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Circuits II: Parallel Circuits

Slide Duration:

Section 1: Electricity
Electric Charge & Coulomb's Law

30m 48s

Intro
0:00
Objective
0:15
Electric Charges
0:50
Matter is Made Up of Atoms
0:52
Most Atoms are Neutral
1:02
Ions
1:11
Coulomb
1:18
Elementary Charge
1:34
Law of Conservation of Charge
2:03
Example 1
2:39
Example 2
3:42
Conductors and Insulators
4:41
Conductors Allow Electric Charges to Move Freely
4:43
Insulators Do Not Allow Electric Charges to Move Freely
4:50
Resistivity
4:58
Charging by Conduction
5:32
Conduction
5:37
Balloon Example
5:40
Charged Conductor
6:14
Example 3
6:28
The Electroscope
7:16
Charging by Induction
7:57
Bring Positive Rod Near Electroscope
8:08
Ground the Electroscope
8:27
Sever Ground Path and Remove Positive Rod
9:07
Example 4
9:39
Polarization and Electric Dipole Moment
11:46
Polarization
11:54
Electric Dipole Moment
12:05
Coulomb's Law
12:38
Electrostatic Force, Also Known as Coulombic Force
12:48
How Force of Attraction or Repulsion Determined
12:55
Formula
13:08
Coulomb's Law: Vector Form
14:18
Example 5
16:05
Example 6
18:25
Example 7
19:14
Example 8
23:21
Electric Fields

1h 19m 22s

Intro
0:00
Objectives
0:09
Electric Fields
1:33
Property of Space That Allows a Charged Object to Feel a Force
1:40
Detect the Presence of an Electric Field
1:51
Electric Field Strength Vector
2:03
Direction of the Electric Field Vector
2:21
Example 1
3:00
Visualizing the Electric Field
4:13
Electric Field Lines
4:56
E Field Due to a Point Charge
7:19
Derived from the Definition of the Electric Field and Coulomb's Law
7:24
Finding the Electric Field Due to Multiple Point Charges
8:37
Comparing Electricity to Gravity
8:51
Force
8:54
Field Strength
9:09
Constant
9:19
Charge Units vs. Mass Units
9:35
Attracts vs. Repel
9:44
Example 2
10:06
Example 3
17:25
Example 4
24:29
Example 5
25:23
Charge Densities
26:09
Linear Charge Density
26:26
Surface Charge Density
26:30
Volume Charge Density
26:47
Example 6
27:26
Example 7
37:07
Example 8
50:13
Example 9
54:01
Example 10
1:03:10
Example 11
1:13:58
Gauss's Law

52m 53s

Intro
0:00
Objectives
0:07
Electric Flux
1:16
Amount of Electric Field Penetrating a Surface
1:19
Symbol
1:23
Point Charge Inside a Hollow Sphere
4:31
Place a Point Charge Inside a Hollow Sphere of Radius R
4:39
Determine the Flux Through the Sphere
5:09
Gauss's Law
8:39
Total Flux
8:59
Gauss's Law
9:10
Example 1
9:53
Example 2
17:28
Example 3
22:37
Example 4
25:40
Example 5
30:49
Example 6
45:06
Electric Potential & Electric Potential Energy

1h 14m 3s

Intro
0:00
Objectives
0:08
Electric Potential Energy
0:58
Gravitational Potential Energy
1:02
Electric Potential Energy
1:11
Electric Potential
1:19
Example 1
1:59
Example 2
3:08
The Electron-Volt
4:02
Electronvolt
4:16
1 eV is the Amount of Work Done in Moving an Elementary Charge Through a Potential Difference of 1 Volt
4:26
Conversion Ratio
4:41
Example 3
4:52
Equipotential Lines
5:35
Topographic Maps
5:36
Lines Connecting Points of Equal Electrical Potential
5:47
Always Cross Electrical Field Lines at Right Angles
5:57
Gradient of Potential Increases As Equipotential Lines Get Closer
6:02
Electric Field Points from High to Low Potential
6:27
Drawing Equipotential Lines
6:49
E Potential Energy Due to a Point Charge
8:20
Electric Force from Electric Potential Energy
11:59
E Potential Due to a Point Charge
13:07
Example 4
14:42
Example 5
15:59
Finding Electric Field From Electric Potential
19:06
Example 6
23:41
Example 7
25:08
Example 8
26:33
Example 9
29:01
Example 10
31:26
Example 11
43:23
Example 12
51:51
Example 13
58:12
Electric Potential Due to Continuous Charge Distributions

1h 1m 28s

Intro
0:00
Objectives
0:10
Potential Due to a Charged Ring
0:27
Potential Due to a Uniformly Charged Desk
3:38
Potential Due to a Spherical Shell of Charge
11:21
Potential Due to a Uniform Solid Sphere
14:50
Example 1
23:08
Example 2
30:43
Example 3
41:58
Example 4
51:41
Conductors

20m 35s

Intro
0:00
Objectives
0:08
Charges in a Conductor
0:32
Charge is Free to Move Until the
0:36
All Charge Resides at Surface
2:18
Field Lines are Perpendicular to Surface
2:34
Electric Field at the Surface of a Conductor
3:04
Looking at Just the Outer Surface
3:08
Large Electric Field Where You Have the Largest Charge Density
3:59
Hollow Conductors
4:22
Draw Hollow Conductor and Gaussian Surface
4:36
Applying Gaussian Law
4:53
Any Hollow Conductor Has Zero Electric Field in Its Interior
5:24
5:35
Electric Field and Potential Due to a Conducting Sphere
6:03
Example 1
7:31
Example 2
12:39
Capacitors

