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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 1 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3 01 – DC Theory

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Page 1: Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : # 01 – DC TheoryAuthor

Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 1 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

01 – DC Theory

Page 2: Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : # 01 – DC TheoryAuthor

Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 2 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

The intent of this presentation is to present enough information to provide the reader with a fundamental knowledge of Direct Current fundamentals used within Michelin and to better

understand basic system and equipment operations.

By enrolling in this self-study course, you have demonstrated a desire to improve yourself and Michelin Manufacturing. However, this self-study course is only one part of the total Michelin training program. Practical experience, IMS (AP) school, selected reading, and your desire to

succeed are also necessary to successfully round out a fully meaningful training program.

Although the words “he,” “him,” and “his” are used sparingly in this course to enhance communication, they are not intended to be gender driven or to affront or discriminate

against anyone.

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 3 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Basic Electrical Theory

Simply put, electricity is nothing more than the flow of electrons through a conductor.

Electric Charge

Since some atoms can lose electrons and other atoms can gain electrons, it is possible to cause a transfer of electrons from one object to another. When this takes place, the equal distribution of the positive and negative charges in each object no longer exists. Therefore, one object will contain an excess number of electrons and its charge must have a negative (-) electric polarity. The other object will contain an excess number of protons and its charge must have a positive (+) polarity.

When a pair of objects contains the same charge (both are positive or both are negative), the objects are said to have like charges. When a pair of bodies contains different charges, they are said to have

unlike or opposite charges. The law of electric charges may be stated as follows.

Like charges repel each other and unlike charges attract each other.

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 4 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

The Coulomb

The magnitude of electric charge a body possesses is determined by the number of electrons compared with the number of protons within the body. The symbol for the magnitude of the electric charge is Q, expressed in units of coulomb (C). The charge of one coulomb means a body contains a charge of 6.28 x 1018 more electrons than protons.

The Electrostatic Field

The fundamental characteristic of an electric charge is its ability to exert a force. This force is present within the electrostatic field surrounding every charged object. When two objects of opposite polarity are brought near each other, the electrostatic field is concentrated in the area between them. Lines of force drawn between the two objects indicate the electric field. If an electron is released at point A in this field, it will be repelled by the negative charge and will be attracted to the positive one. Thus both charges will tend to move the electron in the direction of the lines of force between the two objects. The arrowheads in this drawing indicate the direction of motion that would be taken by the electron if it were in different areas of the electrostatic field.

01 - DC Theory

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Potential Difference

Because of the force of its electrostatic field, an electric charge has the ability to do the work of moving another charge by attraction and repulsion. The ability of a charge to do work is called its potential. When one charge is different from the other, there must be a difference in potential between them.

The sum of the differences of potential of all the charges in the electrostatic field is referred to as electromotive force (emf).

The basic unit of potential difference is the volt (V). The symbol for potential difference is also V, indicating the ability to do the work of forcing electrons to move. Because the volt unit is used, potential difference is called voltage.

01 - DC Theory

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Current

The movement or the flow of electrons is called current. To produce current, the electrons must be moved by a potential difference. Current is represented by the letter symbol I. The basic unit in which current is measured is the ampere (A). One ampere of current is defined as the movement of one coulomb (6.28 x 1018 electrons) past any point of a conductor during one second of time.

Current Flow

In a conductor, such as copper wire, the free electrons are charges that can be forced to move with relative ease by a potential difference. If a potential difference is connected across two ends of a copper wire, the applied voltage forces the free electrons to move. This current is a drift of electrons from the point of negative charge moving through the wire, and returning to the positive charge at the other end. The direction of the electron drift is from the negative side of the source, through the wire, and back to the positive side of the source. The direction of electron flow is from a point of negative potential to a point of positive potential.

01 - DC Theory

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Current Summary

Static electricity is dealing with electrons at rest, whereas electrical current deals with electric charges in motion. As shown before, an electrical current will flow when a conductor joins two points. One point is more negative than the other point. Hence, as long as there is a conductor to carry them, electrons must move from negative regions to more positive regions (or less negative regions). As long as there is some relative difference in the states of the charges and a conduction path, electrons will flow.

