  Top-Rated Free Essay # Unit 7 Ch 23 Electric Circuits

Powerful Essays
Unit 7 Ch 23 Electric Circuits
Due: 7:20am on Friday, December 5, 2014
To understand how points are awarded, read the Grading Policy for this assignment.

Problem 23.1
Draw a circuit diagram for the circuit of .

Part A
Choose the correct diagram.

Correct

Problem 23.2
Draw a circuit diagram for the circuit of .

Part A
Choose the correct diagram.

Correct

Problem 23.3
Draw a circuit diagram for the circuit of .

Part A
Choose the correct diagram.

Correct

Kirchhoff's Rules and Applying Them
Learning Goal:
To understand the origins of both of Kirchhoff's rules and how to use them to solve a circuit problem.
This problem introduces Kirchhoff's two rules for circuits:
Kirchhoff's loop rule: The sum of the voltage changes across the circuit elements forming any closed loop is zero.
Kirchhoff's junction rule: The algebraic sum of the currents into (or out of) any junction in the circuit is zero.
The figure shows a circuit that illustrates the concept of loops, which are colored red and labeled loop 1 and loop 2. Loop 1 is the loop around the entire circuit, whereas loop 2 is the smaller loop on the right. To apply the loop rule you would add the voltage changes of all circuit elements around the chosen loop. The figure contains two junctions (where three or more wires meet)--they are at the ends of the resistor labeled R3 . The battery supplies a constant voltage Vb , and the resistors are labeled with their resistances. The ammeters are ideal meters that read I1 and I2 respectively.

The direction of each loop and the direction of each current arrow that you draw on your own circuits are arbitrary. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get correct equations. If the actual current is in the opposite direction from your current arrow, your answer for that current will be negative. The direction of any loop is even less imporant: The equation obtained from a counterclockwise loop is the same as that from a clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overall sign of the equation because it equals zero).

Part A
The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state.

Hint 1. At the junction
Think of the analogy with water flow. If a certain current of water comes to a split in the pipe, what can you say (mathematically) about the sum of the three water currents at this junction? If this were not true, water would accumulate at the junction.
current voltage resistance

Correct

Part B
Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2 ).
Answer in terms of given quantities, together with the meter readings I1 and

I2

and the current I3 .

Hint 1. Elements in series
The current through resistance R1 is not labeled. You should recognize that the current

I1

passing through the ammeter also passes through

resistance R1 because there is no junction in between the resistor and the ammeter that could allow it to go elsewhere. Similarly, the current passing through the battery must be I1 also. Circuit elements connected in a string like this are said to be in series and the same current must pass through each element. This fact greatly reduces the number of independent current values in any practical circuit.
ΣI = 0 =

=

I3 + I2 − I1

Correct
If you apply the juncion rule to the junction above R2 , you should find that the expression you get is equivalent to what you just obtained for the junction labeled 1. Obviously the conservation of charge or current flow enforces the same relationship among the currents when they separate as when they recombine.

Part C
Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the

direction of the arrow. Remember that the current meter is ideal.
Express the voltage drops in terms of Vb ,

I2

,

I3

, the given resistances, and any other given quantities.

Hint 1. Elements in series have same current
The current through the ammeter is

I1

, and this current has to go through the resistor of resistance R1 because there is no junction in between

that could add or subtract current. Similarly, the current passing through the battery must be I1 also. Circuit elements connected in a string like this are said to be in series and the same current must pass through each element. This fact greatly reduces the number of independent current values in any practical circuit.

Hint 2. Sign of voltage across resistors
In determining the signs, note that if your chosen loop traverses a particular resistor in the same direction as the current through that resistor, then the end it enters through will have a more positive potential than the end from which it exits by the amount I R. Thus the voltage change across that resistor will be negative. Conversely, if your chosen loop traverses the resistor in the opposite direction from its current arrow, the voltage changes across the resistor will be positive. Let these conventions govern your equations (i.e., don't try to figure out the direction of current flow when using the Kirchhoff loop--decide when you put the current arrows on the resistors and stick with that choice).

Hint 3. Voltage drop across ammeter
An ideal ammeter has zero resistance. Hence there is no voltage drop across it.
Σ(ΔV ) = 0 =

=

I3 R3 − I2 R2

Correct

Part D
Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.
Express the voltage drops in terms of Vb ,

I1

,

I3

, the given resistances, and any other given quantities.

Σ(ΔV ) = 0 =

=

Vb − I1 R1 − I3 R3

Correct
There is one more loop in this circuit, the inner loop through the battery, both ammeters, and resistors

R1

and R2 . If you apply Kirchhoff's loop

rule to this additional loop, you will generate an extra equation that is redundant with the other two. In general, you can get enough equations to solve a circuit by either
1. selecting all of the internal loops (loops with no circuit elements inside the loop) or
2. using a number of loops (not necessarily internal) equal to the number of internal loops, with the extra proviso that at least one loop pass through each circuit element.

Problem 23.4

Part A
In the figure, what is the current in the wire above the junction?

