This document provides an overview of fundamental electrical engineering concepts including: - Independent and dependent voltage/current sources and ideal sources that maintain constant voltage/current. - Kirchhoff's laws for circuits - KVL states the net voltage around any closed loop is zero and KCL states the algebraic sum of currents at any node is zero. - Series and parallel resistor circuits and how to calculate equivalent resistance and current/voltage in each component. - Ohm's law relating voltage, current, and resistance and the power formula. - Examples are provided to demonstrate applying concepts like nodal analysis, mesh analysis, and voltage divider rule to solve for values in circuits.

1. CHAPTER 2
FUNDAMENTALS OF
ELECTRIC CIRCUITS
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

2. INDEPENDENT SOURCES
• The voltage/current sources that have the capability of
generating a prescribed voltage or current independent of
any other element within the circuit.
• These sources may output a constant voltage/current, or
they may output voltage/current that varies with time.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
PRINCIPAL ELEMENTS OF ELECTRICAL
CIRCUITS

3. 1) Ideal Voltage Sources
An ideal voltage source is a two-terminal element that
maintains the same voltage across its terminals regardless
of the current flowing through it.
• Vt = constant, no matter what the load current is.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Vt
IL
Vo
L
t
+
-
Vo

4. 2) Ideal Current Sources
An ideal current source is a two-terminal element that
maintains the same current regardless of the voltage
across its terminals.
• IS = constant, no matter what the load voltage is.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
IO
VO
IS

5. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
DEPENDENT (CONTROLLED) SOURCES
• Dependent sources are whose output (current or voltage)
is a function of some other voltage or current in a circuit.
• The symbols typically used to represent dependent
sources are in the shape of a diamond.

6. BRANCH, NODE, LOOP, MESH
• Branch : any portion of a circuit with two terminals
connected to it.
• A branch may consist of one or more circuit elements.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

7. • Node : the point of connection between two or more
branches.
• A node usually indicated by a dot in a circuit.
• Loop : any closed path through the circuit in which no
node is encountered more than once.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

8. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
• Mesh : a loop that does not contain other loops.

9. Electric Current
• Electric current is defined as the time rate of change of
charge passing through a predetermined area.
𝑖 =
∆𝑞
∆𝑡
𝑜𝑟 𝑖 =
𝑑𝑞
𝑑𝑡
• The units of current are called Amperes, where
1 Ampere (A) = 1 Coulomb/second (C/s).
• In order for current to flow, there must exist a closed
circuit.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
CURRENT AND KIRCHHOFF’S CURRENT
LAW

10. • In the circuit of this figure, the current i flowing from the
battery to the light bulb is equal to the current flowing from
the light bulb to the battery.
 no current (and therefore no charge) is “lost” around
the closed circuit. This principle is known as
Kirchhoff’s current law (KCL).
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

11. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Kirchhoff’s Current Law (KCL)
• One of the fundamental laws of circuit analysis.
• Establish in 1874 by G.R. Kirchhoff.
• “The sum of the currents at a node must equal zero.”
𝑛=1
𝑁
𝑖 𝑛 = 0 𝑜𝑟
(𝐸𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠) = (𝐿𝑒𝑎𝑣𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠)

12. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
• Example of Kirchhoff’s current law:
At node 1:
−𝑖 + 𝑖1 + 𝑖2 + 𝑖3 = 0
𝑖 = 𝑖1 + 𝑖2 + 𝑖3
• In this illustration, currents
entering a node are defined as
negative and currents leaving
the node as positive.

13. Voltage
• The total work per unit charge associated with the motion
of charge between two points.
• The units of voltage are called Volts, where
1 Volts (V) = 1 Joule (J)/Coulomb (C).
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
VOLTAGE AND KIRCHHOFF’S VOLTAGE
LAW

14. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Kirchhoff’s Voltage Law (KVL)
• The second fundamental laws of circuit analysis introduced
by G.R. Kirchhoff.
• The principle underlying KVL is that no energy is lost or
created in an electric circuit.
• In circuit terms, the sum of all voltages associated with
source must equal the sum of the load voltages.
• “The net voltage around a closed circuit is zero.”
𝑛=1
𝑁
𝑣 𝑛 = 0

15. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
• Example of Kirchhoff’s voltage law:
𝑣1 = 𝑣2 where
𝑣2 = 𝑣 𝑎𝑏 = 𝑣 𝑎 − 𝑣 𝑏
• In general, elements that provide
energy are referred as sources
and elements that dissipate energy
as loads.

