened.
Part B
What happens to the brightness of bulb A?
Hint
B.1
How to approach this part
Light bulbs D and E are, in effect, disconnected from the circuit. The current exists only in light
bulbs A, B, and C. How would the change affect the currents in A, B, and C?
Hint
Consider changes in resistance
Hint
B.1
How to approach this part
Light bulbs D and E are, in effect, disconnected from the circuit. The current exists only in light
bulbs A, B, and C. How would the change affect the currents in A, B, and C?
Hint
B.2
Consider changes in resistance
How does the resistance of bulb C alone compare with the resistance of bulb C in parallel with
bulbs D and E?
ANSWER:
It gets dimmer.
It gets brighter.
There is no change.
Part C
What happens to bulb C?
Hint
C.1
How to approach this part
This question can be answered by calculating the "before" and "after" currents through C in terms
of the emf provided by the battery and the resistance of each bulb.
Hint
C.2
Find the current in bulb C earlier
Before the switch was open, the total resistance of the earlier circuit was
the resistance of one bulb. What was the current
be
, where
is
in bulb C? Let the emf produced by the battery
.
Express your answer in terms of
and
.
ANSWER:
=
Hint C.3
Find the current in bulb C now
What is the current
battery be
through bulb C after opening the switch? Let the emf produced by the
.
Express your answer in terms of
ANSWER:
=
and
.
Express your answer in terms of
and
.
ANSWER:
=
ANSWER:
It gets dimmer.
It gets brighter.
There is no change.
This is why appliances in your home are always connected in parallel. Otherwise, turning some of
them on or off would cause the current in others to change, which could damage them.
Equivalent Resistance
Description: Find the equivalent resistance of a network of resistors with series and parallel
connections. The network geometry gets progressively more complicated by adding more resistors.
Consider the network of four resistors shown in the diagram, where
= 1.00
.
Part A
, and
= 7.00
= 2.00
,
= 5.00
,
. The resistors are connected to a constant voltage of magnitude
Find the equivalent resistance
Hint
A.1
of the resistor network.
How to reduce the network of resistors
The network of resistors shown in the diagram is a combination of series and parallel connections.
To determine its equivalent resistance, it is most convenient to reduce the network in successive
stages. First compute the equivalent resistance of the parallel connection between the resistors
and
, and imagine replacing the connection with a resistor with such resistance. The resulting
network will consist of three resistors in series. Then find their equivalent resistance, which will
also be the equivalent resistance of the original network.
Hint
A.2
Find the resistance equivalent to
Find the equivalent resistance
Hint
A.2.1
and
of the parallel connection between the resistors
and
Two resistors in parallel
Consider two resistors of resistance
and
equivalent to a resistor with resistance
that are connected in parallel. They are
, which satisfies the following relation:
.
Express your answer in ohms.
ANSWER:
=
If you replace the resistors
and
with an equivalent resistor with resistance
, the
resulting network will consist of three resistors
,
, and
connected in series. Their
equivalent resistance is also the equivalent resistance of the original network.
Hint
A.3
Three resistors in series
Consider three resistors of resistance
equivalent to a resistor with resistance
,
, and
that are connected in series. They are
, which is given by
.
Express your answer in ohms.
ANSWER:
=
.
Consider three resistors of resistance
,
equivalent to a resistor with resistance
, and
that are connected in series. They are
, which is given by
.
Express your answer in ohms.
ANSWER:
=
Part B
Two resistors of resistance
= 3.00
additional resistor of resistance
and
= 3.00
= 3.00
are added to the network, and an
is connected by a switch, as shown in the diagram..
Find the equivalent
resistance
Hint
B.1
of the new resistor network when the switch is open.
How to reduce the extended network of resistors
Since the switch is open, no current passes through the resistor
, which can be ignored then. As
you did in Part A, reduce the network in successive stages. Note that the new resistor
series with the resistors
Hint
B.2
and
, while the new resistor
Find the resistance equivalent to
Find the resistance
,
is in series with
is in
.
, and
equivalent to the resistor connection with
,
, and
.
series with the resistors
Hint
B.2
, while the new resistor
Find the resistance equivalent to
Find the resistance
Hint
B.2.1
and
,
.
, and
equivalent to the resistor connection with
Find the resistance equivalent to
Find the resistance
is in series with
,
, and
.
and
equivalent to the connection between
and
.
Hint
Two resistors in series
B.2.1.1
Consider two resistors of resistance
, and
equivalent to a resistor with resistance
that are connected in series. They are
, which is given by
.
Express your answer in ohms.
ANSWER:
=
If you replace the resistors
resistor
Hint
B.2.2
and
with their equivalent resistor (of resistance
will result in parallel with
), the
.
Two resistors in parallel
Consider two resistors of resistance
and
equivalent to a resistor with resistance
that are connected in parallel. They are
, which satisfies the following relation:
.
