BEKG1113PRINCIPLE OF ELECTRICAL AND ELECTRONICS

CHAPTER 2 (WEEK 3)

FACULTY OF ELECTRONIC AND COMPUTER ENGINEERING

2

Chapter 2

• The topics for this chapter:– Voltage and current sources– Resistor– Resistor Color code– Ohm’s Law– Types of circuit - series, parallel and series-

parallel circuit– Circuit ground, KVL, KCL, Power dissipation

3

Voltage source

– Ideal voltage source

Can provide a constant voltage for any current required by a circuit [1].

– Voltage source can either be AC (Alternating current) or DC (Direct current) [2].

4

Voltage source

• The symbols for voltage sources are represented here [1]:

• VI characteristics for an ideal voltage source is shown by the graph [1]

VI characteristics

0

1

2

3

4

5

6

1 2 3 4 5 6 7

I (Ampere)

V (

Vo

lts)

V

5

Voltage source

• Six categories of voltage sources [1]:– Batteries– Solar cells– Generator– The electronic power supply– Thermocouples– Piezoelectric sonsors

6

• Thermocouples– It is a thermoelectric type of voltage source

that is commonly used to sense temperature. The operation is based on Seebeck effect.

– The types of thermocouples are categorized by letters (J,K,E,N, B and R) that depends on the range of temperature.

7

• The graph of thermocouple characteristics.• Note that K is the most common

thermocouple used.

8

• A few types of thermocouples which depend on their specification requirements.

9

• Piezoelectric sensors– They act as voltage sources and are based

on the piezoelectric effect where a voltage is generated when a piezoelectric material is mechanically deformed by an external force.

– Types of piezoelectric material; quartz & ceramic

– Applications: pressure sensor, force sensor

Chapter 2: D

irect Current (D

C)

10

Chapter 2: Direct Current (DC) Circuit

• The current source– Definition: The ideal current source can

provide a constant voltage for any load.– The symbol for a current source is shown

below

11

Current Source

• Constant current sources – a type of power supply

12

Current Source

• Constant current sources from most transistor circuits

• Constant current battery chargers

Basic DC Circuit

• It consists of a source of electrical energy, some sort of load to make use of that energy, and electrical conductors connecting the source and the load.

13

SOURCE LOAD

+

-

Electrical Circuit Requirements

• Control Device: Allows the user control to turn the circuit on or off. Switches are the devices commonly used for controlling the oppening or closing of circuits.

• Protection Device: Current must be monitored and not allowed to exceed a safe level as to protect users from shock, to protect the equipment from damage, and to prevent fire hazards.

14

Protective Devices

• Fuses: Fuses use a metallic element that melts when current exceeds a preset value. It will blow if more current passes through it.

• Circuit breaker: Circuit breaker works on the different principle. When the current exceeds the rated value of breaker, the magnetic field produced by excessive current operates a mechanism that trips open a switch. After the fault or overload condition has been cleared, the breaker can reset and used again.

15

16

Current Direction

17

Circuit Ground

• Grounding is achieved in an electrical system when one of the conductive wires serving as part of the circuit path is intentionally given a direct path to the earth. This method of grounding is called earth ground.

• In most electronic equipment, a large conductive area on printed circuit board or the metal housing is used as the circuit ground or chassis ground.

18

• Ground is the reference point in electric circuits and has a potential of 0 V with respect to other points in the circuit. All of the ground points in a circuit are electrically the same and therefore common points.

19

20

Resistors

• Definition: – A component that is specifically designed to

have a certain amount of resistance is called resistor.

Color bands

Resistance material(carbon composition)

Insulation coating

Leads

21

Resistors

• Fixed resistors: provides a specific constant value of resistance– Carbon-composition – Chip resistor– Film resistor– Wirewound resistor

• Variable resistor: values can be changed easily with manual or an automatic adjustment– Potentiometer– Rheostat

Carbon-composition

• Mixture of finely ground carbon, insulating filler and resin binder.