I

= 5

A

Correct

Part B
Consider the wire above the junction: we are looking from the top of the figure. Does charge flow toward or away from the junction?
toward away Correct

Problem 23.5
The lightbulb in the circuit diagram of has a resistance of 1.2 Ω . Consider the potential difference between pairs of points in the figure. Suppose that E = 3.5 V .

Part A
What is the magnitude of

ΔV12

?

ΔV12

= 2.2 V

Correct

Part B
What is the magnitude of

ΔV23

?

ΔV23

= 1.3 V

Correct

Part C
What is the magnitude of

ΔV34

?

ΔV34

= 0V

Correct

Part D
What is the magnitude of

ΔV12

if the bulb is removed from the socket (i.e. the circuit is not closed)?

ΔV12

= 0V

Correct

Part E
What is the magnitude of

ΔV23

if the bulb is removed from the socket (i.e. the circuit is not closed)?

ΔV23

= 3.5 V

Correct

Part F
What is the magnitude of

ΔV34

if the bulb is removed from the socket (i.e. the circuit is not closed)?

ΔV34

= 0V

Correct

Problem 23.8
Consider the circuit shown in .

Part A
What is the potential difference across the 10 Ω resistor?
ΔV10

= 5.0 V

Correct

Part B
What is the potential difference across the 20 Ω resistor?
ΔV20

= 10 V

Correct

Part C
Choose the correct graph of the potential as a function of the distance s traveled through the circuit, traveling clockwise from corner. ANSWER:

V

= 0 V at the lower left

Correct

Tactics Box 23.1 Using Kirchhoff's Loop Law
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.

TACTICS BOX 23.1

Using Kirchhoff’s loop law

1. Draw a circuit diagram. Label all known and unknown quantities.
2. 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. 3. "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 element, ΔV is interpreted to mean
ΔV = Vdownstream − Vupstream

For a battery with current in the negative-to-positive direction:

ΔVbat = +E

.

.

For a battery with current in the positive-to-negative direction:
ΔVbat = −E .
For a resistor: ΔVR = −I R

.

4. Apply the loop law:
Σ(ΔV )

i

= 0

.

Part A
The current in the circuit shown in the figure is 0.20 A. What is the potential difference ΔVbat across the battery traveling in the direction shown in the figure? Express your answer in volts.

Hint 1. Find the potential difference across the resistor
ΔVR

What is the potential difference ΔVR across the resistor in the direction assigned to the current in the figure?

ΔVR

= -6.0

V

ΔVbat

= 6.0

V

Correct

Part B
Find the current

I

in the circuit shown in the figure.

Hint 1. Find the potential difference across the battery
If we assign the counterclockwise direction to the current, what is the potential difference across the battery,

ΔVbat

?

ΔVbat = −9.0 V
ΔVbat = +9.0 V
ΔVbat = −4.5 V
ΔVbat = +4.5 V

Hint 2. Find the potential difference across the 40-ohm resistor
If we assign the counterclockwise direction to the current

I

I

in the circuit, what is the potential difference ΔVR1 across the 40-ohm resistor?

ΔVR1

=

−40I

Hint 3. Find the potential difference across the 50-ohm resistor
If we assign the counterclockwise direction to the current

I

in the circuit, what is the potential difference ΔVR2 across the 50-ohm resistor?

ΔVR2

=

−50I

I

= 0.10

A

Correct

Part C
What is the potential difference ΔV 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 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.
Σ(ΔV )

i

=

6 − 4 − 3 + ΔV

= 0

ΔV

= 1.0

V

Correct

Throw the Switch
E

R

In this problem

E

denotes the emf provided by the source, and R is the resistance of each bulb.

Part A
Bulbs A, B, and C in the figure are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference?
Check all that apply.

Hint 1. How to approach the problem
When the switch is open, bulbs B and C form a loop, so the voltage across bulb B must equal the voltage drop across bulb C. When the switch is closed, the switch and bulb C form a loop. We assume that the switch is an ideal conductor, so the voltage drop across it is zero. This affects the potential difference across bulb C. Consequently, the voltage across bulb B will also change. Once you have determined the new potential difference across bulb B, you can calculate the voltage across bulb A and compare it to its value when the switch was open.

Hint 2. Find the potential difference across bulb C when the switch is closed
What is the potential difference ΔVC across bulb C when the switch is closed?
Express your answer in terms of some or all of the variables E and

R

.

ΔVC

= 0

Hint 3. Find the potential difference across bulb B when the switch is closed
What is the potential difference ΔVB across bulb B when the switch is closed?
Express your answer in terms of some or all of the variables E and

R

.

ΔVB

= 0

Hint 4. Find the potential difference across bulb A when the switch is closed
What is the potential difference ΔVA across bulb A when the switch is closed?
Express your answer in terms of some or all of the variables E and

R

.

ΔVA

=

E

Hint 5. Find the potential difference across bulb A when the switch is open
What is the potential difference ΔVA across bulb A when the switch is open?
Express your answer in terms of some or all of the variables E and

R

.