Express your answer in ohms.
ANSWER:
=
If you replace the resistors
,
, and
with an equivalent resistor with resistance
,
the resulting network will consist of four resistors—
,
,
, and
—all connected in
series. Their equivalent resistance is also the equivalent resistance of the original network.
If you replace the resistors
,
, and
with an equivalent resistor with resistance
,
the resulting network will consist of four resistors—
,
,
, and
—all connected in
series. Their equivalent resistance is also the equivalent resistance of the original network.
Hint
B.3
Four resistors in series
Consider four resistors of resistance
,
,
equivalent to a resistor with resistance
, and
that are connected in series. They are
, which is given by
.
Express your answer in ohms.
ANSWER:
=
Part C
Find the equivalent resistance
closed.
Hint
C.1
of the resistor network described in Part B when the switch is
How to reduce the network of resistors when the switch is closed
When the switch is closed, current passes through the resistor
; therefore the resistor must be
included in the calculation of the equivalent resistance. Also when the switch is closed, the resistor
is no longer connected in series with the resistors
switch was open. Instead, now
series with
Hint
C.2
and
, as was the case when the
and their equivalent resistor will be in
.
Find the resistance equivalent to
Find the equivalent resistance
Hint
C.2.1
is in parallel with
and
and
of the parallel connection between the resistors
and
Two resistors in parallel
Consider two resistors of resistance
equivalent to a resistor with resistance
and
that are connected in parallel. They are
, which satisfies the following relation:
.
Express your answer in ohms.
.
Consider two resistors of resistance
and
equivalent to a resistor with resistance
that are connected in parallel. They are
, which satisfies the following relation:
.
Express your answer in ohms.
ANSWER:
=
If you replace the resistors
resistors
,
and
and
with their equivalent resistor (of resistance
with their equivalent resistor (of resistance
), and the
), calculated in Part
B, the resulting network will consist of four resistors—
,
,
, and
—all
connected in series. Their equivalent resistance is also the equivalent resistance of the original
network.
Hint
C.3
Four resistors in series
Consider four resistors of resistance
,
equivalent to a resistor with resistance
,
, and
that are connected in series. They are
, which is given by
.
Express your answer in ohms.
ANSWER:
=
Resistance and Wire Length
Description: Cut a long resistor of known resistance into n identical shorter resistors. Find the
resistance of the short resistors reconnected in parallel.
You have been given a long length of wire. You measure the resistance of the wire, and find it to be
. You then cut the wire into
identical pieces
.
Part A
If you connect the
pieces in parallel as shown
, what is the total
resistance
Hint
A.1
of the
wires connected in parallel?
Find the resistance of the wire segments
A wire's resistance is proportional to its length. Find the resistance
segments of wire.
Hint A.1.1
of each of the
short
Formula for the resistance of a wire
The formula for the resistance
of a wire in terms of its length
and cross-sectional area
A wire's resistance is proportional to its length. Find the resistance
segments of wire.
Hint A.1.1
of each of the
short
Formula for the resistance of a wire
The formula for the resistance
is
of a wire in terms of its length
and cross-sectional area
,
where
is the resistivity of the material of the wire.
ANSWER:
=
Hint
A.2
Resistors in parallel
For resistors in parallel of resistance
be determined from the equality
for
, the total resistance
.
If all resistors have the same resistance
resistors?
Express your answer in terms of
and
, what is the total resistance of the
parallel
.
ANSWER:
=
Express your answer in terms of
and
.
ANSWER:
=
Measuring the EMF and Internal Resistance of a Battery
can
Description: Find the emf and internal resistance of a battery using a simple circuit. Could be used as a
pre-lab.
When switch S in the figure is open, the voltmeter V of the battery reads 3.12
. When the switch is
closed, the voltmeter reading drops to 2.95
, and the ammeter A reads 1.68
meters are ideal, so they do not affect the circuit.
. Assume that the two
Part A
Find the emf
.
Express your answer in volts to three significant digits.
ANSWER:
=
Part B
Find the internal resistance
Hint
B.1
of the battery.
How to approach the problem
The voltmeter reading is equal to
minus the voltage drop across the internal resistor. If there is
no current flowing, there is no voltage drop across the internal resistor, but once current starts to
flow, there will be a voltage drop across it.
Express your answer in ohms to four significant digits.
ANSWER:
=
ANSWER:
=
Part C
Find the circuit resistance
Hint
C.1
.
Find the voltage drop across the circuit resistor
When the switch is closed, what is the voltage drop
across the circuit resistor (of resistance
)?
Express your answer in volts to three significant digits.
ANSWER:
=
You now know the voltage drop across the resistor and the current through it. Use Ohm's law to
find
.
Express your answer in ohms to three significant digits.