41m 23s

Intro
0:00
Objectives
0:08
What is a Capacitor?
0:42
Electric Device Used to Store Electrical Energy
0:44
Place Opposite Charges on Each Plate
1:10
Develop a Potential Difference Across the Plates
1:14
Energy is Stored in the Electric Field Between the Plates
1:17
Capacitance
1:22
Ratio of the Charge Separated on the Plates of a Capacitor to the Potential Difference Between the Plates
1:25
Units of Capacitance
1:32
1:37
Formula
1:52
Calculating Capacitance
1:59
Assume Charge on Each Conductor
2:05
Find the Electric Field
2:11
Calculate V by Integrating the Electric Field
2:21
Utilize C=Q/V to Solve for Capitance
2:33
Example 1
2:44
Example 2
5:30
Example 3
10:46
Energy Stored in a Capacitor
15:25
Work is Done Charging a Capacitor
15:28
Solve For That
15:55
Field Energy Density
18:09
Amount of Energy Stored Between the Plates of a Capacitor
18:11
Example
18:25
Dielectrics
20:44
Insulating Materials Place Between Plates of Capacitor to Increase The Devices' Capacitance
20:47
Electric Field is Weakened
21:00
The Greater the Amount of Polarization The Greater the Reduction in Electric Field Strength
21:58
Dielectric Constant (K)
22:30
Formula
23:00
Net Electric Field
23:35
Key Take Away Point
23:50
Example 4
24:00
Example 5
25:50
Example 6
26:50
Example 7
28:53
Example 8
30:57
Example 9
32:55
Example 10
34:59
Example 11
37:35
Example 12
39:57
Section 2: Current Electricity
Current & Resistance

17m 59s

Intro
0:00
Objectives
0:08
Electric Current
0:44
Flow Rate of Electric Charge
0:45
Amperes
0:49
Positive Current Flow
1:01
Current Formula
1:19
Drift Velocity
1:35
Constant Thermal Motion
1:39
Net Electron Flow
1:43
When Electric Field is Applied
1:49
Electron Drift Velocity
1:55
Derivation of Current Flow
2:12
Apply Electric Field E
2:20
Define N as the Volume Density of Charge Carriers
2:27
Current Density
4:33
Current Per Area
4:36
Formula
4:44
Resistance
5:14
Ratio of the Potential Drop Across an Object to the Current Flowing Through the Object
5:19
5:23
Resistance of a Wire
6:05
Depends on Resistivity
6:09
Resistivity Relates to the Ability of a Material to Resist the Flow of Electrons
6:25
Refining Ohm's Law
7:22
Conversion of Electric Energy to Thermal Energy
8:23
Example 1
9:54
Example 2
10:54
Example 3
11:26
Example 4
14:41
Example 5
15:24
Circuits I: Series Circuits

29m 8s

Intro
0:00
Objectives
0:08
Ohm's Law Revisited
0:39
Relates Resistance, Potential Difference, and Current Flow
0:39
Formula
0:44
Example 1
1:09
Example 2
1:44
Example 3
2:15
Example 4
2:56
Electrical Power
3:26
Transfer of Energy Into Different Types
3:28
Light Bulb
3:37
Television
3:41
Example 5
3:49
Example 6
4:27
Example 7
5:12
Electrical Circuits
5:42
Closed-Loop Path Which Current Can Flow
5:43
Typically Comprised of Electrical Devices
5:52
Conventional Current Flows from High Potential to Low Potential
6:04
Circuit Schematics
6:26
Three-dimensional Electrical Circuits
6:37
Source of Potential Difference Required for Current to Flow
7:29
Complete Conducting Paths
7:42
Current Only Flows in Complete Paths
7:43
Left Image
7:46
Right Image
7:56
Voltmeters
8:25
Measure the Potential Difference Between Two Points in a Circuit
8:29
Can Remove Voltmeter from Circuit Without Breaking the Circuit
8:47
Very High Resistance
8:53
Ammeters
9:31
Measure the Current Flowing Through an Element of a Circuit
9:32
Very Low Resistance
9:46
Put Ammeter in Correctly
10:00
Example 8
10:24
Example 9
11:39
Example 10
12:59
Example 11
13:16
Series Circuits
13:46
Single Current Path
13:49
Removal of Any Circuit Element Causes an Open Circuit
13:54
Kirchhoff's Laws
15:48
Utilized in Analyzing Circuits
15:54
Kirchhoff's Current Law
15:58
Junction Rule
16:02
Kirchhoff's Voltage Law
16:30
Loop Rule
16:49
Example 12
16:58
Example 13
17:32
Basic Series Circuit Analysis
18:36
Example 14
22:06
Example 15
22:29
Example 16
24:02
Example 17
26:47
Circuits II: Parallel Circuits

39m 9s

Intro
0:00
Objectives
0:16
Parallel Circuits
0:38
Multiple Current Paths
0:40
Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
0:44
Draw a Simple Parallel Circuit
1:02
Basic Parallel Circuit Analysis
3:06
Example 1
5:58
Example 2
8:14
Example 3
9:05
Example 4
11:56
Combination Series-Parallel Circuits
14:08
Circuit Doesn't Have to be Completely Serial or Parallel
14:10
Look for Portions of the Circuit With Parallel Elements
14:15
Lead to Systems of Equations to Solve
14:42
Analysis of a Combination Circuit
14:51
Example 5
20:23
Batteries
28:49
Electromotive Force
28:50
Pump for Charge
29:04
Ideal Batteries Have No Resistance
29:10
Real Batteries and Internal Resistance
29:20
Terminal Voltage in Real Batteries
29:33
Ideal Battery
29:50
Real Battery
30:25
Example 6
31:10
Example 7
33:23
Example 8
35:49
Example 9
38:43

34m 3s

Intro
0:00
Objectives
0:17
Capacitors in Parallel
0:51
Store Charge on Plates
0:52
Can Be Replaced with an Equivalent Capacitor
0:56
Capacitors in Series
1:12
Must Be the Same
1:13
Can Be Replaced with an Equivalent Capacitor
1:15
RC Circuits
1:30
Comprised of a Source of Potential Difference, a Resistor Network, and Capacitor
1:31
RC Circuits from the Steady-State Perspective
1:37
Key to Understanding RC Circuit Performance
1:48
Charging an RC Circuit
2:08
Discharging an RC Circuit
6:18
The Time Constant
8:49
Time Constant
8:58
By 5 Time Constant
9:19
Example 1
9:45
Example 2
13:27
Example 3
16:35
Example 4
18:03
Example 5
19:39
Example 6
26:14
RC Circuits: Transient Analysis

1h 1m 7s

Intro
0:00
Objectives
0:13
Charging an RC Circuit
1:11
Basic RC Circuit
1:15
Graph of Current Circuit
1:29
Graph of Charge
2:17
Graph of Voltage
2:34
Mathematically Describe the Charts
2:56
Discharging an RC Circuit
13:29
Graph of Current
13:47
Graph of Charge
14:08
Graph of Voltage
14:15
Mathematically Describe the Charts
14:30
The Time Constant
20:03
Time Constant
20:04
By 5 Time Constant
20:14
Example 1
20:39
Example 2
28:53
Example 3
27:02
Example 4
44:29
Example 5
55:24
Section 3: Magnetism
Magnets