16. Power
• The electric power generated by an active element, or
that dissipated or stored by a passive element, is equal to
the product of the voltage across the element and the
current flowing through it.
• The units of power are called Watts (Joules/second).
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
ELECTRIC POWER AND SIGN
CONVENTION
𝑷 = 𝑽𝑰

17. Passive Sign Convention
• State that if current flows from a higher to a lower voltage
(plus to minus), the power is dissipated and will be a positive
quantity.
• Example:
• Power generated (supplied) always equals power dissipated.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Power dissipated = vi Power dissipated = - vi
Power generated = vi

18. • An ideal resistor is a device that exhibits linear resistance
properties according to Ohm’s law,
which states that the
voltage across a
resistance is directly
proportional to the current
flowing through it.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
RESISTANCE AND OHM’S LAW
𝑽 = 𝑰𝑹
FUNDAMENTAL OF ELECTRICAL ENGINEERING, First Edition, by Giorgio Rizzoni, © 2009 McGraw-Hill Companies, Inc.

19. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
• The value of the resistance R is measured in units of
ohm’s (), where
1  = 1 V/A
• For a resistor R, the power dissipated can be expressed
by
𝑷 = 𝑽𝑰 = 𝑰 𝟐 𝑹 =
𝑽 𝟐
𝑹

20. Open and Short Circuits
• Open circuit : a circuit element whose resistance
approaches infinity.
• Short circuit : a circuit element with resistance
approaching zero.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

21. Series Circuit
• Two or more circuit elements are said to be in series if the
current from one element exclusively flows into the next
elements.
• All series elements have the same current.
Series Resistors
• Equivalent series resistance:
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
𝑹 𝑬𝑸 =
𝒏=𝟏
𝑵
𝑹 𝒏 = 𝑹 𝟏 + 𝑹 𝟐 + ⋯ + 𝑹 𝑵

22. • Example 2.1:
For the circuit shown,
a) Find the equivalent resistance seen by the source.
b) Find the current I .
c) Calculate the voltage drop in each resistor.
d) Calculate the power dissipated by each resistor.
e) Find the power output of the source.
Given: V = 24 V, R1 = 1 , R2 = 3 , and R3 = 4 .
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

23. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
b) 𝐼 =
𝑉
𝑅 𝐸𝑄
=
24 𝑉
8 Ω
= 3 𝐴
c) 𝑉1 = 𝐼𝑅1 = 3 𝐴 1 Ω = 3 𝑉
𝑉2 = 𝐼𝑅2 = 3 𝐴 3 Ω = 9 𝑉
𝑉3 = 𝐼𝑅3 = 3 𝐴 4 Ω = 12 𝑉
Solution:
a) 𝑅 𝐸𝑄 = 𝑅1 + 𝑅2 + 𝑅3
= 1 + 3 + 4
= 8 
REQ
Note: 𝑉1 + 𝑉2 + 𝑉3 = 3 + 9 + 12 = 24 𝑉
The total voltage drop is equal to the voltage output of the source.

24. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Note: 𝑃1 + 𝑃2 + 𝑃3 = 9 + 27 + 36 = 72 𝑉
The total power dissipated by the resistors is the same as the power
output by the source.
e) 𝑃 = 𝐼𝑉 = 3 𝐴 24 𝑉 = 72 𝑊
d) 𝑃1 = 𝐼2 𝑅1
= 3 𝐴 2 1 Ω
= 9 𝑊
𝑃2 = 𝐼2
𝑅2
= 3 𝐴 2
3 Ω
= 27 𝑊
𝑃3= 𝐼2 𝑅3
= 3 𝐴 2 4 Ω
= 36 𝑊

