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BEKG1113
PRINCIPLE 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 seriesparallel 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
6
5
V (Volts)
• VI characteristics for
an ideal voltage
source is shown by
the graph [1]
4
3
V
2
1
0
1
2
3
4
I (Ampere)
5
6
7
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
10
Chapter 2: Direct Current (DC)
Circuit
Chapter 2: Direct Current (DC)
• 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
13
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.
+
SOURCE
-
LOAD
14
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.
15
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.
16
17
Current Direction
18
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.
19
• 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.
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
22
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.
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
28
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.
29
Rheostat
Chapter 2: Direct Current (DC)
• Used to control current
• Potentiometer can be used as a rheostat
by connecting terminal 3 to either terminal
1 or terminal 2
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)
33
• 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.
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.
10V
R
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.
38
The Inversely Proportional
Relationship
Chapter 2: Direct Current (DC) Circuit
• The inversely proportional can be
explained by this graph
39
Example 3
How many amperes of current are in the circuit
of figure below?
100V
R
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.
50V
R
1.0kΩ
41
Example 5
– How many milliamperes are in the circuit
below?
30V
R
5.6kΩ
42
Example 6
– In the circuit below, how much voltage is
needed to produce 5 A of current?
5A
V
5A
R
100Ω
43
Example 7
– How much voltage will be measured across
the resistor below?
V
5 mA
R
56Ω
V
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
3.08 A
R
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
R1
1k
R2
1k
R3
1k
R4
1k
B
• A series circuit provides only one path for
current between two points so that the
current is the same through each series
resistor [1].
49
• The equation of a series resistors circuit
RT  R1  R2  R3  ....  RN
• The current flow is the same through each
element of the series circuit.
IT  I1  I 2  I 3  ...  I N
• The voltage across the source or power supply is
equal to the sum of the voltage drops across the
separate resistors in series.
VT  V1  V2  V3  ....  VN
50
Power Distribution
• Power distribution in series circuit
– It is represented by this equation [2]:
PE  PR1  PR2  PR3
– The power delivered by the supply can be
determined using the equation [2]:
PE  EI s
51
Voltage Sources
• Voltage sources in series [2]:
52
Example 11
• In figures below, find out the total resistance
between point A and point B
A
R1
20
R2
220
R3
1.2k
RT
B
R4
5.6k
A
A
R2
3.3k
R1
3.3k
R3
3.3k
RT
R1
1k
R2
1k
RT
R3
1k
R4
3.3k
B
RT
B
B
A
R5
1k
R6
1k
R7
1k
R8
1k
R9
1k
R10
1k
R4
1k
53
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].
VR X
RX

VT
RT
55
Example 12
a. Find the total resistance RT.
b. Calculate the resulting source current Is.
c. Determine the voltage across each resistor.
2
2
Is
E
R1
+
R2
20V
5
5
R3
1
1
56
Example 13
E
+
36V
1k
3k
R1
R2
R3
2k
57
Example 13
a.
b.
c.
d.
e.
Determine the total resistance RT.
Calculate the source current IS.
Determine the voltage across each resistor.
Find the power supplied by the battery.
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
+ V1
10V
R1
1k
R2
1k
R3
1k
R4
1k
R5
1k
59
• The equation of a parallel resistors circuit
1
1 1
1
1
    ... 
Req R1 R2 R3
RN
• Voltage across each resistor is the same as the
voltage across the parallel combination.
VT  V1  V2  V3  ....  VV
• The current flowing through the parallel
combination is the sum of the current in the
separate branches.
IT  I1  I 2  I 3  .....  I N
60
Power in Parallel Circuits
BENG 1113: CHAPTER 2 WEEK 5
• 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.
PT  P1  P2  P3  ....  PN
61
Exercise 14
• What is the total resistance between point
A and point B
B
A
R5
1k
R6
1k
R7
1k
R8
1k
R9
1k
R10
1k
62
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.
RT
IX 
IT
RX
63
Series-Parallel
64
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
65
E  VR1  VR2  VR3
2
R1
E
+
R2
20V
5
R3
1
66
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
67
I T  I R1  I R2  I R3  I R4  I R5
BENG 1113: CHAPTER 2 WEEK 5
+ V1
10V
R1
1k
R2
1k
R3
1k
R4
1k
R5
1k
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
+ 5Ω -
+ 2Ω -
a
I3
+
10Ω
I1
20V
-
b
I2
8V
70
• Currents I1, I2 and I3 are assigned to the
branches a as shown.
• Applying KCL at node a
I1  I 2  I 3
• Appling KVL at both loops resulting;
 20  5I1  10 I 3  0
2 I 2  10 I 3  8  0
71
• The following equations can be written
5I 2  15I 3  20
2 I 2  10 I 3  8
• Solving all three equations will give us
I1  2 A
I 2  1A
I 3  1A
72
The Mesh Current Method
BENG 1113: CHAPTER 2 WEEK 5
• In the loop current method, you will work
with loop current instead of branch current.
73
The Mesh Current Method
5Ω
20V
10Ω
I1