8m 38s

Intro
0:00
Objectives
0:08
Magnetism
0:35
Force Caused by Moving Charges
0:36
Dipoles
0:40
Like Poles Repel, Opposite Poles Attract
0:53
Magnetic Domains
0:58
Random Domains
1:04
Net Magnetic Field
1:26
Example 1
1:40
Magnetic Fields
2:03
Magnetic Field Strength
2:04
Magnets are Polarized
2:16
Magnetic Field Lines
2:53
Show the Direction the North Pole of a Magnet Would Tend to Point if Placed on The Field
2:54
Direction
3:25
Magnetic Flux
3:41
The Compass
4:05
Earth is a Giant Magnet
4:07
Earth's Magnetic North Pole
4:10
Compass Lines Up with the Net Magnetic Field
4:48
Magnetic Permeability
5:00
Ratio of the magnetic Field Strength Induced in a Material to the Magnetic Field Strength of the Inducing Field
5:01
Free Space
5:13
Permeability of Matter
5:41
Highly Magnetic Materials
5:47
Magnetic Dipole Moment
5:54
The Force That a Magnet Can Exert on Moving Charges
5:59
Relative Strength of a Magnet
6:04
Example 2
6:26
Example 3
6:52
Example 4
7:32
Example 5
7:57
Moving Charges In Magnetic Fields

29m 7s

Intro
0:00
Objectives
0:08
Magnetic Fields
0:57
Vector Quantity
0:59
Tesla
1:08
Gauss
1:14
Forces on Moving Charges
1:30
Magnetic Force is Always Perpendicular to the Charged Objects Velocity
1:31
Magnetic Force Formula
2:04
Magnitude of That
2:20
Image
2:29
Direction of the Magnetic Force
3:54
Right-Hand Rule
3:57
Electron of Negative Charge
4:04
Example 1
4:51
Example 2
6:58
Path of Charged Particles in B Fields
8:07
Magnetic Force Cannot Perform Work on a Moving Charge
8:08
Magnetic Force Can Change Its Direction
8:11
Total Force on a Moving Charged Particle
9:40
E Field
9:50
B Field
9:54
Lorentz Force
9:57
Velocity Selector
10:33
Charged Particle in Crosses E and B Fields Can Undergo Constant Velocity Motion
10:37
Particle Can Travel Through the Selector Without Any Deflection
10:49
Mass Spectrometer
12:21
Magnetic Fields Accelerate Moving Charges So That They Travel in a Circle
12:26
Used to Determine the Mass of An Unknown Particle
12:32
Example 3
13:11
Example 4
15:01
Example 5
16:44
Example 6
17:33
Example 7
19:12
Example 8
19:50
Example 9
24:02
Example 10
25:21
Forces on Current-Carrying Wires

17m 52s

Intro
0:00
Objectives
0:08
Forces on Current-Carrying Wires
0:42
Moving Charges in Magnetic Fields Experience Forces
0:45
Current in a Wire is Just Flow of Charges
0:49
Direction of Force Given by RHR
4:04
Example 1
4:22
Electric Motors
5:59
Example 2
8:14
Example 3
8:53
Example 4
10:09
Example 5
11:04
Example 6
12:03
Magnetic Fields Due to Current-Carrying Wires

24m 43s

Intro
0:00
Objectives
0:08
Force on a Current-Carrying Wire
0:38
Magnetic Fields Cause a Force on Moving Charges
0:40
Current Carrying Wires
0:44
How to Find the Force
0:55
Direction Given by the Right Hand Rule
1:04
Example 1
1:17
Example 2
2:26
Magnetic Field Due to a Current-Carrying Wire
4:20
Moving Charges Create Magnetic Fields
4:24
Current-Carrying Wires Carry Moving Charges
4:27
Right Hand Rule
4:32
Multiple Wires
4:51
Current-Carrying Wires Can Exert Forces Upon Each Other
4:58
First Right Hand Rule
5:15
Example 3
6:46
Force Between Parallel Current Carrying Wires
8:01
Right Hand Rules to Determine Force Between Parallel Current Carrying Wires
8:03
Find Magnetic Field Due to First Wire, Then Find Direction of Force on 2nd Wire
8:08
Example
8:20
Gauss's Law for Magnetism
9:26
Example 4
10:35
Example 5
12:57
Example 6
14:19
Example 7
16:50
Example 8
18:15
Example 9
18:43
The Biot-Savart Law

21m 50s

Intro
0:00
Objectives
0:07
Biot-Savart Law
0:24
Brute Force Method
0:49
Draw It Out
0:54
Diagram
1:35
Example 1
3:43
Example 2
7:02
Example 3
14:31
Ampere's Law

26m 31s

Intro
0:00
Objectives
0:07
Ampere's Law
0:27
Finds the Magnetic Field Due to Current Flowing in a Wire in Situations of Planar and Cylindrical Symmetry
0:30
Formula
0:40
Example
1:00
Example 1
2:19
Example 2
4:08
Example 3
6:23
Example 4
8:06
Example 5
11:43
Example 6
13:40
Example 7
17:54
Magnetic Flux

7m 24s

Intro
0:00
Objectives
0:07
Magnetic Flux
0:31
Amount of Magnetic Field Penetrating a Surface
0:32
Webers
0:42
Flux
1:07
Total Magnetic Flux
1:27
Magnetic Flux Through Closed Surfaces
1:51
Gauss's Law for Magnetism
2:20
Total Flux Magnetic Flux Through Any Closed Surface is Zero
2:23
Formula
2:45
Example 1
3:02
Example 2
4:26

1h 4m 33s

Intro
0:00
Objectives
0:08
0:44
0:46
Direction of the Induced Current is Given by Lenz's Law
1:09
Formula
1:15
Lenz's Law
1:49
Lenz's Law
2:14
Lenz's Law
2:16
Example
2:30
Applying Lenz's Law
4:09
If B is Increasing
4:13
If B is Decreasing
4:30
Maxwell's Equations
4:55
Gauss's Law
4:59
Gauss's Law for Magnetism
5:16
Ampere's Law
5:26
5:39
Example 1
6:14
Example 2
9:36
Example 3
11:12
Example 4
19:33
Example 5
26:06
Example 6
31:55
Example 7
42:32
Example 8
48:08
Example 9
55:50
Section 4: Inductance, RL Circuits, and LC Circuits
Inductance