ΔVA

=

2E
3

The potential difference across A is unchanged.
The potential difference across B drops to zero.
The potential difference across A increases by 50%.
The potential difference across B drops by 50%.

Correct
Every time the ends of a resistor are joined together, or connected through an ideal conductor, the voltage across the resistor drops to zero and the resistor is said to be short-circuited.

Part B
One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown in the figure. Bulbs A, B, C, and D are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference?
Check all that apply.

Hint 1. How to approach the problem
When the switch is closed, bulb C is short-circuited, so the voltage across it drops to zero and no current passes through the bulb. This decreases the equivalent resistance of the circuit, resulting in a greater current drawn by the circuit. If you estimate the current in the circuit, you can then calculate the potential difference across bulb A, as well as the voltage drops across bulbs B and D. Keep in mind that when the switch is closed, bulbs B and D form a loop.

Hint 2. Find the equivalent resistance of the circuit when the switch is closed
When the switch is closed, bulb C is short-circuited, so the voltage across it drops to zero and no current passes through the bulb. What is the equivalent resistance Req of the circuit in this case?
Express your answer in terms of some or all of the variables E and

R

.

Hint 1. Parallel and series connections
Note that bulbs B and D are connected in parallel and their equivalent resistance is connected in series with bulb A.
Req

=

3R
2

Hint 3. Find the voltage across bulb A when the switch is closed
What is the potential difference across bulb A when the switch is closed?
Express your answer in terms of some or all of the variables E and

R

.

Hint 1. Ohm's law
Ohm's law tells us that the voltage V across a resistor is equal to the product of the current
R of the resistor:
V = IR

I

that passes through it and the resistance

.

Hint 2. Find the current that passes through bulb A
Let

Req

denote the equivalent resistance of the circuit. Apply Ohm's law to the circuit and find the current

IA

that passes through bulb

A.
Express your answer in terms of some or all of the variables E ,

R

, and

Req

.

IA

=

E
R eq

ΔVA

=

2E
3

Hint 4. How to determine whether choice D is correct
Consider the loop formed by the emf source and bulbs A and D. According to the loop rule, the algebraic sum of the potential differences of bulbs A and D and the voltage delivered by the emf source must equal zero. Therefore, once you have determined whether the voltage across bulb A increases or decreases when the switch is closed, since the voltage provided by the emf source is constant, you can also determine how the voltage across bulb D changes.

Hint 5. Find the voltage across bulb B when the switch is closed
Consider the loop formed by the emf source and bulbs A and B when the switch is closed. Apply the loop rule and find the potential difference
ΔVB across bulb B.
Express your answer in terms of some or all of the variables E and

R

.

Hint 1. Loop rule
Recall that the algebraic sum of all the potential differences in a loop, including those associated with voltage sources, must equal zero.

Hint 2. How to use your previous results
Apply the loop rule and use your results found in Hint B.3, where you calculated the voltage across bulb A, to solve for ΔVB .

ΔVB

=

E
3

Hint 6. Find the voltage across bulb B when the switch is open
What is the potential difference ΔVB across bulb B when the switch is open?
Express your answer in terms of some or all of the variables E and

R

.

Hint 1. How to approach the problem
When the switch is open, bulbs B, C, and D form a loop, so the algebraic sum of their potential differences must equal zero. Moreover, bulbs B and C are connected in series, so the same current passes through them. Because they also have the same resistance, the voltages across them are also equal:
ΔVB = ΔVC = I R .
Therefore, the voltage across bulb B must be half of the voltage ΔVD across bulb D. Find ΔVD first. To do that, consider the loop

formed by the emf source and bulbs A and D, and apply the loop rule.

Hint 2. Find the equivalent resistance of the circuit when the switch is open
What is the equivalent resistance Req of the circuit when the switch is open? Note that bulbs B and C are connected in series, and their equivalent resistance is in parallel with bulb D.
Express your answer in terms of some or all of the variables E and

R

.

R

.

Req

5R

=

3

Hint 3. Find the voltage across bulb A when the switch is open
What is the potential difference across bulb A when the switch is open?
Express your answer in terms of some or all of the variables E and
ΔVA

=

3E
5

ΔVB

=

E
5

The potential difference across A increases.
The potential difference across B doubles.
The potential difference across B drops to zero.
The potential difference across D is unchanged.

Correct

Power Dissipation in Resistive Circuit Conceptual Question
A single resistor is wired to a battery as shown in the diagram below. Define the total power dissipated by this circuit as P0 .

Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram to the left .

Part A
What is the power, in terms of

P0

, dissipated by this circuit?

Hint 1. How to find the power dissipated by a circuit
The power dissipated by a circuit (or by an element in a circuit) is defined by the relation
P = IV .
If the circuit consists of resistors, we can combine this relation with Ohm's law,
V = IR

,

to yield two alternate versions of the power formula:
P = I

2

R

and
P =

V

2

R

.