ANSWER:
=
This is the kind of circuit you would use in real life to measure the emf and internal resistance of a
battery. You need the second resistor
to increase the resistance in the circuit so that the current
flowing through the ammeter is not too large. In fact, you would need to figure out roughly how big a
resistance to use once you had determined the emf of the battery, depending on the range of the
ammeter you were using.
Kirchhoff's Current Rule Ranking Task
Description: Short conceptual problem about currents through resistors in various circuits. (ranking
task)
The placement of resistors in a circuit is one factor that can determine the current passing through the
resistor. You will be given three circuits, and for each circuit you will be asked to compare the current
through the various resistors.
In each of the circuits in Parts A to C, all resistors are identical.
Part A
Rank the resistors in the figure below (A to C) on the basis of the current that flows through them.
Hint
A.1
Kirchhoff's current rule for circuit junctions
Kirchhoff’s current rule states that the current flowing into a junction (a point at which the number
of paths available for current flow changes) must equal the current flowing out of the junction. The
portion of the current that flows through each available path depends on the resistance of each path.
Paths with less resistance will receive a larger share of the current.
Rank from largest to smallest. To rank items as equivalent, overlap them.
ANSWER:
View
Part B
Rank the resistors in the figure below (A to C) on the basis of the current that flows through them.
Hint
B.1
Kirchhoff's current rule for circuit junctions
Kirchhoff’s current rule states that the current flowing into a junction (a point at which the number
of paths available for current flow changes) must equal the current flowing out of the junction. The
portion of the current that flows through each available path depends on the resistance of each path.
Paths with less resistance will receive a larger share of the current.
Rank from largest to smallest. To rank items as equivalent, overlap them.
ANSWER:
View
Part C
Rank the resistors in the figure below (A to D) on the basis of the current that flows through them.
Hint
C.1
Kirchhoff's current rule for circuit junctions
Kirchhoff’s current rule states that the current flowing into a junction (a point at which the number
of paths available for current flow changes) must equal the current flowing out of the junction. The
portion of the current that flows through each available path depends on the resistance of each path.
Paths with less resistance will receive a larger share of the current.
Rank from largest to smallest. To rank items as equivalent, overlap them.
ANSWER:
View
Kirchhoff's Loop Rule Conceptual Question
Description: Short conceptual problem related to Kirchhoff's Loop (Voltage) Rule.
The circuit shown below
consists of four
different resistors and a battery. You don't know the strength of the battery or the value any of the four
resistances.
Part A
Select the expressions that will be equal to the voltage of the battery in the circuit, where
example, is the potential drop across resistor A.
Hint
A.1
, for
Kirchhoff's voltage rule for closed circuit loops
Kirchhoff’s loop rule states that in any closed circuit loop, the voltage supplied by a battery must
be used by the devices in the loop. Therefore, the voltage drop across all of the resistors in a single
closed circuit loop must add up to the voltage of the battery. Carefully identify all of the closed
loops in this circuit.
Check all that apply.
ANSWER:
Capacitors in Parallel
Description: A series of questions on basic calculations for capacitors connected in parallel; there is a
similar skill-builder ("Capacitors in Series") about capacitors connected in series. (Knight potential
difference notation)
Learning Goal: To understand how to calculate capacitance, voltage, and charge for a parallel
combination of capacitors.
Frequently, several capacitors are connected together to form a collection of capacitors. We may be
interested in determining the overall capacitance of such a collection. The simplest configuration to
analyze involves capacitors connected in series or in parallel. More complicated setups can often
(though not always!) be treated by combining the rules for these two cases. Consider the example of a
parallel combination of capacitors: Three capacitors are connected to each other and to a battery as
shown in the figure.
The individual capacitances are
,
, and
, and the battery's voltage is
.
Part A
If the potential of plate 1 is
, then, in equilibrium, what are the potentials of plates 3 and 6?
Assume that the negative terminal of the battery is at zero potential.
Hint A.1
Electrostatic equilibrium
When electrostatic equilibrium is reached, all objects connected by a conductor (by wires, for
example) must have the same potential. Which plates on this diagram are at the same potential?
ANSWER:
and
and
and
and
and
and
Part B
If the charge of the first capacitor (the one with capacitance
the second and third capacitors?
Hint
B.1
, then what are the charges of
Definition of capacitance
Capacitance
is given by
voltage across it.
Hint
B.2
) is
, where
is the charge of the capacitor and
is the
Voltages across the capacitors
As established earlier, the voltage across each capacitor is
for capacitors connected in parallel.
. The voltage is always the same
ANSWER:
and
and
and
and
Part C
Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the
charges of the three individual capacitors calculated in the previous part, find the total charge
for this equivalent capacitor.
Express your answer in terms of
ANSWER:
=
and
.