6m 41s

Intro
0:00
Objectives
0:08
Self Inductance
0:25
Ability of a Circuit to Oppose the Magnetic Flux That is Produced by the Circuit Itself
0:27
Changing Magnetic Field Creates an Induced EMF That Fights the Change
0:37
Henrys
0:44
Function of the Circuit's Geometry
0:53
Calculating Self Inductance
1:10
Example 1
3:40
Example 2
5:23
RL Circuits

42m 17s

Intro
0:00
Objectives
0:11
Inductors in Circuits
0:49
Inductor Opposes Current Flow and Acts Like an Open Circuit When Circuit is First Turned On
0:52
Inductor Keeps Current Going and Acts as a Short
1:04
If the Battery is Removed After a Long Time
1:16
Resister Dissipates Power, Current Will Decay
1:36
Current in RL Circuits
2:00
Define the Diagram
2:03
Mathematically Solve
3:07
Voltage in RL Circuits
7:51
Voltage Formula
7:52
Solve
8:17
Rate of Change of Current in RL Circuits
9:42
Current and Voltage Graphs
10:54
Current Graph
10:57
Voltage Graph
11:34
Example 1
12:25
Example 2
23:44
Example 3
34:44
LC Circuits

9m 47s

Intro
0:00
Objectives
0:08
LC Circuits
0:30
Assume Capacitor is Fully Charged When Circuit is First Turned On
0:38
Interplay of Capacitor and Inductor Creates an Oscillating System
0:42
Charge in LC Circuit
0:57
Current and Potential in LC Circuits
7:14
Graphs of LC Circuits
8:27
Section 5: Maxwell's Equations
Maxwell's Equations

3m 38s

Intro
0:00
Objectives
0:07
Maxwell's Equations
0:19
Gauss's Law
0:20
Gauss's Law for Magnetism
0:44
1:00
Ampere's Law
1:18
Revising Ampere's Law
1:49
Allows Us to Calculate the Magnetic Field Due to an Electric Current
1:50
Changing Electric Field Produces a Magnetic Field
1:58
Conduction Current
2:33
Displacement Current
2:44
Maxwell's Equations (Complete)
2:58
Section 6: Sample AP Exams
1998 AP Practice Exam: Multiple Choice Questions

32m 33s

Intro
0:00
0:11
Multiple Choice 36
0:36
Multiple Choice 37
2:07
Multiple Choice 38
2:53
Multiple Choice 39
3:32
Multiple Choice 40
4:37
Multiple Choice 41
4:43
Multiple Choice 42
5:22
Multiple Choice 43
6:00
Multiple Choice 44
8:09
Multiple Choice 45
8:27
Multiple Choice 46
9:03
Multiple Choice 47
9:30
Multiple Choice 48
10:19
Multiple Choice 49
10:47
Multiple Choice 50
12:25
Multiple Choice 51
13:10
Multiple Choice 52
15:06
Multiple Choice 53
16:01
Multiple Choice 54
16:44
Multiple Choice 55
17:10
Multiple Choice 56
19:08
Multiple Choice 57
20:39
Multiple Choice 58
22:24
Multiple Choice 59
22:52
Multiple Choice 60
23:34
Multiple Choice 61
24:09
Multiple Choice 62
24:40
Multiple Choice 63
25:06
Multiple Choice 64
26:07
Multiple Choice 65
27:26
Multiple Choice 66
28:32
Multiple Choice 67
29:14
Multiple Choice 68
29:41
Multiple Choice 69
31:23
Multiple Choice 70
31:49
1998 AP Practice Exam: Free Response Questions

29m 55s

Intro
0:00
0:14
Free Response 1
0:22
Free Response 2
10:04
Free Response 3
16:22

• ## Related Books

### Circuits II: Parallel Circuits

• Ohm’s Law is an empirical law which states that the potential drop across a resistor is equal to the product of the current through the resistor and the resistance.
• Electrical power is the rate at which electrical energy is transformed into other types of energy.
• Conventional current flows from high potential to low potential in a circuit. Electron current flows in the opposite direction.
• Current only flows in complete paths.
• Kirchhoff’s Voltage Law states that the sum of all potential drops around a close loop must be zero. This is a restatement of the law of conservation of energy.
• Kirchhoff’s Current Law states that the sum of all current entering any point in a circuit must equal the sum of all current leaving that point. This is a restatement of the law of conservation of charge.
• Voltmeters are connected in parallel with the element to be measured. They have a very high resistance.
• Ammeters are connected in series with the element to be measured. They have a very low resistance.
• Real batteries have some amount of inherent resistance, known as the internal resistance of the battery. The terminal voltage is slightly lower than the battery’s emf due to the potential drop across the internal resistance of the battery in an operating circuit.

### Circuits II: Parallel Circuits

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:16
• Parallel Circuits 0:38
• Multiple Current Paths
• Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
• Draw a Simple Parallel Circuit
• Basic Parallel Circuit Analysis 3:06
• Example 1 5:58
• Example 2 8:14
• Example 3 9:05
• Example 4 11:56
• Combination Series-Parallel Circuits 14:08
• Circuit Doesn't Have to be Completely Serial or Parallel
• Look for Portions of the Circuit With Parallel Elements
• Lead to Systems of Equations to Solve
• Analysis of a Combination Circuit 14:51
• Example 5 20:23
• Batteries 28:49
• Electromotive Force
• Pump for Charge
• Ideal Batteries Have No Resistance
• Real Batteries and Internal Resistance
• Terminal Voltage in Real Batteries
• Ideal Battery 29:50
• Real Battery 30:25
• Example 6 31:10
• Example 7 33:23
• Example 8 35:49
• Example 9 38:43

### Transcription: Circuits II: Parallel Circuits

Hello, everyone, and welcome back to www.educator.com.0000

I’m Dan Fullerton, in this lesson we are going to continue to try0003

and make the world a better place by helping folks understand how circuits work,0006

specifically focusing on parallel circuits and ideal and real batteries in this lesson.0010