25. Parallel Circuit
• Two or more circuit elements are said to be in parallel if
the elements share the same terminals.
• All parallel elements have the same voltage.
Parallel Resistors
• Equivalent parallel resistance:
or
where
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
𝟏
𝑹 𝑬𝑸
=
𝟏
𝑹 𝟏
+
𝟏
𝑹 𝟐
+ ⋯ +
𝟏
𝑹 𝑵
𝑹 𝑬𝑸 =
𝟏
𝟏
𝑹 𝟏
+ 𝟏
𝑹 𝟐
+ ⋯ + 𝟏
𝑹 𝑵
𝑉1 = 𝑉2 = 𝑉3 = 𝑉4

26. Various Parallel Resistors Networks
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

27. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
• Example 2.2:
For the circuit shown,
a) Find the equivalent resistance seen by the source.
b) Find the total current I .
c) Calculate the currents in each resistor.
d) Calculate the power dissipated by each resistor.
e) Find the power output of the source.
Given: V = 24 V, R1 = 1 , R2 = 3 , and R3 = 4 .

28. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Solution:
a)
1
𝑅 𝐸𝑄
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
=
1
1
+
1
3
+
1
4
= 1.583 Ω
∴ 𝑅 𝐸𝑄 =
1
1.583
= 0.632 Ω
b) 𝐼 =
𝑉
𝑅 𝐸𝑄
=
24 𝑉
0.632 Ω
= 37.975 𝐴 ≈ 38 𝐴
REQ
I

29. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
c) 𝐼1 =
𝑉
𝑅1
=
24 𝑉
1 Ω
= 24 𝐴
REQ
I
𝐼2 =
𝑉
𝑅2
=
24 𝑉
3 Ω
= 8 𝐴
𝐼3 =
𝑉
𝑅3
=
24 𝑉
4 Ω
= 6 𝐴
Note: 𝐼1 + 𝐼2 + 𝐼3 = 24 + 8 + 6 = 38 𝐴
The sum of the individual current is equal to the current output of the
source.

30. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Note: 𝑃1 + 𝑃2 + 𝑃3 = 576 + 192 + 144 = 912 𝑉
The total power dissipated by the resistors is the same as the power
output by the source.
e) 𝑃 = 𝐼𝑉 = 38 𝐴 24 𝑉 = 912 𝑊
d) 𝑃1 = 𝐼1
2 𝑅1
= 24 𝐴 2 1 Ω
= 576 𝑊
𝑃2 = 𝐼2
2
𝑅2
= 8 𝐴 2
3 Ω
= 192 𝑊
𝑃3= 𝐼3
2 𝑅3
= 6 𝐴 2 4 Ω
= 144 𝑊

31. Series and Parallel Resistor Combinations
• Example 2.3:
The Wheatstone Bridge consists of two series circuits that are connected
in parallel with each other.
1) Find the value of the voltage
Vab = Vad - Vbd in terms of the four
resistances and the source voltage Vs.
2) If R1 = R2 = R3 = 1 k, Vs = 12 V,
and Vab = 12 mV, what is the value
of Rx.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

32. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Solution:
1) 𝑉𝑎𝑑 =
𝑅2
𝑅1+𝑅2
𝑉𝑠 and 𝑉𝑏𝑑 =
𝑅 𝑥
𝑅3+𝑅 𝑥
𝑉𝑠
Thus,
𝑉𝑎𝑏 = 𝑉𝑎𝑑 − 𝑉𝑏𝑑 =
𝑅2
𝑅1+𝑅2
−
𝑅 𝑥
𝑅3+𝑅 𝑥
𝑉𝑠
2) 0.012 =
1000
1000+1000
−
𝑅 𝑥
1000+𝑅 𝑥
12
𝑅 𝑥 = 996 Ω
R1
R2
R3
RX
Vs
a Vab b
c
d
R1
R2
R3
RX

33. Voltage Divider Rule (VDR)
• VDR is useful in determining the voltage drop across a
resistance within a series circuit.
where VX = the voltage drop across the measured resistor,
RX = the resistance value of the measured resistor,
REQ = the circuit total resistance,
VS = the circuit applied voltage
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
𝑽 𝑿 =
𝑹 𝑿
𝑹 𝑬𝑸
𝑽 𝑺
+ V1 -
+ V3 -
+
V2
-S