Part D
Using the value of
Hint
D.1
, find the equivalent capacitance
for this combination of capacitors.
Using the definition of capacitance
Use the general formula
to find
. The charge on the "equivalent" capacitor is
, and the voltage across this capacitor is the voltage across the battery,
Express your answer in terms of
.
.
ANSWER:
=
The formula for combining three capacitors in parallel is
.
How do you think this formula may be generalized to
capacitors?
Equivalent Capacitance
Description: Find the equivalent capacitance of a network of capacitors with series and parallel
connections.
Consider the combination of capacitors shown in the diagram, where
,
= 3.00
, and
= 5.00
.
= 3.00
,
= 11.0
Part A
Find the equivalent capacitance
Hint
A.1
of the network of capacitors.
How to reduce the network of capacitors
To find the equivalent capacitance of the given network of capacitors, it is most convenient to
reduce the network in successive stages. First, replace the capacitors
,
, and
, which are
in parallel, with a single capacitor with an equivalent capacitance. By doing so, you will reduce the
network to a series connection of two capacitors. At this point, you only need to find their
equivalent capacitance.
Hint
A.2
Find the capacitance equivalent to
Find the capacitance
,
, and
equivalent to the parallel connection of the capacitors
,
, and
.
Hint
A.2.1
Find the capacitance equivalent to
Find the capacitance
and
equivalent to the parallel connection of the capacitors
and
.
Hint
Two capacitors in parallel
A.2.1.1
Consider two capacitors of capacitance
to a capacitor with capacitance
and
connected in parallel. They are equivalent
given by
.
Express your answer in microfarads.
ANSWER:
Consider two capacitors of capacitance
to a capacitor with capacitance
and
connected in parallel. They are equivalent
given by
.
Express your answer in microfarads.
ANSWER:
=
If you replace the capacitors
and
with a capacitor of capacitance
network would be a parallel connection between
and
, the resulting
.
Express your answer in microfarads.
ANSWER:
=
If you replace the capacitors
,
, and
with a capacitor of capacitance
resulting network would be a series connection between
is also the equivalent capacitance of the original network.
Hint
A.3
and
, the
. Its equivalent capacitance
Two capacitors in series
Consider two capacitors of capacitance
capacitor of capacitance
and
connected in series. They are equivalent to a
that satisfies the following relation:
.
Express your answer in microfarads.
ANSWER:
=
Part B
Two capacitors of capacitance
shown in the diagram.
= 6.00
and
= 3.00
are added to the network, as
Find the equivalent
capacitance
Hint
B.1
of the new network of capacitors.
How to reduce the extended network of capacitors
To determine the equivalent capacitance of the extended network of capacitors, it is again
convenient to reduce the network in successive stages. First, determine the equivalent capacitance
of the series connection of the capacitors
and
. Then, combine it with the equivalent
capacitance of the parallel connection of
,
, and
, and replace the five capacitors with
their equivalent capacitor. The resulting network will consist of two capacitors in series. At this
point, you only need to find their equivalent capacitance.
Hint
B.2
Find the equivalent capacitance of
Find the equivalent capacitance
,
,
,
, and
of the combination of capacitors
,
,
,
, and
.
Hint
B.2.1
Find the equivalent capacitance of
Find the equivalent capacitance
and
of the series connection of
and
.
Hint
Two capacitors in series
B.2.1.1
Consider two capacitors of capacitance
to a capacitor of capacitance
and
connected in series. They are equivalent
that satisfies the following relation:
.
Consider two capacitors of capacitance
to a capacitor of capacitance
and
connected in series. They are equivalent
that satisfies the following relation:
.
Express your answer in microfarads.
ANSWER:
=
Hint
B.2.2
Find the equivalent capacitance of
Find the equivalent capacitance
,
, and
of the parallel connection of
,
, and
Hint
Three capacitors in parallel
B.2.2.1
Consider three capacitors of capacitance
,
equivalent to a capacitor with capacitance
, and
connected in parallel. They are
given by
.
Express your answer in microfarads.
ANSWER:
=
Express your answer in microfarads.
ANSWER:
=
If you replace the capacitors
,
,
,
, and
with a capacitor of capacitance
the resulting network would be a series connection between
and
capacitance is also the equivalent capacitance of the original network.
Hint
B.3
Two capacitors in series
. Its equivalent
,
Consider two capacitors of capacitance
capacitor of capacitance
and
connected in series. They are equivalent to a
that satisfies the following relation:
.
Express your answer in microfarads.
ANSWER:
=
RC Circuit and Current Conceptual Question
Description: Short conceptual problem about current through a charging and discharging RC circuit.
In the diagram below,
the two resistors,
open.
and
, are identical and the capacitor is initially uncharged with the switch
Part A
How does the current through
switch is first closed?