Because several circuit parameters can be changing simultaneously, it is easiest to use the formula in which only one of the terms is changing for your situation. This makes it much easier to determine the power dissipated in a resistive circuit.

Hint 2. Effect of adding a resistor in series
Adding a resistor in series affects both the total resistance and total current in a circuit. Combining an understanding of these changes with the appropriate version of the power formula should allow you to answer this question.
0.5P 0

Correct
The second resistor is now removed from the circuit and rewired in parallel with the original resistor as shown in the schematic to the left

Part B
What is the power, in terms of

P0

, dissipated by this circuit?

Hint 1. How to find the power dissipated by a circuit
The power dissipated by a circuit (or by an element in a circuit) is defined by the relation
P = IV .
If the circuit consists of resistors, we can combine this relation with Ohm's law,
V = IR

,

to yield two alternate versions of the power formula:
P = I

2

R

and
P =

V

2

R

.

Because several circuit parameters can be changing simultaneously, it is easiest to use the formula in which only one of the terms is changing for your situation. This makes it much easier to determine the power dissipated in a resistive circuit.

Hint 2. Effect of adding a resistor in parallel
Adding a resistor in parallel affects both the total resistance and current in a circuit. Combining an understanding of these changes with the appropriate version of the power formula should allow you to answer this question.
2P 0

Correct

Problem 23.11

Part A
What is the equivalent resistance of group (a) of resistors shown in the figure?

R

eq

= 1.0

Correct

Part B

Ω

What is the equivalent resistance of group (b) of resistors shown in the figure?
R

eq

= 1.0

Ω

Correct

Part C
What is the equivalent resistance of group (c) of resistors shown in the figure?
R

eq

= 0.50

Ω

Correct

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 (R =

V
I

), 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:
I = I1 = I2 = I3 =. . .
V = V1 + V2 + V3 +. . .
Req−series = R1 + R2 + R3 +. . .

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:
V = V1 = V2 = V3 =. . .

V = V1 = V2 = V3 =. . .
I = I1 + I2 + I3 +. . .
1
R eq−parallel

=

1
R1

+

1
R2

+

1
R3

+. . .

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 Req for various combinations of resistors.

Part A
For the combination of resistors shown, find the equivalent resistance between points A and B.

Req

= 9

Ω

Correct
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.

Req

= 2

Ω

Correct
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 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 2. What kind of connection is this?
The 2-Ohm resistor is connected:
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

Req

= 5

Ω

Correct
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 "sub-combinations" 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 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 2. Find R eq for the "4-6-12" combination
What is the equivalent resistance for the "4 ohm - 6 ohm - 12 Ohm" combination?
= 2

Req−4−6−12

Ω

Hint 3. Find R eq for the top branch
What is the equivalent resistance for the top branch of the circuit (between C and D)?
Req−top

= 6

Ω

Hint 4. Find R eq for the bottom branch
What is the equivalent resistance for the bottom branch of the circuit (between E and F)?
Req−bottom

= 6

Ω

Req

= 3

Ω

Correct
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.

Problem 23.10

Part A
What is the equivalent resistance of group (a) of resistors shown in the figure?

R

eq

= 11

Ω

Correct

Part B
What is the equivalent resistance of group (b) of resistors shown in the figure?
R

eq

= 9.0

Ω

Correct

Problem 23.14
You have a collection of 1.0 kΩ resistors.

Part A
How can you connect four of them to produce an equivalent resistance of 0.25 kΩ ?
2 parallel and 2 parallel and all it in series
2 in series and 2 in series and all it in parallel all in parallel all in series

Correct

Measuring the EMF and Internal Resistance of a Battery
When switch S in the figure is open, the voltmeter V of the battery reads 3.10V . When the switch is closed, the voltmeter reading drops to 2.94V , and the ammeter A reads 1.63A . Assume that the two meters are ideal, so they do not affect the circuit.

Part A
Find the emf

E

.

E

= 3.10

V

Correct

Part B
Find the internal resistance r of the battery.

Hint 1. How to approach the problem
The voltmeter reading is equal to E 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.
r

= 9.816×10−2

Ω

Correct

Part C
Find the circuit resistance R.

Hint 1. Find the voltage drop across the circuit resistor
When the switch is closed, what is the voltage drop VR across the circuit resistor (of resistance R)?
VR

= 2.94

V

R

= 1.80

Ω

Correct
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 R 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 Loop Rule Conceptual Question
The circuit shown belowconsists 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 VA , for example, is the potential drop across resistor A.
Check all that apply.

Hint 1. 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.
VA + VB
VA + VC
VB + VC
VA + VB + VC
VA + VB + VC + VD
VA + VD
VD

Correct

Problem 23.22
Consider the circuit shown in . Suppose that

ΔV

= 5.5

V

.

Part A
What is the value of resistor R?
Express your answer to two significant figures and include the appropriate units.
R

= 30 Ω

Correct

Problem 23.25

Part A
Find the current through resistor a) in the figure.
Express your answer to two significant figures and include the appropriate units.