Definition:

An electric current is a movement of electrons in a general direction through a conductor .

01 - DC Theory

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The strength or magnitude of the current will be a measure of the number of electrons that pass through any part of the conductor in one second. The magnitude of a current is measured in AMPERES. When a conductor is carrying a current of 1 ampere, it means that 6.28 x 1018 electrons are passing any point of the conductor in one second. This difference of 6.28 x 1018 electrons is represented by a coulomb. Hence, if 1 Ampere flows for 1 second, a quantity of 1 Coulomb of electricity has been used. 10 Coulombs/second is a current of 10 amperes. A current of 30 amperes is equal to: 30coulombs/second. To find the current flowing in a circuit, the quantity in coulombs is divided by the time in seconds.

1 Ampere = 1 Coulomb / 1 SECOND

Current in AMPERES = I

Quantity in COULOMBS = Q

Time in SECONDS = t Therefore,

Q = I x t coulombs

I = Q/t amperes

01 - DC Theory

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The Electric Circuit Concept

1. One of the first requirements for an electrical circuit is a source of electromotive force (emf). Emf is a force that causes electrons in a circuit to flow and is measured in volts. The Volt may be defined as that difference in potential which, when applied across a resistance of 1 OHM it will produce a current flow of 1 AMPERE.

2. The second requirement for an electrical circuit is a complete path from one side of the emf source to the other side of the emf source for current flow. Look at the following circuit:

01 - DC Theory

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The Electric Circuit Concept

The battery is the emf source. The switch and copper wire will provide a complete path for current flow when the switch is closed. When the switch is closed, the battery supplies energy to electrons in the copper wire and causes them to flow from the negative side to the positive side of the battery.

01 - DC Theory

S1

V

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Resistance

Resistance is the opposition to current flow. In the same way that friction resists the motion of mechanical devices and the flow of fluids through pipes, resistance opposes the flow of electrons through a conductor. In order to maintain a constant flow in a fluid system, pressure must be maintained to overcome the friction in the system.

COMPARISON OF FLUID AND ELECTRICAL SYSTEMS

FLUID ELECTRICAL

Pressure (PSI or BARS) Potential Difference - (Volts)

Flow (gallons/min. or liters/min.) Current (Amperes)

Resistance/Restriction Resistance (Ohms)

(Dynamic Pressure Drop)

Resistance is always present in an electrical circuit. To add resistance to a circuit, electrical components called resistors are used. A resistor is a device whose resistance to current flow is a known or specified value.

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Resistance

Resistance is measured in ohms () and is represented by the symbol R in equations. One ohm can be defined, as that amount of resistance that will limit the current in a conductor to one ampere when the voltage applied to the conductor is one volt. Resistors are common components of many electrical and electronic circuits. Some frequent uses for resistors are to establish the proper value of circuit voltage, to limit current, and to provide a load.

Ohm's Law

In the early part of the 19th century Georg Simon Ohm proved by experiment that a precise relationship exists between current, voltage, and resistance. This relationship is called Ohm's Law and is stated as follows:

The current in a circuit is DIRECTLY proportional to the applied voltage and INVERSELY proportional to the circuit resistance.

Ohm's Law written as a mathematical relationship:

01 - DC Theory

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Look at the following circuit:

The current flowing through the circuit is 1 Ampere. Let's see what the current is if we use different voltages for the battery.

01 - DC Theory

R BATTERY CORRESPONDING CURRENT

10 5 v 0.5 A

10 20 v 2.0 A

10 100 v 10.0 A

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Example 1:

For the circuit shown, find the Voltage:Given: I = 1.5A R = 10

Ohm’s Law:

Solution: Transpose the equation; V = I * R

(1.5) (10) = 15 Volts

Example 2:

For the circuit shown, find the Resistance:

I = 28 mA V = 56v

Solution:

I = 56 / 0.028 = 2000 = 2K

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 15 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Example 3:

For the circuit shown, find the Current:Given: V = 24Volts R = 10

Ohm’s Law:

Solution:

24 / 100 = .24Amps = 240mA

01 - DC Theory

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Series Circuits

The above circuit is classified as a series circuit.