Hint A.1
immediately after the
Using Kirchhoff's junction rule for currents
At all times, the current through
through
compare with the current through
(entering the junction) must be equal to the sum of the currents
and the capacitor’s branch (exiting the junction). The relative sizes of the currents
Hint A.1
Using Kirchhoff's junction rule for currents
At all times, the current through
(entering the junction) must be equal to the sum of the currents
through
and the capacitor’s branch (exiting the junction). The relative sizes of the currents
through
branches.
and the capacitor’s branch are determined by the effective resistances of these
ANSWER:
The current through
the current through
.
Part B
How does the current through
switch has been closed?
Hint
B.1
compare with the current through
a very long time after the
Using Kirchhoff's junction rule for currents
At all times, the current through
(entering the junction) must be equal to the sum of the currents
through
and the capacitor’s branch (exiting the junction). The relative sizes of the currents
through
branches.
and the capacitor’s branch are determined by the effective resistances of these
Hint
B.2
Current associated with a fully charged capacitor
When a capacitor is fully charged, the current in the branch of the circuit containing the capacitor is
zero.
ANSWER:
The current through
the current through
.
Part C
How does the current through
compare with the current through
is opened (after being closed a very long time)?
Hint C.1
immediately after the switch
Effect of a discharging capacitor
The charge is "held" on the plates of the capacitor by the potential difference of the battery. When
the battery is removed from the system, the negative charge on one plate of the capacitor will flow
toward the positive charge on the other plate if a conducting path (circuit loop) exists. During this
discharging phase, the capacitor behaves in a way very analogous to a battery.
ANSWER:
The current through
the current through
.
The charge is "held" on the plates of the capacitor by the potential difference of the battery. When
the battery is removed from the system, the negative charge on one plate of the capacitor will flow
toward the positive charge on the other plate if a conducting path (circuit loop) exists. During this
discharging phase, the capacitor behaves in a way very analogous to a battery.
ANSWER:
The current through
the current through
.
Ionic Potentials across Cell Membranes Conceptual Question
Description: Short conceptual problem dealing with ionic potentials across cell membranes.
In its resting state, the membrane surrounding a neuron is permeable to potassium ions but only slightly
permeable to sodium ions. Thus, positive K ions can flow through the membrane in an attempt to
equalize K concentration, but Na ions cannot as quickly. This leads to an excess of Na ions outside of
the cell. If the space outside the cell is defined as zero electric potential, then the electric potential of the
interior of the cell is negative. This resting potential is typically about
situation is shown in the figure.
80
. A schematic of this
In response to an electrical stimulus, certain channels in the membrane can become permeable to Na
ions. Due to the concentration gradient, Na ions rush into the cell and the interior of the cell reaches an
electric potential of about 40
. This process is termed depolarization. In response to
depolarization, the membrane again becomes less permeable to Na ions, and the K ions flow out of the
interior of the cell through channels established by the positive electric potential inside of the cell. This
then reestablishing the resting potential. This is termed repolarization. Only a small percentage of the
available Na and K ions participate in each depolarization/repolarization cycle, so the cell can respond
to many stimuli in succession without depleting its "stock" of available Na and K ions. A graph of an
electric potential inside a cell vs. time is shown in the next figure
for a single
depolarization/repolarization cycle.
Part A
During the resting phase, what is the electric potential energy of a typical Na ion outside of the cell?
Hint A.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
.)
ANSWER:
40
+40
80
. Thus,
ANSWER:
40
+40
80
+80
0
Part B
During the resting phase, what is the electrical potential energy of a typical K ion inside of the cell?
Hint B.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
that
. Thus,
ANSWER:
40
+40
80
+80
0
.)
and recalling
ANSWER:
40
+40
80
+80
0
Part C
During depolarization, what is the work done (by the electric field) on the first few Na ions that enter
the cell?
Hint
C.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
. Thus,
.)
Hint
C.2
Algebraic sign of the work
In general, work is defined as the product of the force applied parallel (or antiparallel) to the
displacement of an object. Thus,
.
The work done by a force is positive if the force and the displacement are parallel; it is negative if
the force and displacement are opposite in direction.
Hint C.3
Magnitude of the work
Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers,
the magnitude of the work done on an object is equal to the magnitude of the object’s change in
energy. Since the primary form of energy present in this example is electric potential energy, the
The work done by a force is positive if the force and the displacement are parallel; it is negative if
the force and displacement are opposite in direction.
Hint C.3
Magnitude of the work
Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers,
the magnitude of the work done on an object is equal to the magnitude of the object’s change in
energy. Since the primary form of energy present in this example is electric potential energy, the
magnitude of the work done is equal to the change in the ion’s electric potential energy.
ANSWER:
40
+40
80
+80
120
+120
0
Part D
During repolarization, what is the work done (by the electric field) on the first few K ions that exit the
cell?