Ia

= 1.5 A

Correct

Part B
Find the potential difference across resistor a) in the figure.
Express your answer to two significant figures and include the appropriate units.

ΔVa

= 7.5 V

Correct

Part C
Find the current through resistor b) in the figure.
Express your answer to two significant figures and include the appropriate units.
Ib

= 0.50 A

Correct

Part D
Find the potential difference across resistor b) in the figure.
Express your answer to two significant figures and include the appropriate units.
ΔVb

= 2.5 V

Correct

Part E
Find the current through resistor c) in the figure.
Express your answer to two significant figures and include the appropriate units.
Ic

= 0.50 A

Correct

Part F
Find the potential difference across resistor c) in the figure.
Express your answer to two significant figures and include the appropriate units.
ΔVc

= 2.5 V

Correct

Part G
Find the current through resistor d) in the figure.
Express your answer to two significant figures and include the appropriate units.

Id

= 0.50 A

Correct

Part H
Find the potential difference across resistor d) in the figure.
Express your answer to two significant figures and include the appropriate units.
ΔVd

= 2.5 V

Correct

Comparing brightness of light bulbs
Consider five identical light bulbs connected to a battery as shown in each of the identical circuits below in Part A.

Part A
Rank the brightness of all five bulbs (A-E) from brightess to dimmest.
Rank the bulbs from brightest to dimmest. To rank items as equivalent, overlap them.

Hint 1. Brightness
The brightness of a light bulb is related to the power P dissipated by that bulb. Recall that power depends upon the square of the current I that passes through the bulb and the resistance R of the bulb. Since each light bulb in this circuit has the same resistance, the lightbulb with the largest current passing through it will be the brightest.

Hint 2. Compare the brightness of bulbs A and B
greater than
The brightness of bulb A is

less than

the brightness of bulb B.

equal to

Hint 3. Compare the brightness of bulbs D and E
greater than
The brightness of bulb D is

less than

the brightness of bulb E.

equal to

Hint 4. Compare the brightness of bulbs C and D
Hint 1. Compare the current through bulbs C and D

greater than
The current that passes through bulb C is

less than

the current that passes through bulb D.

equal to

The brightness of bulb C is

less than

the brightness of bulb D.

equal to

Hint 5. Compare the brightness of bulbs C and A
Hint 1. Compare the current through bulbs C and A
Hint 1. How to approach the problem
A current I leaves the battery and flows around this circuit. Consider how much of this current passes through light bulb A and light bulb B. Then consider how much of this current flows through light bulb C and bulbs D and E. If you quantify the current in each bulb in terms of a fraction of the total current I , you will be able to compare the amount of current that passes through light bulb C to that through light bulb A.

Hint 2. Find the amount of current that passes through bulb A
A current

I

leaves the battery and flows around this circuit. How much current flows through light bulb A?

I
2
3
1
2
1
3

I
I
I

Hint 3. Find the amount of current that passes through bulb C
A current

I

leaves the battery and flows around this circuit. How much current flows through light bulb C?

I
2
3
1
2
1
3

I
I
I

The current that passes through bulb C is

less than equal to

the current that passes through bulb A.

greater than
The brightness of bulb C is

less than

the brightness of bulb A.

equal to

Hint 6. Compare the brightness of bulbs D and A
Hint 1. Compare the current through bulbs C and A
Hint 1. How to approach the problem
A current I leaves the battery and flows around this circuit. Consider how much of this current passes through light bulb A and light bulb B. Then consider how much of this current flows through light bulb C and bulbs D and E. If you quantify the current in each bulb in terms of a fraction of the total current I , you will be able to compare the amount of current that passes through light bulb D to that through light bulb A.

Hint 2. Find the amount of current that passes through bulb A
A current

I

leaves the battery and flows around this circuit. How much current flows through light bulb A?

I
2
3
1
2
1
3

I
I
I

Hint 3. Find the amount of current that passes through bulb D
A current

I

leaves the battery and flows around this circuit. How much current flows through light bulb D?

I
2
3
1
2
1
3

I
I
I

The current that passes through bulb D is

less than

the current that passes through bulb A.

equal to

The brightness of bulb D is

less than equal to

the brightness of bulb A.

Correct

How does this change effect the circuit?
Suppose bulb E is unscrewed and removed from its socket. (The empty socket remains in the circuit.)

Part B
Does bulb A get brighter, dimmer, or stay the same brightness?

Hint 1. How to approach the problem
When light bulb E is unscrewed and removed from the circuit, the current I that leaves the battery changes. This will change the amount of current that flows through light bulb A. Keep in mind that the brightness of a light bulb is related to the square of the current that passes through the bulb and the resistance of the bulb.

Hint 2. Determine how the current changes
When light bulb E is unscrewed and removed from the circuit, will the current that leaves the battery increase or decrease?