• The ratio of carbon to insulating filler sets the resistance value.

• Commonly use fixed resistor.

22

23

Chip Resistors

• SMT (Surface Mount Technology) component

• Very small in size ,suitable for compact assemblies.

24

Film Resistors

• The resistive material could be carbon film or metal film.

• The desired resistance value is obtained by removing part of resistive material in a spiral pattern along the rod

25

• Another type of film type resistor

26

Wirewound Resistors

• Constructed with resistive wire wound around an insulating rod and then sealed.

• Used in application that require higher power rating resistance

27

• Several types of wirewound resistors

Potentiometer

• used to divide voltage

• Has 3 terminal; terminal 1&2 have a fixed resistance between them, which is the total resistance. Terminal 3 is connected to moving contact.

28

Rheostat

• Used to control current

• Potentiometer can be used as a rheostat by connecting terminal 3 to either terminal 1 or terminal 2

Chapter 2: D

irect Current (D

C)

29

30

Resistors

• Resistor color code

31

32

Ohm’s Law

• Ohm’s law states that current is directly proportional to voltage and inversely proportional to resistance. The formula given are:

V = IR where: I = current in amperes (A)

V = voltage in volts (V)

R = resistance in ohms (Ω)

• Voltage - is not affected by either current or resistance. It is either too low, normal, or too high. If it is too low, current will be low. If it is normal, current will be high if resistance is low, or current will be low if resistance is high. If voltage is too high, current will be high.

• Current - is affected by either voltage or resistance. If the voltage is high or the resistance is low, current will be high. If the voltage is low or the resistance is high, current will be low.

• Resistance - is not affected by either voltage or current. It is either too low, okay, or too high. If resistance is too low, current will be high at any voltage. If resistance is too high, current will be low if the voltage is fix.

33

34

The linear relationship of current and voltage.

• The relationship can be portrayed by the graph below:

35

Example 1

Show that if the voltage in the circuit is increased to three times its present value, the current will triple in value.

R 10V 4.7kΩ

36

Example 2

Assume that you are measuring the current in a circuit that is operating with 25V. The ammeter reads 50mA. Later, you notice that the current has dropped to 40mA. Assuming that the resistance did not change, you must conclude that the voltage source has changed. How much has the voltage changed, and what is its new value?

37

The inverse relationship of current and resistance.

• As you have seen, current varies inversely with resistance as expressed by Ohm’s law, I = V/R. When the resistance is reduced, the current goes up; when the resistance is increased, the current goes down. For example, if the source voltage is held constant and the resistance is halved, the current doubles in value; when the resistance is doubled, the current is reduced by half.

Chapter 2: D

irect Current (D

C) C

ircuit

38

The Inversely Proportional Relationship

• The inversely proportional can be explained by this graph

39

Example 3How many amperes of current are in the circuit of figure below?

R

100V 22Ω

If the resistance in above figure is changed to 47 Ω and the voltage to 50 V, what is the new value of current?

40

Example 4

• Calculate the current in figure below.

R

50V 1.0kΩ

41

Example 5

– How many milliamperes are in the circuit below?

R

30V 5.6kΩ

42

Example 6

– In the circuit below, how much voltage is needed to produce 5 A of current?

R V 100Ω

5 A

5 A

43

Example 7

– How much voltage will be measured across the resistor below?

V

V 5 mA

R 56Ω

44

Example 8

– Suppose that there is a current of 8 µA through a 10 Ω resistor. How much voltage is across the resistor?

45

Example 9

– In the circuit of figure below, how much resistance is needed to draw 3.08 A of current from battery?

12 V R

3.08 A

46

Example 10

– Suppose that the ammeter in figure below indicates 4.55 mA of current and the voltmeter across the supply reads 150V. What is the value of R?

47

Types of Circuit

• Series

• Parallel

• Series-parallel circuit

48

Series Circuit

• Resistor in series

• A series circuit provides only one path for current between two points so that the current is the same through each series resistor [1].