In this circuit there is only one current path. This distinguishes the series circuit from the parallel circuit. The battery is the voltage source and is usually referred to as the applied or total voltage of the

circuit.

Important Properties

The amount of current flowing in one part of the circuit is equal to the amount flowing in any other part of the circuit. In other words, the same current flows through R1, R2, and R3.

IT = IR1 = IR2 = IR3

As seen from Ohm's Law, a current flowing through a resistor produces a potential difference across the terminals of the resistor. (V = I x R) This potential difference is called a Voltage drop.

01 - DC Theory

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Series Circuits

Kirchhoff's Voltage Law

VT = VR1 + VR2 + VR3

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RT IT = (R1 IT) + (R2 IT) + (R3 IT) (Substitute Ohm’s Law into Kirchhoff’s Law)

Divide both sides of the equation by IT

IT cancels and the following equation is the result:

Rule for Series Circuits: Total resistance for any series resistive circuit is the sum of the individual resistances.

RT = R1 + R2 + R3

01 - DC Theory

VT = VR1 + VR2 + VR3 (Kirchhoff's Voltage Law for Series Circuits)

IT = IR1 = IR2 = IR3 (Current common in series circuits)

VR1 = R1 IT

VR2 = R2 IT

(Ohm's Law) VR3 = R3 IT VT = RT IT

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Unit of Power

Power is the rate of doing work, and the unit of electrical power is the watt.

Power is expressed as VOLTS X AMPS or as an equation. P = V x I

Note that: I = V R and V = I X R (Ohm's Law).

By substituting Ohm's Law into the expression P = V x I we get:

01 - DC Theory

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Unit of Power

Example:

A device is connected across a supply of 230V. There is a current flow of 5 AMPS. What is the power dissipated by the circuit?

P = V x I

P = 230V x 5A = 1150W or 1.15KW

Power for electric motors is sometimes rated in watts or kilowatts, but more usually in Horsepower (hp).

1 hp = 746 watts

Energy and work are essentially the same and are expressed in identical units. Power is different, however, because it is the time rate of doing work. The Kilowatt-hour (KWh) is the practical commercial unit of electrical energy or work performed. If a device, as in the above example, were connected to the supply for a period of ten (10) hours, then 11.5 KWh would be used (1.15 KW x 10 Hrs.).

Work = Power x time

01 - DC Theory

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Practice Exercises

Find the Power PT:

1. VT = 48v

IT = 2A

PT =

2. R1 = 10

IT = 8A

PT =

3. VT = 40v

R1 = 100

PT =

01 - DC Theory

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Practice Exercises

Given:

R1 = 20

R2 = 30

IT = 2A

Find:

RT =

PT =

VT =

VR1 =

PR1 =

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Practice Exercises

Given:

IT = 60 A

R1 = 5

R2 = 3

R3 = 9

R4 = 5

R5 = 8

Find:

VT =

VR1 =

VR2 =

VR3 =

VR4 =

PT =

01 - DC Theory

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Practice Exercises

Given:

VT = 210 V

R1 = 6

R2 = 19

R3 = 5

Find:

IT =

01 - DC Theory

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Practice Exercises

Given:

VT = 250 V

IT = 25 A

R1 = 5

R2 = 2.5

Find:

RT =

R3 =

State Ohm’s Law: __________

01 - DC Theory

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Resistor Color Code

01 - DC Theory

COLOR BAND 1, 1st # BAND 2, 2nd #BAND 3

BLACK

BROWN

RED

ORANGE

YELLOW

GREEN

BLUE

VIOLET

GRAY

WHITE

GOLD

SILVER

NONE

0

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9 1,000,000,000

100,000,000

10,000,000

1,000,000

100,000

10,000

1,000

100

10

1

20%

10%

5%

+-

+-

+-

0.01

0.1

MULTIPLIERBAND 4

TOLERANCE

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The Digital Multi-Meter (DMM)

Digital multi-meters (DMM) have generally replaced the analog-type multi-meter (VOM) as the test device of choice for maintainers because they are easier to read, are often more compact and have greater accuracy. The DMM performs all standard VOM measurement functions of AC and DC. Some offer frequency and temperature measurement. Many have such features as peak-hold display that provides short-term memory for capturing the peak value of transient signals as well as audible and visual indications for continuity testing and level detection.