Our objectives begin to understand the behavior of series in parallel combinations of resistors0017

to find current voltage resistance and power.0023

Applying ohms law and Kirchhoff's rules to DC circuits.0025

Understanding the properties of voltmeters and ammeter, and ideal and real batteries, understanding how they work and what the difference is.0029

Let us talk about parallel circuits.0038

Parallel circuits have more than 1 current path.0040

The removal of a circuit element from a parallel circuit may allow the other elements to continue operating.0043

Think of the lights in a room in your house, if 1 light bulb goes out, typically everything in the house does not go off.0049

Instead, you go switch the 1 light bulb, it comes back on but everything else continues operating.0055

That is a feature of a parallel circuit.0060

To draw a simple one, we will start off with a source of potential difference and0062

then we will draw some resistors but we will make sure that we now have multiple current paths.0067

There is an example of a parallel circuit.0081

Before, we found an equivalent resistance for series elements, we added up those resistances.0084

For a parallel circuit 1/the equivalent resistance is equal to 1/your first resistor + 1/your second resistor and so on.0090

So that your equivalent resistance is always smaller than your smallest individual resistor.0100

You could also look at the total current flow is going to be equal to the sum of the individual currents through each branch of the circuit.0106

So if we call this R1, R2, R3, and R4, the current flow through here I1, I2, I3, I4, and so on.0116

These all have to come together to give you your total current flow.0131

That is really just restatement of Kirchhoff's current law .0135

The potential is going to be the same across each of these elements0139

because they are connected opposite sides of that source of potential difference.0145

The voltages are the same in a parallel circuit, for however many elements you might have.0149

Now this formula can get a little cumbersome when you are dealing with a lot of resistors.0157

A quick way, a little cheer formula if you only have 2 resistors in parallel to find the equivalent resistance,0161

you can say that the equivalent resistance is equal to your first resistor × your second resistor/ the sum of those 2 resistances.0167

It only works for 2 resistors in parallel.0176

You got more than 2, you have got to go back to your original formula here.0179

Let us take a look at basic parallel circuit analysis and we are going to do this with a VIRP table again.0184

We have a 12 V source of potential difference and three 2000 ohm resistors.0189

We will make our VIRP table and label our elements.0195

Let us call this R1, R2, and R3.0201

We will have a line for R1, R2, or R3, and our total.0206

Maybe might not need this table but I think it will play.0221

We will start by filling in what we know again.0225

We know our total voltage is 12V, we know our 3 resistances 2000, 2000, 2000.0228

And because it is a parallel circuit, right away I can look at it and tell the potential drop across each of these resistors is also 12V.0240

This ensures we have 12V on either side of our battery at any point on the wire has the same potential.0247

Therefore, these are all 12.0253

Once I have done that, I can find my current flow I = V/ R.0257

In that case, 12/ 2000 is going to be about 0.006.0262

R2 will be the same and the current through R2 will be the same and R3.0267

Because all of these currents are going to add together, our total current flow is going to be 3 × that0272

or the sum of those which is 0.018 amps or 18 milliamps.0282

We can find our equivalent resistance.0288

Before I do that though, we could have used the 1/ R equivalent = 1/ R1 + 1/ R2, that would work.0290

And I can also look by inspection and say you know I have to be certain my total resistance0298

because it is a parallel circuit has to be less than 2000 ohms,0304

because it has to be smaller than the smallest individual resistor in that parallel configuration.0307

R = V/ I 12/ 0.018 is going to give me a resistance of 667 ohm which is less than 2000.0313

That checks out.0323

To find our power again, any of your favorite power formulas V × I, I² R, V²/ R.0325

If you happen to be in that kind of mood to fill in your power.0331

0.072 W or 72 mw, same here, same here.0334

Find our total power, I could add all those up or I can do the same thing with the information from these rows.0341

Any of my favorite power formulas to get the same answer, 216 W or 0.216 W.0347

Basic parallel circuit analysis.0355

Let us do another one, a 15 ohm resistor R1 and a 30 ohm resistor R20360

are to be connected in parallel between points A and B, in a circuit containing a 90V battery.0363

Complete the diagram to show the 2 resistors connected in parallel.0369

Let us do that first, we want to connect these in parallel.0373

We need to make 2 current paths.0378

We will call this one R1, this will be R2.0382

Determine the potential difference across resistor R1 and calculate the current resistor R2.0387

Et us go to our VRIP table and our elements are R1, R2, R total.0392

Our total potential 90V.0412

We know that R1 is 15 ohms, it tells us R2 is 30 ohms.0415

As we look at this, because the battery is connected directly to either side of those resistors,0420

we can tell that the potential across each one is also 90V.0427

To find current here, that is V/ R 90/ 15 that is just going to be 6 amps, 90/ 3 that is going to be 3 amps.0430

Which means our total current is the sum of those in a parallel circuit or 9 amps.0438

Now, I can find my equivalent resistance R = V/ I or 10 ohms .0443

Which checks out, it is less than my smallest individual resistor in the parallel configuration.0448

All I have to do is find my powers to finish up my table.0454

V × I, this will give me 540W, 270W, and 90 × 9, 810W.0458

Or I could just added does up.0466

Once I have my VIRP table, I can answer the question that is asking me.0468

Determine the potential difference across resistor R1.0471

There it is, 90V.0475

V R1 = 90V.0478

And calculate the current in resistor R2.0482

Current through R2, I look up right on my table, there it is 3 amps.0485

That is straightforward.0492

Looking more at this equivalent resistance.0494

Let us take an example with 3 identical lamps connected in parallel with each other.0497

If the resistance of each lamp is X, what is the equivalent resistance of this parallel combination?0501

Right away, we know the answer has to be less than X because they are in parallel.0506

We can get rid of that one right away.0510

1/ our equivalent resistance is going to be 1/ X + 1/ X + 1/ X which is going to be equal to 3/ X,0518

which implies then that 1/ R equivalent of 3/ X then R equivalent must be equal to X/ 3 ohms.0528

The correct answer must be B.0538

How about a circuit that has an extra resistor?0544

Let us look at one that has 4 parallel resistors.0547

We got a 12V battery, we want to know the current measure by ammeter A which is the current flowing through this resistor,0551

the circuit’s equivalent resistance and how much power is dissipated in the 36 ohm resistor.0559

Let us label our resistors R1, R2, R3, R4 and we can make our VIRP table.0564

Those are R1, R2, R3, R4, and total.0579

Now that masterpiece of artwork is on the paper, we can start filling in the things that we know.0602