Hint 1. Relationship between resistance and current
When the total resistance of a circuit increases while the battery remains unchanged, the current that leaves the battery will decrease.
Similarly, when the total resistance of a circuit decreases while the battery remains unchanged, the current that leaves the battery will increase. Hint 2. Determine how the total resistance changes
When light bulb E is unscrewed, current can no longer flow across that branch of the circuit. This means that all the current that leaves the battery must flow across light bulb C. Does the removal of light bulb E cause the total resistance increase or decrease?
It will increase.
It will decrease.

It will increase.
It will decrease.

Light bulb A gets brighter.
Light bulb A gets dimmer.
Light bulb A stays the same brightness.

Correct

Problem 23.26
Consider the circuit shown in . Suppose that

E

= 10

Ω

.

Part A
Find the current through the resistor a.
Express your answer to two significant figures and include the appropriate units.
Ia

= 1.0 A

Correct

Part B
Find the potential difference across the resistor a.
Express your answer to two significant figures and include the appropriate units.
ΔVa

= 5.0 V

Correct

Part C
Find the current through the resistor b.

Express your answer to two significant figures and include the appropriate units.
Ib

= 0.50 A

Correct

Part D
Find the potential difference across the resistor b.
Express your answer to two significant figures and include the appropriate units.
ΔVb

= 5.0 V

Correct

Part E
Find the current through the resistor c.
Express your answer to two significant figures and include the appropriate units.
Ic

= 0.50 A

Correct

Part F
Find the potential difference across the resistor c.
Express your answer to two significant figures and include the appropriate units.
ΔVc

= 2.5 V

Correct

Part G
Find the current through the resistor d.
Express your answer to two significant figures and include the appropriate units.
Id

= 0.50 A

Correct

Part H
Find the potential difference across the resistor d.
Express your answer to two significant figures and include the appropriate units.

ΔVd

= 2.5 V

Correct

PSS 23.1: Resistor Circuits
Learning Goal:
To practice Problem-Solving Strategy 23.1 Resistor Circuits.
Find the currents through and the potential difference across each resistor in the circuit shown on the diagram . Use the following values:
= 15.0Ω , R2 = 45.0Ω , R3 = 20.0Ω , and R4 = 25.0Ω .

PROBLEM-SOLVING STRATEGY 23.1

E

= 12.0V ,

R1

Resistor Circuits

We can analyze any resistor circuit by sequentially reducing parallel and series resistor combinations to their equivalent resistors until only the battery and a single equivalent resistor are left.
PREPARE
SOLVE

Draw a circuit diagram. Label all known and unknown quantities.

Base your mathematical analysis on Kirchhoff's laws and on the rules for series and parallel resistors.
Step by step, reduce the circuit to the smallest possible number of equivalent resistors.
Determine the current through and potential difference across the equivalent resistors.
Rebuild the circuit, using the facts that the current is the same through all resistors in series and the potential difference is the same for all parallel resistors.

ASSESS

Use two important checks as you rebuild the circuit:

Verify that the sum of the potential differences across series resistors matches ΔV for the equivalent resistor.
Verify that the sum of the currents through parallel resistors matches I for the equivalent resistor.

Prepare
The circuit drawing is already given in the problem introduction. The battery voltage and all resistances are known. The current through and voltage across each resistor are unknown.

Solve
Use the information and the insights that you have accumulated to construct the necessary mathematical expressions and to derive the solution. You will need to solve parts C and D together.

Part A
Step by step, reduce the circuit to the smallest possible number of equivalent resistors in order to find the equivalent resistance Req of the entire circuit. Express you answer in ohms to three significant figures.

Hint 1. Equivalent resistance
N

R1

R2

RN

N

If

N

resistors of resistance R1 ,

R2

,...,

RN

were connected in series, the equivalent resistance is the sum of the N individual resistances:

If

N

resistors of resistance R1 ,

R2

,...,

RN

were connected in parallel, the equivalent resistance is the inverse of the sum of the inverses of

Req = R1 + R2 + ⋯ + RN

the N individual resistances:
−1

R eq = (

1
R1

+

1
R2

+⋯+

1
RN

)

Hint 2. Reduce R 3 and R 4
This diagram shows the first step in the reduction of the circuit. What is the value of the unknown resistance ReqA ?

Hint 1. Determine how R 3 and R 4 are connected

Resistors

R3

and R4 are connected in

ReqA

= 45.0

Ω

Hint 3. Reduce R 2 and R eqA
This diagram shows the next step in the reduction of the circuit. What is the value of the unknown resistance ReqB ?

Hint 1. Determine how R 2 and R eqA are connected

Resistors

R2

and ReqA are connected in

ReqB

= 22.5

Ω

Hint 4. Determine how R 1 and R eqB are connected

Resistors

R1

and ReqB are connected in

Req

= 37.5

Correct

Ω

Part B
Find Ieq , the current through the equivalent resistor.

Hint 1. Ohm's law
Recall that by Ohm's law the potential drop ΔV across a resistor of resistance R is
ΔV = I R

where I is the current flowing through the resistor.