The display shows other functions as well. • Low battery indicator• Annunciators show what is being measured (volts, ohms, amps, etc.)• Autopolarity indicates negative readings with a minus sign when the • leads are connected incorrectly without any damage • Autoranging automatically selects proper measurement range• One selector switch makes it easy to select measurement functions• Overload protection prevents damage to the meter and the circuiit, and protects the user

01 - DC Theory

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Measuring Resistance

The figure below shows the steps that should be followed when measuring resistance. Remember that resistance measurements are carried out without the power being applied to the component under test, and resistance values can vary by as much as 20% due to the tolerance of certain resistors. Do not be misled if your meter reading is slightly different from the color band on the resistor. If a resistor’s value is off and exceeds the tolerance, the resistor should be replaced. A resistor will rarely short, but typically will open. If a resistor does open, the DMM display will flash on and off or display OL (open line) because the resistor has an infinite resistance.

• Turn off power to the circuit• Select resistance • Plug the black test lead into the COM jack and the red test lead into the jack• Connect the probe tips across the component or portion of the circuit for which

you want to determine the resistance• View the reading and be sure to note the unit of measure: , K, M, etc.

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Measuring Voltage

The figure below shows the steps that should be followed when measuring voltage. The measurement of both voltage and resistance is where the DMM finds its greatest utilization. For voltage and resistance measurement, the red lead is inserted into the V – Ω (volt or ohm) meter jack.

Select volts AC (V~), volts DC (V---), milli-volts (V---) as desired

Plug the black test lead into the COM jack and the red test lead into the V jack

Touch the probe tips to the circuit across a load or power source as shown

(parallel to the circuit to be tested) View the reading being sure to note the

unit of measure

Note: For DC readings of the correct polarity (+ or -), touch the red test probe

to the positive side of the circuit, and the black test probe to the negative side

of the circuit ground. If you reverse the connections, a DMM with auto-polarity

will merely display a minus sign indicating negative polarity. With an analog

meter you risk damaging the meter.

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Measuring Current

The figure below shows the steps that should be followed when measuring current. The measurement of current is rarely performed when troubleshooting, as the circuit path has to be opened to insert the DMM in series with the current flow. However, if current is to be measured, the red lead is inserted into one of the ampere jacks, 10 amp (10A) or 300 milliamp (300 mA) input jack depending on the expected value of the reading.• Turn off the power to the circuit• Disconnect, cut or unsolder the circuit, creating a place where the meter probes

can be inserted• Select amps AC (A~), or amps DC (A---) as desired• Plug the black test lead into the COM jack and the red test lead into 10 amp

(10A) or 300 milliamp (300mA) jack depending on the expected value of the reading• Connect the probe tips to the circuit across the break as shown so that all current

will flow through the meter ( a series connection)• Turn the circuit power back on• View the reading being sure to note the unit of measure

Note: If test leads are reversed a negative (-) sign will be displayed

01 - DC Theory

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01 - DC Theory

Parallel Circuits

The above circuit is classified as a parallel circuit. In this circuit a total current (IT) leaves the negative terminal of the battery. When the current reaches the junction labeled (A), it splits into two branch currents (IR1) and (IR2).

When (IR1) and (IR2) reach the junction labeled B, they join back together as IT and continue on to the positive side of the battery. Since no current can be lost, the sum of IR1 and IR2 must be equal to IT.

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Parallel Circuits

Important Properties:The voltage across one branch of a parallel circuit is equal to the voltage across any other parallel branch of the circuit.

VT = VR1 = VR2

The sum of the currents entering a junction must equal the sum of currents leaving the junction.

Kirchhoff's Current Law

IT = IR1 + IR2

01 - DC Theory

The sum of the currents entering a junction must equal the sum of currents leaving the junction.