Our total potentially we have 1 battery here 12V.0607

We know R1 is 6, R2 is 12, R3 is 36, and R4 is 18 ohms.0611

We also can tell just by looking at these that our equivalent resistance is going to be less than 6 ohms,0619

has to be less than our smallest resistor.0626

Now it is a parallel circuit so all of the potentials have to be the same.0630

We can fill those in 12, 12, 12, 12.0634

And since I know 2 things in these rows, I can solve for the others.0639

Current I = V/ R, that is going to be 2 amps through R1.0642

By the way, there is the answer.0648

What is the current measured by ammeter A?0650

It is the current through R1 which is going to be 2 amps.0651

I through R2 is going to be 12/ 12 or 1, 12/ 36 that is going to be 1 /3 .33, 12/ 18 2/ 3 0.67.0657

And all of our currents add up to our total current here which is going to give us 4 amps.0667

The equivalent resistance then, R = V/ I, 3 ohms it is less than 6.0672

Everything is good.0678

Let us calculate our powers while we are here, it is good practice.0681

12 × 2 gives us 24 W, 12 W that will be 4 W.0683

12 × 2/3 8 W, and we can at all those up or just 12 × 4 should give us 48 W.0689

I will check, 24, 12, 4, and 8 is 48 W.0695

We got our VIRP table, we can answer the rest of the questions.0699

What is the circuits equivalent resistance?0702

There it is, 3 ohms.0705

And how much power is dissipated in a 36 ohm resistor?0707

That would be 4 W.0710

Let us take a look at an ammeter in a parallel circuit.0716

The circuit diagram shown below, ammeter A1 reads 10 amps, what is the reading of ammeter A2?0719

To analyze this, it is a parallel circuit, try and vision those ammeters are not there and you can see it is simple, 2 resistor parallel circuits.0726

Let us make our VIRP table and we have R1, R2, and total.0735

Let us fill in what we know.0756

We know our total current flow must be 10 amps because that is what the ammeter tells us.0758

We have got 10 amps there, we will call that R1 and that R2.0764

R1 is 20 ohms, R2 is 30 ohms.0769

Let us see, what else do we know here?0775

We have got 10 amps through the entire circuit, 20 and 30.0778

Let us do this one a little bit differently.0787

If this is 20 and 30, let us find out our equivalent resistance just calculating it the old fashioned way.0789

Our equivalent is going to be R1 R2/ R1 + R2 which will be 20 × 30/ 20 + 30 which is 50, that is going to be 600/ 50.0795

For a total equivalent resistance of 12 ohms.0809

V = I × R, that means it is going to be 120V.0815

This will also be 120V, it is a parallel circuit.0818

I = V/ R, 6 and 4 amps.0822

The question we are asked is what is the reading of ammeter A2?0827

Ammeter A2 is the current flowing through R1 which is 6 amp.0831

Not all circuits are purely series or parallel, some× we have combination circuits.0842

Let us take a look at an example there.0847

A circuit can be series in parallel and have portions of each.0850

First, look for portions of the circuit that have parallel elements.0855

Replace the parallel resistors with the equivalent single resistor to simplify the circuit down and down,0858

until you get something it is a little bit more recognizable that you can deal with.0863

You can analyze with a VIRP table if you want, to help keep track of things.0868

But it is not quite as straightforward with a VIRP table anymore.0871

Then, work back to your original circuit using Kirchhoff's laws until you know the current voltage and0875

resistance of all the individual circuit elements.0881

Oftentimes, you are going to end up with a system of equations to solve those.0882

Sometimes the math can be a little bit involved with these combination circuits.0886

Let us take a look here.0891

We have a combination circuit here, R1, R2, R3, and R4.0893

I'm going to start by making my VIRP table just to help track of what is going on.0899

We will start off VIRP and we have our 4 resistors, R1, R2, R3, R4, and our total.0904

I always like to start by filling in what I know, what we are certain about.0933

Only 1 battery, so our total potential is 10V.0937

We know our resistances, we have 20 ohms, 30 ohms, R3 is 50 ohms, and R4 is 20 ohms.0941

And I think that about covers what we know to begin with.0952

As I look at the circuit, I see we have almost a series circuit except we have these 2 elements in parallel.0956

What I'm going to do is, I'm going to redraw the circuit replacing these 2 resistors0962

with their equivalent single resistors so that I have a series circuit to analyze, to help me understand what is going on.0966

The way I will do that is, we still have our battery 10V, we have our 20 ohm resistor up here R1.0974

We have our equivalent 2 and 3, which we are going to call R2, 3.0985

We do not know its value yet.0990

And we have R4 which is 20 ohms.0993

To find R2, 3, I’m going to find the equivalent resistance of R2 and R3.0998

R2, 3 is going to be equal to the product of the resistances.1004

30 × 50 ohms/ the sum of their resistances, 30 + 50 is 80.1009

This trick only works when you have 2 resistors in parallel.1015

And I come up with about 18.75 ohms.1019

I can find our total resistance here if I know R1 and R4, 20 and 20.1025

The equivalent of these two is 18.75.1032

Our total will be 20 + 20 + 18.75 or 58.75 ohms.1035

Current is V/ R so 10/ 58.75 that is going to be about 0.170 amps.1043

Which means we have 0.170 amps here, that has to be our current flowing through R1.1051

There is no other place for it to go by Kirchhoff's current law.1058

That will be 0.170.1061

We also have to have that 0.170 going through R4.1065

When we get here, the current is going to branch and the total between R2 and 3 has to equal 0.17.1070

We do not know exactly what those values are yet.1076

We will see what else we can learn from the circuit for now.1079

I know 2 things in these rows, I can find the potential drop across R1 and R4.1083

Let us do that, V = IR, 20 × 0.17 that is going to be a potential drop of 3.4V.1088

The same math down here for R4, 3.4.1095

That means that we are left with, that is 6.8 V so we are left with 3.2 V to drop across these.1100

Let us just draw the same here.1107

I'm going to call this point, ground.1108

Usually not a physical connection, just a point we are going to call 0V to simplify our analysis.1111

That means this side of the battery must be 10V higher.1118

We dropped 3.4 across R1, that means we are left with 6.6 V here.1122

We also had dropped 3.4V across R4 so that means if this side is 0, this side must have been 3.4.1130