Ieq

= 0.320

A

Correct
Since you have successfully found Req and Ieq , you should be ready to find the currents through and the potential differences across each resistor. Part C
Find the following currents.
The current

I1

through the resistor of resistance R1 = 15.0Ω .

The current

I2

through the resistor of resistance R2 = 45.0Ω .

The current

I3

through the resistor of resistance R3 = 20.0Ω .

The current

I4

through the resistor of resistance R4 = 25.0Ω .

Hint 1. Identify I1
The emf source E and R1 are connected in series. This means that the current that leaves the positive terminal of the battery must pass through R1 . What is the current

I1

through resistor R1 ?

I1

= 0.320

A

Hint 2. Find I2 by first finding V 2
If you know the potential difference ΔV2 across

R2

, you can find the current

I2

through that resistor using Ohm's law. You can find ΔV2 by

applying Kirchhoff's loop law to the loop shown. Take the direction of the current to be the same as that of the loop.

Write an equation for ΔV2 .

ΔV1

, where ΔV1 is the magnitude of the potential difference across resitor

R1

.

ΔV2

=

E − ΔV1

Hint 3. Find I3 and I4
Because R3 and R4 are in series, the same current must flow through both resistors. To find this current, apply Kirchhoff's junction law at the junction indicated.

Write an equation for I3 .

I2

.

I3 = I4 =

=

I1 − I2

I1

,

I2

,

I3

,

I4

= 0.320,0.160,0.160,0.160

A

Correct

Part D
Find the following potential differences.
The magnitude of the potential difference ΔV1 across the resistor of resistance R1 = 15.0Ω .
The magnitude of the potential difference ΔV2 across the resistor of resistance R2 = 45.0Ω .
The magnitude of the potential difference ΔV3 across the resistor of resistance R3 = 20.0Ω .
The magnitude of the potential difference ΔV4 across the resistor of resistance R4 = 25.0Ω .
ΔV1

,

ΔV2

,

ΔV3

,

ΔV4

= 4.80,7.20,3.20,4.00

V

Correct

Assess
When you work on a problem on your own, without the computer-provided feedback, only you can assess whether your answer seems right. The following questions will help you practice the skills necessary for such an assessment.

Part E
Which of the following relationships must be true according to the laws of series and parallel connections? (Include in your answer only relationships that must be true according to the laws of series and parallel connections, not those that are true in this problem only because of the particular resistances given.)

Check all that apply.
ΔV2 = ΔV3 + ΔV4
ΔV2 = ΔV1 + ΔV3 + ΔV4
ΔV1 = ΔV2 + ΔV3 + ΔV4
ΔV3 = ΔV4
ΔV2 = ΔV3
ΔV1 = ΔV2
ΔV1 = ΔV3 + ΔV4

Correct
Now go back to your answers in the solve portion of the problem. If you substitute them into the correct formulas identified here, would they become true identities? They should (but because of rounding errors, the left and the right sides of the identities may be slightly different).
Take your time to check your answers. This is a very reliable way of screening and it is especially important in problems when a single mistake may cause all your answers to be incorrect.

Capacitors in Series
Learning Goal:
To understand how to calculate capacitance, voltage, and charge for a combination of capacitors connected in series.
Consider the combination of capacitors shown in the figure. Three capacitors are connected to each other in series, and then to the battery. The values of the capacitances are C , 2C , and
3C , and the applied voltage is ΔV . Initially, all of the capacitors are completely discharged; after the battery is connected, the charge on plate 1 is Q .

Part A
What are the charges on plates 3 and 6?

Hint 1. Charges on capacitors connected in series
When the plates of two adjacent capacitors are connected, the sum of the charges on the two plates must remain zero, since the pair is isolated from the rest of the circuit; that is, Q 2 + Q 3 = 0 and Q 4 + Q 5 = 0 , where Q i is the charge on plate i .

Hint 2. The charges on a capacitor's plates
When electrostatic equilibrium is reached, the charges on the two plates of a capacitor must have equal magnitude and opposite sign.

+Q

and +Q

−Q

and −Q

+Q

and −Q

−Q

and +Q

0 and +Q
0 and −Q

Correct

Part B
If the voltage across the first capacitor (the one with capacitance C ) is

ΔV1

, then what are the voltages across the second and third capacitors?

Hint 1. Definition of capacitance
The capacitance C is given by

Q
ΔV

, where Q is the charge of the capacitor and Δ

V

is the voltage across it.

Hint 2. Charges on the capacitors
As established earlier, the absolute value of the charge on each plate is

Q

: It is the same for all three capacitors and thus for all six plates.

2ΔV1
1
2

and 3ΔV1

ΔV 1

ΔV1

and

1
3

ΔV 1

and ΔV1

0 and ΔV1

Correct

Part C
Find the voltage ΔV1 across the first capacitor.

Hint 1. How to analyze voltages
According to the law of conservation of energy, the sum of the voltages across the capacitors must equal the voltage of the battery.
ΔV1

=

6ΔV
11

Correct

Part D
Find the charge Q on the first capacitor.