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Parallel Circuits

When two resistors are connected in parallel, the electrons can move more easily. In a parallel circuit the electrons can move through one branch of the circuit without having to go through the other branch. The total current flowing through a parallel circuit splits up, some going one way and some going the other. The branch currents will sum together.

01 - DC Theory

IT = IR1 + IR2 (Kirchhoff's Current Law for Parallel Circuits

VT = VR1 = V R2 (Voltage common in parallel circuits)

IR1 = VT R1

IR2 = VT R2 (Ohm's Law)

IT = VT RT

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Parallel Circuits

(Substitute Ohm’s Law into Kirchhoff’s Current Law)

(Divide both sides of the equation by VT)

And then,

VT now cancels out and the following equation is the result:

Use the reciprocal to get

01 - DC Theory

V

1 )

R

V + R

V( = )V

1(

R

V

T2

T

1

T

TT

T

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 35 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Parallel Circuits

Given:

VT = 120 V

R1 = 15

R2 = 10

Find:

RT =

IT =

IR1 =

01 - DC Theory

Page 36: Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : # 01 – DC TheoryAuthor

Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 36 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Parallel Circuits

Given:

IT = 3 A

R1 = 12

R2 = 15

R3 = 10

Find:

RT =

VT =

IR1 =

IR2 =

IR3 =

01 - DC Theory

-

+

VT

IT

R2IR2R1IR1

R3IR3

Page 37: Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : # 01 – DC TheoryAuthor

Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 37 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Parallel Circuits

Given:

VT = 12 V

R3 = 1

Find:

PR3 =

01 - DC Theory

-

+

VT

IT

R2IR2R1IR1

R3IR3

Page 38: Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : # 01 – DC TheoryAuthor

Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 38 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Parallel Circuits

Given:

IT = 6 A

R1 = 4

R2 = 4

R3 = 4

Find:

RT =

VT =

IR1 =

IR3 =

PR2 =

01 - DC Theory

-

+

VT

IT

R2IR2R1IR1

R3IR3

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 39 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Series and Parallel Circuits

If resistors are arranged in both series and parallel in the same circuit, then solving the parallel section first and then the entire circuit is the simplest solution. In the circuit shown, R2 and R3 are in parallel and this combination is in series with R1. We may call the resistance of the parallel section REQ and find it by the reciprocal method:

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 40 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Series and Parallel Circuits

We may then redraw the circuit as shown below, where the branch with resistance REQ has replaced the parallel section R2 and R3.

Now the resistance of the entire circuit is simply: RT = R1 + REQ.

Note: REQ = Equivalent resistance for R2 in parallel with R3 and is shown as a broken line.

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 41 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Series and Parallel Circuits

Given:

VT = 150 V

R1 = 6

R2 = 4

R3 = 14

R4 = 16

Find:

IT =

IR1 =

IR3 =

VR4 =

PR1 =

PR2 =

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 42 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Series and Parallel Circuits

Given:

VT = 30 V

R1 = 10

R2 = 15

R3 = 20

Find:

RT =

IT =

VR1 =

IR3 =

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 43 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Series and Parallel Circuits

Given:

VT = 60 V

R1 = 30

R2 = 20

R3 = 10

R4 = 5

Find:

RT =

IT =

PR2 =

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 44 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Series and Parallel Circuits

Given:

IT = 1 A

R1 = 2

R2 = 10

R4 = 4

VR2 = 6 V

Find:

R3 =

RT =

VT =

VR4 =

PT =

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 45 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Supplemental Problems

Known: Pt = 3600 W

Calculate :

Rt = ______

01 - DC Theory

R 3R 2R 1

V 1 = 60V

V T

V 3 = 20 VV 2 = 20V

P 2 = 400W

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 46 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Supplemental Problems

Known: R1 = 6Ω I2 = 3A P3 = 432W

Calculate :

Rt = ______

01 - DC Theory

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Presentation : IMS – Tech Managers Conference Author : IMS Staff Creation date : 08 March 2012 Classification : D3 Conservation : Page : 47 01 – DC Theory Author : IMS Stafff Creation date : 31 Oct 2012 Classification : D3

Supplemental Problems

Calculate :

IR1

IR5

IR2

VR4

01 - DC Theory