R41k

R31k

R21k

R11k

A B

49

• The equation of a series resistors circuit

• The current flow is the same through each element of the series circuit.

• The voltage across the source or power supply is equal to the sum of the voltage drops across the separate resistors in series.

NT IIIII ...321

NT VVVVV ....321

NT RRRRR ....321

50

Power Distribution

• Power distribution in series circuit – It is represented by this equation [2]:

– The power delivered by the supply can be determined using the equation [2]:

1 2 3E R R RP P P P

E sP EI

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• Voltage sources in series [2]:

Voltage Sources

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Example 11

• In figures below, find out the total resistance between point A and point B

RT

B

A

R45.6k

R31.2k

R2220

R120

53

B

A

R43.3k

R33.3k

R23.3kR1

3.3k

RT

B

A

R41k

R31k

R21k

R11k

RT

BA

R101k

R91k

R81k

R71k

R61k

R51k

RT

54

Voltage Divider

• The voltage across a resistor in a series circuit is equal to the value of that resistor times the total applied voltage divided by the total resistance of the series configuration [2].

TT

XR V

R

RV

X

55

Example 12a. Find the total resistance RT.

b. Calculate the resulting source current Is.c. Determine the voltage across each resistor.

R3

5

R2 1

R1

2

+E 20V

Is

2

5

1

56

Example 13

R3 2k

R2

3k

R1

1k

+E 36V

57

Example 13

a. Determine the total resistance RT.

b. Calculate the source current IS.

c. Determine the voltage across each resistor.

d. Find the power supplied by the battery.

e. Determine the power dissipated by each resistor.

f. Comment on whether the total power supplied equals the total power dissipated.

58

Parallel Circuit

• The parallel resistors circuit

+ V110V

R51k

R41k

R31k

R21k

R11k

59

• The equation of a parallel resistors circuit

• Voltage across each resistor is the same as the voltage across the parallel combination.

• The current flowing through the parallel combination is the sum of the current in the separate branches.

1 2 3

1 1 1 1 1...

eq NR R R R R

VT VVVVV ....321

NT IIIII .....321

Power in Parallel Circuits

• Total power in a parallel circuit is found by adding up the powers of all the individual resistors, the same as for the series circuits.

BE

NG

1113: C

HA

PT

ER

2 WE

EK

5

60

NT PPPPP ....321

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Exercise 14

• What is the total resistance between point A and point B

R51k

R61k

R71k

R81k

R91k

R101k

A B

Current Divider

• A parallel circuit acts as a current divider because the current entering the junction of parallel branches “divides” up into several individual branch currents.

62

TX

TX I

R

RI

Series-Parallel

63

KVL

• KVL is the abbreviation of Kirchoff’s Voltage Law

• The sum of the voltage drops around a closed loop is equal to the sum of the voltage sources of that loop

• Total voltage = 0

64

65

R3

5

R2 1

R1

2

+E 20V

321 RRR VVVE

KCL

• Kirchoff’s Current Law

• The current arriving at any junction point in a circuit is equal to the current leaving that junction

• Current in = current out

66

BE

NG

1113: C

HA

PT

ER

2 WE

EK

5

67

+ V110V

R51k

R41k

R31k

R21k

R11k

54321 RRRRRT IIIIII

68

The Branch Current Method

• The branch current method is a circuit analysis method using KVL and KCL to find the current in each branch of a circuit by generating simultaneous equation.

69

Example: The Branch Current Method

20V 8V

5Ω 2Ω

10ΩI1 I2

I3

a

b

+

-

++ - -

70

• Currents I1, I2 and I3 are assigned to the branches a as shown.

• Applying KCL at node a

• Appling KVL at both loops resulting;

1 2 3I I I

08102

010520

32

31

II

II

71

• The following equations can be written

• Solving all three equations will give us

1

2

3

2

1

1

I A

I A

I A

8102

20155

32

32

II

II

The Mesh Current Method

• In the loop current method, you will work with loop current instead of branch current.