34. • Example 2.4:
Determine the voltage across the R2 and the R3.
Solution:
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
𝑉2 =
𝑅2
𝑅 𝐸𝑄
𝑉𝑆 =
20
10 + 20 + 30
60 = 20 V
𝑉3 =
𝑅3
𝑅 𝐸𝑄
𝑉𝑆 =
30
10 + 20 + 30
60 = 30 V

35. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Current Divider Rule (CDR)
• CDR is useful in determining the current flow through one
branch of a parallel circuit.
where IX = the current flow through any parallel branches,
RX = the resistance of the branch through which the
current is to be determined,
REQ = the total resistance of the parallel branch,
IS = the circuit applied current
𝑰 𝑿 =
𝑹 𝑬𝑸
𝑹 𝑿
𝑰 𝑺

36. • Example 2.5:
Find each of the branch currents in the figure shown below.
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Solution:
𝑅1 𝑅2 𝑅3
1
𝑅 𝐸𝑄
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
=
1
3 𝑘Ω
+
1
8 𝑘Ω
+
1
24 𝑘Ω
=
1
2 𝑘Ω
𝑅 𝐸𝑄 = 2 𝑘Ω

37. Thus,
𝐼 𝑅1 =
𝑅 𝐸𝑄
𝑅1
∙ 𝐼 =
2 𝑘Ω
3 𝑘Ω
12 𝑚𝐴 = 8 𝑚𝐴
𝐼 𝑅2 =
𝑅 𝐸𝑄
𝑅2
∙ 𝐼 =
2 𝑘Ω
8 𝑘Ω
12 𝑚𝐴 = 3 𝑚𝐴
𝐼 𝑅3 =
𝑅 𝐸𝑄
𝑅3
∙ 𝐼 =
2 𝑘Ω
24 𝑘Ω
12 𝑚𝐴 = 1 𝑚𝐴
𝐼 𝑅1 + 𝐼 𝑅2 + 𝐼 𝑅3 = 8 + 3 + 1 = 12 𝑚𝐴 ( KCL  )
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING

38. • For the particular case of two parallel resistors,
and, the current passing through
R1 and R2 are
𝐼1 =
𝑅 𝐸𝑄
𝑅1
∙ 𝐼 =
𝑅1 𝑅2
𝑅1+𝑅2
𝑅1
∙ 𝐼 
𝐼2 =
𝑅 𝐸𝑄
𝑅2
∙ 𝐼 =
𝑅1 𝑅2
𝑅1+𝑅2
𝑅2
∙ 𝐼 
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
𝑹 𝑬𝑸 = 𝑹 𝟏 𝑹 𝟐 =
𝑹 𝟏 𝑹 𝟐
𝑹 𝟏 + 𝑹 𝟐
I
I1 I2
𝑰 𝟏 =
𝑹 𝟐
𝑹 𝟏 + 𝑹 𝟐
∙ 𝑰
𝑰 𝟐 =
𝑹 𝟏
𝑹 𝟏 + 𝑹 𝟐
∙ 𝑰
Note: It only works for two parallel resistors.

39. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
MEASURING DEVICES
Ohmmeter
• The ohmmeter is a device that, when
connected across a circuit element,
can measure the resistance of the
element.
• The resistance of an element can be
measured only when the element is
disconnected from any other circuit.

40. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Ammeter
• The ammeter is a device that, when connected in series
with a circuit element, can measure the current flowing
through the element.
1. The ammeter must be placed in series with the
element whose current is to be measured (e.g.,
resistor R2).
2. The ammeter should not restrict the flow of current
(i.e., cause a voltage drop), or else it will not be
measuring the true current flowing in the circuit. An
ideal ammeter has zero internal resistance.

41. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Measurement of current

42. EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING
Voltmeter
• The voltmeter is a device that can measure the voltage
across a circuit element.
1. The voltmeter must be placed in parallel with the
element whose voltage it is measuring.
2. The voltmeter should draw no current away from the
element whose voltage it is measuring, or else it will not
be measuring the true voltage across that element.
Thus, an ideal voltmeter has infinite internal resistance.

43. Measurement of voltage
EEE 1012 INTRODUCTION TO ELECTRICAL ENGINEERING