ECEG-1351 Fundamentals of Electricity and
Digital Electronics
Addis Ababa Institute of Technology (AAIT) School of Electrical and
Computer Engineering
Learning Outcomes

At the end of this lecture, students should get
familiarized with:
Basic Circuit Concepts.
 Charge, Current, Voltage, Power, Sources.
 Basic Circuit Elements and Circuit Laws.
 Identify Nodes, Branches, Loops and Mesh.
 Mesh and Node analysis (KVL and KCL).

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Introduction: Electric Circuit
An electric circuit is defined as “An
interconnection of electrical elements linked
together in a closed path so that an electric
current may flow continuously”.
 “The path of flow for charge carriers”



A circuit consists of source (energy) and/or drain
(device) elements.
Electric circuits are broadly classified as Direct
Current (D.C.) circuits and Alternating Current (A.C.)
circuits.
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Introduction: Charge & Current
Charge is the physical property of matter that
causes it to experience a force when placed in
an electromagnetic field, measured in coulombs
(C).
 Electric current is the time rate of change of
charge, measured in amperes (A).

 Mathematically,
the relationship
current i, charge q, and time t is.
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Overview of Solid-State Materials

Charge, q, is measured in Coulomb (C)
1C = 6.24 x 1018 electrons.
-19C.
 1 electron = -1.6 x 10


Conservation of Charges


Charges can neither be created nor destroyed, only
transferred.
Electric Current
dq
i
dt
t
q   i(  ) d
to
DC: direct current, remains constant with time.
 Time varying current, e.g.: alternating current (AC) varies
sinusoidally with time.

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Types of Current

A direct current (dc) is a current that remains
constant with time.

An alternating current (ac) is a current that
varies sinusoidal with time.
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Example 1

How much charge is represented by 4,600
electrons?
Solution:
Each electron has −1.602 × 10−19 C. Hence 4,600
electrons will have;
−1.602 × 10−19 C/electron × 4,600 electrons
= −7.369 × 10−16 C
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Example 2

The total charge entering a terminal is given by
q = 5t sin 4πt m C. Calculate the current at t =
0.5 s.
Solution:
i =dq/dt =d/dt(5t sin 4πt) mC/s = (5 sin 4πt + 20πt
cos 4πt) mA
At t = 0.5,
i = 5 sin 2π + 10π cos 2π = 0 + 10π = 31.42 mA
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Example 3

Determine the total charge entering a terminal
between t = 1s and t = 2s, If the current passing
through a terminal is i = (3t2 − t) A.
Solution:
Answer: Simplify integral of current from 1 to 2
seconds!
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Voltage
This is an emf which is also known as voltage
or potential difference.
 The voltage vab between two points a and b in
an electric circuit is the energy (or work) needed
to move a unit charge from point a to b;
mathematically,
vab = dw/dq

Voltage (or potential difference) is the
energy required to move a unit charge
through an element, measured in volts (V).
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Vab= -Vba
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Power and Energy

Power, p, is the time rate of expending or
absorbing energy, measured in watts (W).
p  vi
p = dw/dt,

Since power depends on the value of current and
voltage, voltage polarity and current direction
play a big role in determining the positive value of
t
t
the power.
w   p dt   vi dt
to

to
Energy, w, is the capacity to do work, measured in
joules (J).
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Power

Power absorbed = −Power supplied.
Absorbed Power = Supplied Power

The law of conservation of energy must be
obeyed in any electric circuit. For this reason, the
algebraic sum of power in a circuit, at any instant
of time, must be zero:

i.e. the total power supplied to the circuit = total
power absorbed.
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Energy

The energy absorbed or supplied by an element
from time t0 to time t is given by;
t
t
to
to
w   p dt   vi dt
Example 4: An energy source forces a constant
current of 2A for 10s to flow through a light bulb. If
2.3 kJ is given off in the form of light and heat
energy, calculate the voltage drop across the bulb.
Answer: The total charge is:
The voltage drop is:

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Circuit Elements
There are two types of elements found in electric
circuits: passive elements and active elements.
 An active element is capable of generating energy
while a passive element is not.
 Passive elements do not require power supply
initially for its operation or the device which electrical
characteristics does not depend on the power supply.
Examples are; resistors, capacitors, and inductors.
 Active elements do require power supply initially for
its operation or the device which electrical
characteristics depend on the power supply. Examples
are; generators, batteries, and operational amplifiers
BJT, FET.