Hint
D.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
.)
Hint
. Thus,
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
. Thus,
.)
Hint
D.2
Algebraic sign of the work
In general, work is defined as the product of the force applied parallel (or antiparallel) to the
displacement of an object. Thus,
.
The work done by a force is positive if the force and the displacement are parallel; the work done is
negative if the force and displacement are opposite in direction.
Hint D.3
Magnitude of the work
Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers,
the magnitude of the work done on an object is equal to the magnitude of the object’s change in
energy. Since the primary form of energy present in this example is electric potential energy, the
magnitude of the work done is equal to the change in the ion’s electric potential energy.
ANSWER:
40
+40
80
+80
120
+120
0
Problems
P23.11. Prepare: Please refer to Figure P23.11. The three resistances in (a), (b), and
(c) are parallel resistors. We will thus use Equation 23.12 to find the equivalent
resistance.
Solve: (a) The equivalent resistance is
(b) The equivalent resistance is
(c) The equivalent resistance is
Assess: We must learn how to combine series and parallel resistors.
P23.13.
Prepare:
For resistors in parallel,
Adding four of the resistors in parallel gives
Solve: We can put four of the resistors in parallel. The total resistance is now
Assess: There are other ways to arrive at the same using more from our collection of
resistors.
P23.15.
Prepare:
The resistance of the three
resistors in parallel is
Solve: That parallel combination in series with the
resistor adds to an equivalent
resistance of
So that’s the answer: the three
resistors in parallel with each other, and then that
combination in series with the
resistor.
Assess: There are typical resistances that one can buy; generally one can’t (cheaply)
buy resistors with every possible value of resistance. So it is common to combine the
resistors you have in combinations of series and parallel to arrive at a different value of
equivalent resistance called for in the circuit.
P23.16. Prepare: When resistors are connected in parallel the combination has less
resistance than any of the individual resistors. Two
resistors in parallel have an
equivalent resistance of
Solve: We can connect three such parallel pairs in series so that the total resistance is
Assess: It is often necessary to combine standard value resistors in creative ways to
arrive at a non-standard required resistance.
P23.17. Prepare: The connecting wires are ideal with zero resistance. We have to
reduce the circuit to a single equivalent resistor by continuing to identify resistors that are
in series or parallel combinations.
Solve:
For the first step, the resistors 30 Ω and 45 Ω are in parallel. Their equivalent resistance
is
Req 1 = 18 Ω
For the second step, resistors 42 Ω and Req 1 = 18 Ω are in series. Therefore,
Req 2 = Req 1 + 42 Ω = 18 Ω + 42 Ω = 60 Ω
For the third step, the resistors 40 Ω and Req 2 = 60 Ω are in parallel. So,
Req 3 = 24 Ω
The equivalent resistance of the circuit is 24 Ω.
Assess: Have a good understanding of how series and parallel resistors combine to
obtain equivalent resistors.
P23.24. Prepare: Please refer to Figure P23.24. The batteries are ideal, the
connecting wires are ideal, and the ammeter has a negligibly small resistance.
Solve: Kirchhoff’s junction law tells us that the current flowing through the 2.0 Ω
resistance in the middle branch is I1 + I2 = 3.0 A. We can therefore determine I1 by
applying Kirchhoff’s loop law to the left loop. Starting clockwise from the lower left
corner,
+ 9.0 V – I1(3.0 Ω) – (3.0 A)(2.0 Ω) = 0 V I1 = 1.0 A
1.0 A) = 2.0 A
I2 = (3.0 A – I1) = (3.0 A –
Finally, to determine the emf we apply Kirchhoff’s loop law to the right loop and start
counterclockwise from the lower right corner of the loop:
− I2(4.5 Ω) − (3.0 A)(2.0 Ω) = 0 V ⇒
– (2.0 A)(4.5 Ω) – 6.0 V = 0 V ⇒
= 15.0 V
Assess: The currents and the emf look reasonable.
P23.27. Prepare: The battery and the connecting wires are ideal. The figure shows
how to simplify the circuit in Figure P23.27 using the laws of series and parallel
resistances. We have labeled the resistors as R1 = 6.0 Ω, R2 = 15 Ω, R3 = 6.0 Ω, and R4 =
4.0 Ω. Having reduced the circuit to a single equivalent resistance Req, we will reverse the
procedure and “build up” the circuit using the loop law and the junction law to find the
current and potential difference of each resistor.
Solve: R3 and R4 are combined to get R34 = 10 Ω, and then R34 and R2 are combined to
obtain R234:
R234 = 6 Ω
Next, R234 and R1 are combined to obtain
Req = R234 + R1 = 6.0 Ω + 6.0 Ω = 12 Ω
From the final circuit,
Thus, the current through the battery and R1 is IR1 = 2.0 A and the potential difference
across R1 is I(R1) = (2.0 A) (6.0 Ω) = 12 V.