ΔV1

.

Q

=

CΔV1

Correct

Part E
Using the value of

Q

just calculated, find the equivalent capacitance Ceq for this combination of capacitors in series.

Hint 1. Using the definition of capacitance
The "equivalent" capacitor has the same charge as each of the individual capacitors:

Ceq

=

6C
11

Correct
The formula for combining three capacitors in series is
1
C series

=

1
C1

+

1
C2

+

1
C3

.

How do you think this formula may be generalized to n capacitors?

Problem 23.39

Part A
What is the equivalent capacitance for the circuit of the figure?
Express your answer to two significant figures and include the appropriate units.

C

= 0.57 μF

Correct

Part B
What is the charge of the capacitor 4.0 μF ?
Express your answer to two significant figures and include the appropriate units.

Q

. Use the general formula C =

Q
ΔV

to find Ceq .

Correct

Part C
What is the charge of the capacitor 2.0 μF ?
Express your answer to two significant figures and include the appropriate units.
q

= 6.9 μC

Correct

Part D
What is the charge of the capacitor 1.0 μF ?
Express your answer to two significant figures and include the appropriate units.
q

= 6.9 μC

Correct

Problem 23.33
A 9.0μF capacitor, a 10μF capacitor, and a 18μF capacitor are connected in series.

Part A
What is their equivalent capacitance?
C

= 3.8

μF

Correct

Problem 23.38

Part A
What is the equivalent capacitance for the circuit of the figure?
Express your answer to two significant figures and include the appropriate units.

C

= 1.9 μF

Correct

Part B
How much charge flows through the battery as the capacitors are being charged?
Express your answer to two significant figures and include the appropriate units.
q

= 22 μC

Correct

Capacitors in Parallel
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
C , 2C , and 3C , and the battery's voltage is ΔV .

Part A
If the potential of plate 1 is

ΔV

, 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 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?
ΔV

and ΔV and 3ΔV

2ΔV
ΔV
ΔV
2

and 0 and ΔV
3

Correct

Part B
If the charge of the first capacitor (the one with capacitance C ) is

Q

, then what are the charges of the second and third capacitors?

Hint 1. Definition of capacitance
Capacitance C is given by

Q
ΔV

, where Q is the charge of the capacitor and ΔV is the voltage across it.

Hint 2. Voltages across the capacitors
As established earlier, the voltage across each capacitor is

ΔV

. The voltage is always the same for capacitors connected in parallel.

2Q
Q

and

2

Q
0

and 3Q
Q
3

and Q

and 0

Correct

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 Q tot for this equivalent capacitor.

C

.

Q

tot

=

6CΔV

Correct

Part D
Using the value of

Q

tot

, find the equivalent capacitance Ceq for this combination of capacitors.

Hint 1. Using the definition of capacitance
Use the general formula C = voltage across the battery,

Q
ΔV

ΔV

to find Ceq . The charge on the "equivalent" capacitor is

.

Ceq

=

6C

Correct
The formula for combining three capacitors in parallel is
Cparallel = C1 + C2 + C3 .
How do you think this formula may be generalized to n capacitors?

Problem 23.32
A 1.0μF capacitor, a 12μF capacitor, and a 20μF capacitor are connected in parallel.

Part A
What is their equivalent capacitance?
C

= 33

μF

Correct

Problem 23.36

Part A
What is the equivalent capacitance of the three capacitors in the figure?

C

= 37

Correct

μF

Q

tot

, and the voltage across this capacitor is the

Problem 23.55
A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure. A typical battery has 1.0 Ω internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.7Ω resistor.

Part A
What is the potential difference between the terminals of the battery?
V

= 1.1

V

Correct

Part B
What fraction of the battery's power is dissipated by the internal resistance?
ΔP
P

= 27

%

Correct

Problem 23.56

Part A
For the real battery shown in the figure, calculate the power dissipated by a resistor

R

connected to the battery when R = 2.5Ω .

P

= 3.6×10−2

W

Correct

Part B
For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 5.1Ω .
P

= 5.0×10−2

W

Correct

Part C
For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 10Ω .
P

= 5.6×10−2

W

Correct

Part D
For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 22Ω . (Your results should suggest that maximum power dissipation is achieved when the external resistance R equals the internal resistance. This is true in general.)
P

= 4.8×10−2

W

Correct

Part E
R

R

Ω

For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 40Ω .
P

= 3.6×10−2

W

Correct

Part A
The symbol shown represents a _____.

Correct

Part A
The bulbs in the circuit below are connected __________.

Correct

Part A

Which terminal of the battery has a higher potential?

ANSWER: the bottom terminal the top terminal

Correct

Part A
When three resistors are combined in series, the total resistance of the combination is __________.
greater than any of the individual resistance values the average of the individual resistance values less than any of the individual resistance values

Correct
Score Summary:
Your score on this assignment is 99.8%.
You received 33.94 out of a possible total of 34 points.

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