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Sources
Sources are active circuit elements which are
either voltage or current generator capable of
supplying energy to a circuit.
 There are two kinds/categories of sources:



Independent Sources and
Dependent Sources.
• Voltage sources
• Current sources
INDEPENDENT
SOURCES
DEPENDENT
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• Voltage-controlled voltage source (VCVS)
• Current-controlled voltage source (CCVS)
• Voltage-controlled current source (VCCS)
• Current-controlled current source (CCCS)
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Ideal Independent Sources

Referred to as ideal sources.

Voltage sources.

An active element, which provides a specified voltage independent
of the current through it and any other circuit variable.

For example, v(t) = 10 cos 100t or v(t) = 9
i(t)
+
v(t)
-
v(t)
Arbitrary
circuit
~
+
-
v(t) = V sin ωt
AC voltage
DC voltage
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~
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Ideal Independent Sources
Referred to as ideal sources.
 Current sources


An active element, which provides a specific current
independent of other circuit variables for example voltage
across the source.

For example, i(t) = 4 sin 100t or i(t) = -0.1
i(t)
D.C. CURRENT SOURCE
A.C. CURRENT SOURCE
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Ideal Dependent Sources

An active element in which the source quantity is controlled by another
voltage or current.
Types are :
Voltage-controlled voltage source (VCVS).
Current-controlled voltage source (CCVS).
Voltage-controlled current source (VCCS).
Current-controlled current source (CCCS).
Current
source
Voltage
source
Note: Dependent sources are useful for modeling
elements such as transistors, operational amplifiers
and integrated circuits.
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Ideal Sources Symbols
Independent
voltage source
Independent DC
voltage source
Dependent voltage
source
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Independent
current source
Dependent
current source
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Class Exercises
Find the current in a element if the charge flowing
through the element is given as;
q(t) = 3t3 + 6t2 +8t –4.
& If the current in an electrical device is given by;
i(t) = 2t + 4, With q(0) = 1.5 C
Solution:
dq d (3t 3  6t 2  8t  4)
i (t )  
dt
dt
t
t
0
0
q (t )   i (t ) dt  q (0)   (2t  4) dt  1.5
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i (t )  9t 2  12t  8
q (t )  t 2  4t  1.5
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R - Voltage Current Relations

Resistor: Conductor is used to carry current from one point to
another. For the conductor, at any instant time ‘t’ the current
i(t) passing through it and the voltage v(t) across it satisfy the
relation defined by the curve in the v-i plane.
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R - Voltage Current Relations

Resistor:
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Voltage Current Relations

Capacitor: Any two conducting surfaces separated by an
insulating medium form a capacitor.

A two terminal element will be called a CAPACITOR, if at
any instant time t, the charge in it q(t) and the voltage across it
v(t) satisfy a relation defined by the curve in the v-q plane.
Capacitance is the property of the capacitor.
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Voltage Current Relations

Capacitor:
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Voltage Current Relations

Inductor: A wire of certain length, when twisted into a coil
becomes a basic inductor.

A two terminal element will be called an INDUCTOR, if at
any instant of time, the flux linkage in it Ψ(t) and the current
passing though it i(t) satisfy a relation defined by the curve in iΨ plane. Inductance is the property of the inductor.
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Voltage Current Relations

Inductor:
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Circuit Basic Law’s: Ohms Law
A limiting device to set voltage and current levels.
 Linear resistor obeys Ohm’s law, V = IR.
V

I
I
R
V = - IR
V = IR
VR
I
VR
Short Circuit: R = 0
2
2
i
v
p  vi  i 2 R    v 2G
G R
Open Circuit: R = ꝏ
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R
R
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Circuit Law’s - Open & Short Circuit
Open Circuit
In open circuit condition, current i(t) through the resistor
will be zero irrespective of whatever be the voltage v(t)
across it. To satisfy this, value of R must be infinity. Thus,
in open circuit condition I (t) = 0, R = ꝏ, and G = 0.
Short Circuit
In short circuit condition, voltage v(t) across the resistor will
be zero irrespective of whatever be the current i(t) flowing
through it. To satisfy this, value of R must be zero. Thus, in
short circuit condition v(t) = 0, R = 0 and G = ꝏ.
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Branch, Node, Loop and Mesh
A Branch represents a single element such as a
voltage source or a resistor etc.
 A Node is the point of connection between two or
more branches. (A point where two or more circuit
elements join).
 A loop is any closed path in an electrical circuit.
 A Mesh is a loop that does not enclose any other
loop or loops inside.