BE

NG

1113: C

HA

PT

ER

2 WE

EK

5

72

73

The Mesh Current Method

20V 8V

5Ω 2Ω

10ΩI1 I2

a

b

74

• Appling KVL at both loops resulting;

• Solving both equations will give us;

08102

010520

122

211

III

III

AI

AI

1

2

2

1

The Node Voltage Method

• Another method of analysis is node voltage method.

• It is based on finding the voltages at each node in the circuit using KCL.

BE

NG

1113: C

HA

PT

ER

2 WE

EK

5

75

The Node Voltage Method

76

20V 8V

5Ω 2Ω

10ΩI1 I2

I3

a

b

+

-

++ - -

77

• Currents I1, I2 and I3 are assigned to the branches a as shown.

• Applying KCL at node a

• Express the currents in terms of circuit voltages using Ohms’s law

1 2 3I I I

10

2

85

20

3

2

1

VaI

VaI

VaI

• Substituting these terms into the current equation yields;

BE

NG

1113: C

HA

PT

ER

2 WE

EK

5

78

VVa

VaVaVa

10102

8

5

20

79

Superposition

• A linear network which contains two or more independent sources can be analyzed to obtain the various voltages and branch currents by allowing the sources to act one at a time, then superposing the results.

• This principle applies because the linear relationship between current and voltage.

• Voltage sources are replaced by short circuits; current sources are replaced by open circuits.

• Superposition can be applied directly to the computation of power.

80

R423ohmR3

27ohm

R247ohm

R14ohm

+

-

Vs1200V

Is120A

81

o.c R423ohmR3

27ohm

R247ohm

R14ohm

+

-

Vs1200V

82

• By applying superposition principle, with the 200V acting alone, the 20A current source is replaced by an open circuit.

AI

AI

R

T

T

65.1

31.3

5.60

23

83

s.c

Is120A

R423ohmR3

27ohm

R247ohm

R14ohm

84

• When the 20-A source acts alone, the 200V source is replaced by a short circuit.

• The total current in the 23Ω resistor is

23 23 23 11.23old newI I I A

AI

RT

58.92023

02.11

02.11

23

85

Source Transformation

b

a

2A 5ohm

5ohm

+

-10V

a

b

Thevenin’s equivalent Norton’s equivalent

86

Example: The Node Voltage Method

VA

RA

RB

RC RE

VBRD

4 1 2 5

3

87

• Assign all currents are going out from node 1 &2.

• From node 1;

• From node 2;

1 1 1 2 0A

A B C

V V V V V

R R R

2 1 2 2 0B

C D E

V V V V V

R R R

88

Example: Mesh Analysis

25V

2Ω

5Ω

10Ω 2Ω

50V4ΩI1 I2 I3

89

• The matrix that we’ll get

• Solving

05024

0410525

02552

323

32212

211

III

IIIII

III

AI

AI

AI

45.10

17.3

31.1

3

2

1

-ve sign in I1 represents that the assigned current direction opposed the actual current direction.

90

• Solve the same circuit using node voltage method

25V

2Ω

5Ω

10Ω 2Ω

50V4ΩI1 I2 I3

1 2

3

91

Solution

• at node 1

• At node 2

1 1 1 2250

2 5 10

V V V V

2 1 2 2 500

10 4 2

V V V V

92

• From these

• The currents can be determined

11

1 22

23

1.312

3.1710

5010.45

2

VI A

V VI A

VI A

VV

VV

1.29

61.2

2

1

93

Test 1 Week 7

• Scientific notation and engineering notation, Resistor color code

• Ohm’s Law, KVL and KCL in Mesh or Node Analysis

• Basic Circuit measurement and measurement equipment

94

References

[1] Thomas L.Floyd; Principle of Electric Circuits 8th Ed; Pearson Education; 2007

[2] Robert L.Boylestad; Introductory Circuit Analysis11th Ed; Pearson Education; 2007