As we rebuild the circuit, we note that series resistors must have the same current I and
that parallel resistors must have the same potential difference ΔV.
In Step 1 of the previous figure, Req = 12 Ω is returned to R1 = 6.0 Ω and R234 = 6.0 Ω in
series. Both resistors must have the same 2.0 A current as Req. We then use Ohm’s law to
find
ΔVR1 = (2 .0A)(6.0 Ω) = 12 V
ΔVR234 = (2.0 A)(6.0 Ω) = 12 V
As a check, 12 V + 12 V = 24 V, which was ΔV of the Req resistor. In Step 2, the
resistance R234 is returned to R2 and R34 in parallel. Both resistors must have the same ΔV
= 12 V as the resistor R234. Then from Ohm’s law,
As a check, IR2 + IR34 = 2.0 A, which was the current I of the R234 resistor. In Step 3, R34
is returned to R3 and R4 in series. Both resistors must have the same 1.2 A as the R34
resistor. We then use Ohm’s law to find
(ΔV)R3 = (1.2 A)(6.0 Ω) = 7.2 V
(ΔV)R4 = (1.2 A)(4.0 Ω) = 4.8 V
As a check, 7.2 V + 4.8 V = 12 V, which was ΔV of the resistor R34.
Potential difference
Current (A)
(V)
R1
12
2.0
R2
12
0.8
R3
7.2
1.2
R4
4.8
1.2
The three steps as we rebuild our circuit are shown.
Resistor
Assess: This problem requires a good understanding of how to first reduce a circuit to a
single equivalent resistance and then to build up a circuit.
P23.28. Prepare: The circuit reduction process shows that
. So the current
through the battery (and the first resistor) is
Solve: The potential difference across the first resistor is
Use the loop law around the left-most loop to deduce that
Moving on, the equivalent resistance of the four right-most resistors (those not in the leftmost loop) is
, so the junction law tells us the current splits evenly at point 1,
giving
through the middle horizontal resistor. Then the potential difference across
that middle horizontal resistor is
Use the loop law around
the loop containing the battery and point 2 (but not point 3) to deduce that
Because both downstream branches from point 3 have the same resistance (
) the
current splits evenly there too, giving the current through the right-most horizontal
resistor of
The potential difference across that resistor is then
The loop law around the outer loop then gives
Assess: Checking with other calculations (such as different applications of the loop law) give
the same results.
P23.36. Prepare: Please refer to figure P23.36. The pictorial representation shows
how to find the equivalent capacitance of the three capacitors shown in the figure.
Solve:
Because C1 and C2 are in series, their equivalent capacitance Ceq 12 is
Ceq 12 = 12 µF
Then, Ceq 12 and C3 are in parallel. So,
Ceq = Ceq 12 + C3 = 12 µF + 25 µF = 37 µF
Assess: We must understand well how to combine series and parallel capacitance.
P23.38. Prepare: Assume ideal battery, wires, and capacitors.
Solve:
(a) The two on the right are in parallel, so we add them to get
gives
. That in series with
(b) The amount of charge that flows while charging is given by
Assess: The values are typical.
P23.39. Prepare: For capacitors in series we know the equivalent capacitance is less
than any of the individual capacitances. The charge on capacitors in series is the same.
Solve:
(a) All three are in series, so
(b) The charge on each capacitor is the same as we would calculate on an equivalent
capacitor.
Assess: Indeed, the equivalent capacitance is less than the smallest capacitor.
P23.44. Prepare: The capacitor discharges through a resistor. The switch in the
circuit in Figure P23.44 is in position a. When the switch is in position b the circuit
consists of a capacitor and a resistor. Current and voltage during a capacitor discharge are
given by Equations 23.22. Because the charge on a capacitor is Q = CΔV, the decay of
the capacitor charge is given by Q = Q0 e t/ .
Solve: (a) The switch has been in position a for a long time. That means the capacitor is
fully charged to a charge
Immediately after the switch is moved to the b position, the charge on the capacitor is Q0
= 18 µC. The current through the resistor is
−
τ
Note that as soon as the switch is closed, the potential difference across the capacitor ΔVC
appears across the 50 Ω resistor.
(b) The charge Q0 decays as Q = Q0 e t / , where
−
τ
Thus, the charge is
= 11 µC.
The resistor current is
(c) Likewise, the charge is Q = 2.4 µC and the current is I = 24 mA.
Assess: All of these values seem reasonable.
P23.45. Prepare: Before the action potential the membrane potential is approximately
and at the peak
of depolarization it is
We’ll use positive values since we
only need the strength of the electric field.
Solve: Before the action potential,
At the peak of depolarization,
Assess: The high values are due to the thinness of the membrane.