6.6 V to 3.4, we must have dropped 3.2V across R2 and across R3.1139

We know their potential difference.1146

We can find the current through those.1148

I = V/ R so 3.2/ 30 is going to give us 0.107 amps, going through our 30 ohm resistor.1151

3.2/ 50 is going to be about 0.064 amps which does not quite add up to 0.17 just by a rounding error,1163

but that is really has to do with tracking our significant figures and how accurately we are calculating these.1176

Roughly these two add up to 0.17.1182

The current splits and then it comes back to be 0.17 amps again.1185

For our powers, your choice of formulas.1190

Just going down, I get 0.578 W, 0.342 W, 0.205 W, 0.578 W.1193

Again, 10 × 0.17 is going to give me 1.7 W, which should be the sum of all of those, and it is.1202

Analysis of a combination circuit.1212

The VIRP table did not tell us everything we needed to know but it was kind of useful for tracking some of the characteristics of our circuit.1215

Let us start getting into some circuits and add a little bit more challenge to them.1224

Here we have 2 voltage sources and we are asked to find the current flowing through R3, if R3 has a value of 6 ohms.1231

What is the power dissipated in R3?1239

R3, let us label that, we know that is 6 ohms.1243

I'm going to start by picking a point on my circuit to call ground or 0V.1249

I’m going to do that right here.1253

That is 0V and if I do that that means the side must be 12V.1256

That is also 0V so this must be 16 V over there.1261

I’m also going to go and start defining some of my currents.1267

I’m going to pick a direction, if I'm wrong we will just a negative value.1270

It really does not matter what direction you write.1273

Let us assume that we have this current through R1, I'm going to call I1 and call that direction positive.1275

I'm going to call this direction positive 4 I2, the current through R2.1282

We will label this is I3, the current through R3.1287

If I analyze the circuit right at this point here, I can use Kirchhoff's current law to give me an equation.1293

It says I1 + I2 is going to equal I3.1299

That can be handy.1305

I'm going to go and I'm going to start making loops using Kirchhoff's voltage law to see if I can get some other equations that will help me solve the system.1307

If I start down here and I make a loop this way, I see the negative sign of the 12V battery first, as I start.1316

I'm going to write -12V + the voltage drop across R1 is 8 I1, 8 I1 + 6 I3, brings me back to 0.1323

That is going to equal 0.1338

I also know that I1 + I2 = I3, which implies then because I1 + I2 = I3, I can write this as -12 + 8 I1 + 6.1341

I’m going to replace I3 with I1 + I2 = 0.1358

Or 14 I1 8 I1 + 6 I1 + 6 I2 = 12.1365

I'm going to call that my equation number 1.1379

14 I1 + 6 I2 = 12.1382

I got that from applying Kirchhoff's voltage law around this loop and also throwing in Kirchhoff's current law1384

from that point where all my currents are coming together.1387

Let us take a different loop and maybe I will start here and we will go around this way.1394

As I go past the battery, I see the negative side first.1400

I'm going to write - 16 + the drop across R2 which is going to be 12 I2 + 12 I2 + 6 I31403

brings me back to my starting points so the entire sum of voltage drop across a closed loop must be 0.1417

Then we are going to use what we learn up above I1 + I2 = I3 again.1423

In order to rewrite this as -16 + 12 I2 + 6 ×, replacing I3 with I1 + 2 = 0,1434

which implies then as I multiply this out, that is 6 I1 + 18 I2 is going to equal 16.1451

We will call that equation 2.1467

We got to find some way to combine those.1471

A bunch of different things you can do but the easiest one I see is as I look at 1, if I happen to multiply the left side and the right side by -3,1474

I keep that equality but then I have got some manipulations I can do pretty easily with number 2.1483

I'm going to come over here to the left and I'm going to multiply this side by -3 and that side by -3.1490

The reason I'm doing that is I want to cancel out the I2.1496

14 × -3 is going to be -42.1500

I want to make sure that I track what I'm doing.1504

I’m going to take 1 and multiply it by -3 to get 14.1507

That is going to be -42 I1 -18 I2 = 12.1513

Right down below it, I'm just going to rewrite my equation 2.1521

6 I1 + 18 I2 = 60.1526

A little bit of math trick here.1536

If the left side = the right side for this equation, the left side = the right side for that equation, which they do that.1539

It is what the equal signs means.1545

That means we should be able to add both left sides, add both right side and maintain an equality.1547

That is what I'm going to do realizing that I'm going to get rid of I2 in this equation.1551

When I do that, the left hand side I end up with the -36 I1 must be equal to.1557

I did not multiply this side by -3.1567

-3 × 12 will be -36.1574

-36 I1 = -20.1577

I can solve for I1 then to find that I1 is 20/ 36 or 0.556 amps.1581

Now that I have got that, I can go plug that back into my equation number 2 again, in order to find out what I2 is.1591

Let us plug that in the 2 and I will do that over here if I have 6 I1 + 18 I2 = 16, that implies then that 6 × I1 0.556, + 18 I2 = 16.1598

A little bit more math there and I come up with 18 I2 = 12.664,1621

which implies then that I2 must be 12.664 divided by 18 or about 0.703 amps.1634

We have got I2, we have got I1, to find I3, that is just the sum of I1 and I2.1644

I3 = I1 0.556 + I2 0.703 and I find that I3 is equal to 1.26 amps.1651

Find the current flowing through R3.1665

We did that, that answer is right here 1.26 amps.1668

Finally, what is the power dissipated in R3?1673

To find the power dissipated in R3, a couple of ways we can do that but the simplest that I see at the moment1676

is the power through 3 is going to be I² R just 1.26² × that 6 ohms, or 9.52 W.1683

As you can see, these couple of voltage source problems can get a little bit messy with systems of equations and you can have more or less.1704

And there is a lot of other ways to solve these besides what I did with this little math trick.1711

Regardless though, you are just doing algebra once you have written these Kirchhoff's current law and1716

Kirchhoff's equivalent equations, eliminating variables and solving for unknowns.1720

Let us take a look at batteries.1729

A solar battery which is a combination of cells, provides a potential difference often× referred to as an electromotive force or EMF.1732

It is important to note, electromotive force is not a force, it is a potential difference.1739

A battery can be thought as a pump for charge raising it from a lower potential to a higher potential.1744

Ideal batteries, these magic batteries we use in so many of our circuits have no internal resistance.1750