b

2Ω
a
I2
8V
74
• Appling KVL at both loops resulting;
 20  5I1  10I1  I 2   0
2 I 2  10I 2  I1   8  0
• Solving both equations will give us;
I1  2 A
I 2  1A
75
The Node Voltage Method
BENG 1113: CHAPTER 2 WEEK 5
• Another method of analysis is node
voltage method.
• It is based on finding the voltages at each
node in the circuit using KCL.
76
The Node Voltage Method
+ 5Ω -
+ 2Ω -
a
I3
+
10Ω
I1
20V
-
b
I2
8V
77
• Currents I1, I2 and I3 are assigned to the
branches a as shown.
• Applying KCL at node a
I1  I 2  I 3
• Express the currents in terms of circuit
voltages using Ohms’s law
20  Va
I1 
5
Va  8
I2 
2
Va
I3 
10
78
• Substituting these terms into the current
equation yields;
BENG 1113: CHAPTER 2 WEEK 5
20  Va Va  8 Va


5
2
10
Va  10V
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
R1
4ohm
R2
47ohm
Is1
20A
R3
27ohm
+
Vs1
200V
-
R4
23ohm
81
R1
4ohm
R2
47ohm
o.c
R3
27ohm
+
Vs1
200V
-
R4
23ohm
82
• By applying superposition principle, with
the 200V acting alone, the 20A current
source is replaced by an open circuit.
RT  60.5
I T  3.31A
I 23  1.65 A
83
R1
4ohm
R2
47ohm
Is1
20A
R3
27ohm
s.c
R4
23ohm
84
• When the 20-A source acts alone, the
200V source is replaced by a short circuit.
RT  11.02
I 23
11.02

 20  9.58 A
23
• The total current in the 23Ω resistor is
I 23  I 23old  I 23new  11.23 A
85
Source Transformation
Norton’s equivalent
Thevenin’s equivalent
a
5ohm
a

+
10V
2A
5ohm
b
b
86
Example: The Node Voltage
Method
4
VA
RA
1
RC
RB
2
RD
3
5
RE
VB
87
• Assign all currents are going out from
node 1 &2.
• From node 1;
V1  VA V1 V1  V2


0
RA
RB
RC
• From node 2;
V2  V1 V2 V2  VB


0
RC
RD
RE
88
Example: Mesh Analysis
10Ω
2Ω
5Ω
2Ω
I1
I2
25V
4Ω
I3
50V
89
• The matrix that we’ll get
2 I1  5I1  I 2   25  0
 25  5I 2  I1   10 I 2  4I 2  I 3   0
4I 3  I 2   2 I 3  50  0
• Solving
I1  1.31A
I 2  3.17 A
I 3  10.45 A
-ve sign in I1 represents that
the assigned current direction opposed
the actual current direction.
90
• Solve the same circuit using node voltage
method
1
10Ω
2
2Ω
5Ω
2Ω
I1
I2
25V
3
4Ω
I3
50V
91
Solution
• at node 1
V1 V1  25 V1  V2


0
2
5
10
• At node 2
V2  V1 V2 V2  50
 
0
10
4
2
92
• From these
V1  2.61V
V2  29.1V
• The currents
can be
determined
V1
I1 
 1.31A
2
V1  V2
I2 
 3.17 A
10
V2  50
I3 
 10.45 A
2
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