P23.54. Prepare: The
internal resistance of the battery is in series with the
external
resistor. So the total resistance of the circuit is
As a preliminary calculation, use Ohm’s law to find the current in the circuit.
Solve:
(a) The potential difference between the terminals of the battery is
(b) The total power dissipated is
The power dissipated
internally in the battery is
or 1/3 of the total.
Assess: While ideal batteries are sources of constant emf, real batteries with internal
resistance have a potential difference between the terminals that depends on the external
resistance (the load).
P23.58. Prepare: Please refer to Fig. P23.58. The connecting wires are ideal, but the
battery is not. We will designate the current in the 5.0 Ω resistor I5 and the voltage drop
ΔV5. Similar designations will be used for the other resistors.
Solve: Since the 10 Ω resistor is dissipating 40 W,
ΔV10 = I10R10 = (2.0 A)(10 Ω) = 20.0 V
The 20 Ω resistor is in parallel with the 10 Ω resistor, so they have the same potential
difference: ΔV20 = ΔV10 = 20.0 V. From Ohm’s law,
The combined current through the 10 Ω and 20 Ω resistors first passes through the 5.0 Ω
resistor. Applying Kirchhoff’s junction law at the junction between the three resistors,
I5 = I10 + I20 = 1.0 A + 2.0 A = 3.0 A
ΔV5 = I5R5 = (3.0 A)(5.0 Ω) = 15 V
Knowing the currents and potential differences, we can now find the power dissipated:
P5 = I5ΔV5 = (3.0 A)(15.0 V) = 45
P20 = I20ΔV20 = (1.0 A)(20.0 V) = 20 W
P23.64. Prepare: Please refer to Figure P23.64. Assume the batteries and the
connecting wires are ideal.
Solve: (a) The two batteries in this circuit are oriented to “oppose” each other. The
direction of the current is counterclockwise because the 12 V battery “wins.”
(b) There are no junctions, so the same current I flows through all circuit elements.
Applying Kirchhoff’s loop law in the counterclockwise direction and starting at the lower
right corner, ∑ΔVi = 12 V − I(12 Ω) − I(6.0 Ω) - 6.0 V – IR = 0.
Note that the IR terms are all negative because we’re applying the loop law in the
direction of current flow, and the potential decreases as current flows through a resistor.
We can easily solve to find the unknown resistance R:
6.0 V − I(18 Ω) – IR = 0
(c) The power is P = I2R = (0.25 A)2(6 Ω) = 0.38 W.
(d)
The potential difference across a resistor is ΔV = IR, giving ΔV6 = 1.5 V, and ΔV12 = 3.0
V. Starting from the lower left corner, the graph goes around the circuit clockwise,
opposite from the direction in which we applied the loop law. In this direction, we speak
of potential as lost in the batteries and gained in the resistors.
P23.74. Prepare: Please refer to Figure P23.74. In an RC circuit, the capacitor
voltage discharge is given by Equation 23.22, i.e., ΔV = ΔV0e t/ .
Solve: From Figure P23.74, we note that ΔV0 = 30 V and ΔV = 10 V at t = 2 ms. So,
−
τ
P23.77. Prepare: The chapter says that when the sodium channels open, the potential
inside the cell changes from
to
relative to outside the cell. This
will be accompanied by a movement of charge
as the ions flow into the cell.
The number of channels is the total charge that moves divided by the amount of charge
per channel (which is given as
ions per channel).
Solve:
Assess: Example 23.15 says there are “a great many channels” in a cell, and our answer
bears this out.
P23.80. Prepare: When the defibrillator is attached to a person, a capacitor is
discharged and the potential difference across the capacitor and current through the
resistor (the heart) should look like that for a capacitor discharging in an RC circuit. For
this case, the potential difference across a capacitor is given by
and the
current through the resistor (the heart) is given by
Solve: The set of graphs labeled A match the above two functions.
Assess: Knowing the general shape of a function is a valuable skill.
P23.81.
Solve:
Prepare:
In an
circuit the time constant is
The correct choice is C.
Assess: We were told that the capacitor discharges quickly, and our answer agrees with
that 3.2 ms is quick, but reasonable.
P23.82. Prepare: The initial current may be determined by
and the time
constant may be determined by
Solve: Examining the above expressions for the initial current and time constant, we
see that the initial current will decrease and the time constant will increase if the
resistance increases. The correct choice is B.
Assess: The ability to look at an expression and answer a “what if ” question is a
valuable skill.
P23.83. Prepare: The time constant does not depend on the voltage; it is still
So lowering the voltage leaves the time constant unchanged.
Solve: The correct choice is B.
Assess: The time constant gives us an idea of how fast the capacitor will discharge, i.e.,
we know the potential difference will decay to 37% of its initial value after one time
constant, and that is true no matter what the value of the initial voltage is.