In practicality, it is really not that simple.1757

Real batteries have some amount of resistance to the flow of charge within the battery itself1759

which we call the internal resistance given the symbol R sub I.1767

In real batteries, the terminal voltage is going to be slightly lower than the battery’s EMF because you1772

lose some of that potential difference due to a voltage drop across the internal resistance of the battery.1778

You love to have that internal resistance as low as possible but it always exist there to some extent.1784

Let us take a look at an ideal battery.1789

We have current flowing from A to B through our battery, positive and negative.1793

There is a resistance, the voltage across our battery we call Δ V which is going to be the voltage of B - the voltage of A,1797

which implies that the voltage of the battery is going to be equal to its EMF script E which is also terminal voltage VT.1810

Pretty straightforward, same sort of battery we have been dealing with.1821

The real battery gets a little bit more complicated because we have this internal resistance inside the battery.1825

Potential difference of our battery here is still Δ V or the voltage at B - the voltage at A.1833

The voltage on our battery is equal to IR as well, which now we have to worry about the EMF1843

of our battery - I × our internal resistance which is what we call the terminal voltage for a real battery.1851

The terminal voltage is the EMF - the voltage drop across that internal resistance of the battery.1860

Let us take a look at a problem involving these.1868

The terminal voltage of a real battery is 15 V.1873

If the battery is an EMF of 18V and supplies 10 W of power to resistor R, find the value of R and RI, the internal resistance.1876

I can analyze this with a VIRP table because really it looks like a simple 2 resistor series circuit.1885

The only difference is RI, one of the resistor is inside the battery.1893

Let us make our VIRP table and our resistors are RI, the internal resistance R, and we have a ρ for our total.1897

Filling in what we know.1918

We know that we have a total EMF of 18V.1921

Let us see, we know that the voltage drop across R, the terminal voltage of the battery is 15V which means the voltage drop across RI must be 3V.1926

Let us see, what else do we know in here.1936

We have 10W of power supplied to resistor R so that must be 10W.1939

Once we know that is 10W, power = V × I, that means our current must be 0.67 amps.1944

It is a series circuit so we have the same current everywhere throughout our circuit.1953

We can start to find our resistances.1957

R = V/ I 3/0.67 is going to give us 4 ½, 15/ 0.67 will give us 22 ½.1960

Their total resistance, we can add these up or 18/0.67.1968

Either way, we get 27.1972

Just finishing our VIRP table to powers.1975

V × I, 3 × 2/3 is going to be 2, and 18 × 2/ 3 is going to give us 12.1978

Of course, our powers add up.1984

The question, find the value of R and RI.1986

I have no problem, I did that a little bit ago.1990

For R and RI, we found that we have 22.5 ohms and 4.5 ohms.1993

Looking at another internal resistance problem.2003

A 50 ohm and 100 ohm resistor are connected as shown to a battery of EMF of 40V and internal resistance of R.2006

Find the value of R if the current in the circuit is 1 amp.2013

What is the battery’s terminal voltage?2017

Let us see here, as we start I'm going to take a look and see if I can simplify this a little bit.2021

I’m going to redraw this with an equivalent resistance here, we have got 100 and 50.2027

Let us draw this as our potential difference of 40 V.2032

We have our R there and we will put in a our R1, 2 there which is going to be equal to 100 ×2038

50/ their sum 150 which is about 33 ohms, which works that is less than either one of those.2050

There is R.2056

What I can do is, I can take a look and use KVL as we go around the loop, the current going this way.2059

As I apply Kirchhoff's voltage law, I see I will get -40 starting at this point, + IR + 33 I, gets me back to my starting 0.2068

Which implies then as I look at this, that 33 + R × I = 40.2084

We know the current in the circuit is 1 amp.2095

If our total is 1 amp, that means I is 1, that means 33 + R = 40.2098

R, our internal resistance must be 7 ohms.2107

We found that is 7 ohms.2111

What is the battery’s terminal voltage now?2114

Our voltage across R is going to be IR which is 1 amp × 7 ohms or 7 V.2120

Our terminal voltage which is the EMF - IR is going to be 40V -7V or 33 V for the terminal voltage of our battery.2129

Let us take a look at another circuit.2145

This one that has some meters in it.2150

We have got 2 voltage sources and given the schematic, determine the reading of both the ammeter and the voltmeter.2152

Let us start by taking a look at our circuits and see what we can define here.2160

I'm going to set this point down here, I'm going to call that my ground or 0V, which means this must be 30, that point must also be 30, it is on the same wire.2164

The ammeter should not affect it.2173

If this is 0 down here too, then this point must be -5V.2176

We can do that.2181

Let us start defining a couple of our currents.2183

Let us call that current 1 through R1, that is going to be.2185

Voltmeter does not affect our circuit, you should have a very high resistance so no current flowing up there.2190

They are minimal.2194

This is I1 as well.2196

This we could call I2 through there, which means I3 heads down through there.2198

We want a reading of both the ammeter and the voltmeter.2208

As I run KVL around the loop, I'm going to start here and I see -30 first + 10 ohms × I1, the voltage drop across R1, + 20 I1.2211

I see a -5 and I get back to 0.2233

I have 30 I1 must be equal to 35 or I1 = 35/ 30 or 1.17 amps.2236

There is the current I1 and let us see here.2249

If I have got 30V here and I1 is 1.17 amps, that means that I’m dropping 10 × I1, which is dropping 11.7V across here.2253

If I had 30 and I dropped 11.7, that means I have got 18.3V here at this point.2265

If I have got 18.3V here and I know I have 0V here, because it is attached to what I called 0,2273

my ground over there, the voltmeter reading must be the difference from 1 side to the other or 18.3V.2283

There we go, that is a little bit near.2305

This was actually a lot more straightforward than solving the entire circuit.2307

Terminal reading of the ammeter and voltmeter.2311

The ammeter must read I1 1.17 amps.2313

Let us take a look at one last problem here.2322

3 resistors 4 ohms, 6 ohms, 8 ohms, are connected in parallel in an electric circuit.2324

The equivalent resistance of the circuit is, we know with parallel resistors,2329

the equivalent resistance is always smaller than the smallest individual resistor so it has to be less than 4 ohms.2334

There is our answer right there, A.2342

Thank you so much for watching www.educator.com.2345

We will see you again real soon, make it a great day.2347

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