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Example 4

Determine the number of branches, nodes and
meshes in the circuit given below.
R
30 V
DC
2A
R
R
R
R
Branches = 7, nodes = 3, meshes = 4
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Resistor in Series
When connected in series, resistors form a ‘string’.
 Resistors are connected to one node only.
 One path for current and same current through all points.
 Find single Equivalent resistor by:

i
One path
One path
a
R1= 3k R2= 7k R3= 15k
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b
i
Req= 25 k
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Resistor in Parallel
• Resistors are connected to both node a and node b.
• Many paths for current, same voltage.
• Single Equivalent resistor by:
i1
a
a
R1= 3k
a
R2= 6k
i2 i3
b
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b
Req= 2k
i1
b
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Exerxise 5
Find Req
a
2
Figure (a)
6
Req
3
1
b
2
a
6
Figure b
3
Req
1
b
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Series Parallel Equivalent Summary
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Voltage Divider Rule
 Calculate voltage for series circuit only;
i
R1= 3k
R2= 7k
v1
v2
a
vS= 10V
b
R1
v1 
vS
R1  R2
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R2
v2 
vS
R1  R2
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Current Divider Rule
 Calculate current for parallel circuit only,
iS = 5A
a
a
vS= 20V
i1
i2
R1= 6 k
R2= 12 k
b
b
R2
i1 
iS
R1  R2
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R1
i2 
iS
R1  R2
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Kirchhoff’s Current Law (KCL)
i2
i5
Node A
i1
i3

The algebraic sum of currents entering a node is 0.
N
 in  0
n 1

i4
Or
N
N
n 1
n 1
 in (entering )   in (leaving )
Assume one direction to be positive, in this case
- Currents entering a node is positive. Eg. i1, i2, i5
- Currents leaving a node is negative. Eg. i3, i4
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Exercise 6
 Find current io and v0 in the following circuit.
i
io
6A
2
io/4
8
V0
Answers
i0  4 A v0  8 V
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Kirchhoff’s Voltage Law (KVL)
v4
v3
v5
N
v2
v n  0
n 1
vn
v1
The algebraic sum of voltages around a loop is 0.
 Assume one direction in the loop to be +ve, in this case
the anti-clockwise direction.
Example : v1 + v2 + v3 + v4 + v5+…+vn = 0

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Exercise 7
 Find v1 and v2 in the following circuit.
4
v1
10 V
i
8V
v2
2
Answers
v1  12 V ,
v 2  6 V
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Class Exercise
 Find the equivalent resistance for the circuit shown below.
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Circuit Analysis Techniques
Nodal Analysis
 Nodal Analysis provides a general procedure for
analyzing circuits using node voltages as the
circuit variables.
 Choosing node voltages instead of element
voltages as circuit variables is convenient and
reduces the number of equations one must solve
simultaneously.
 In nodal analysis, we are interested in finding the
node voltages.
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Nodal Analysis Steps
1.
Find out all the nodes of the circuit and select one as reference node.
2.
Mark the reference node as
1 ,
2
and other independent nodes as
0
,…..etc.
3.
Assume node voltages as V 1 ,V 2 ,V 3 ,…..etc. in all independent nodes.
4.
Apply KCL and write the equations for all independent nodes. For
each node, for all the elements meeting at that node find the element
currents flowing away from the node. Add these currents and equate
to zero.
5.
Solve these equations for the node voltages.
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Example 8
 Consider the circuit shown below. Using node voltage
method, find the power consumed by the resistors and
the power supplied by the current sources.
Solution:
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Example 8, Con.
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Class Exercise
 Calculate the node voltages in the circuit shown below.
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Circuit Analysis Techniques
Mesh Analysis
Mesh Analysis provides another general procedure
for analyzing circuits, using mesh currents as the
circuit variables.
 Using mesh currents instead of element currents as
circuit variables is convenient and reduces the number
of equations that must be solved simultaneously.
Recall that a loop is a closed path with no node passed
more than once. A mesh is a loop that does not
contain any other loop within it.
 Mesh analysis applies KVL to find unknown currents.

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Mesh Analysis Steps
The following procedure can be followed to solve circuits using mesh
current method:
1.
Find the number of independent meshes and identify one set of
independent meshes.
2.
Assume mesh currents in these independent meshes. ( Once mesh
currents are assumed, element currents can be obtained in terms of
mesh currents )
3.
Apply KVL and write the equations for all the independent meshes
traveling along the mesh current direction.
4.
Solve these equations for mesh currents.
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Example 9
 Consider the circuit shown below. Using mesh current
method, find the power consumed by the resistors and
power supplied by the voltage sources.
Solution:
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Example 9, Cont.
Mesh current equations are
- 10 + 5 I 1 + 2 ( I 1 - I 2 ) = 0
10 I 2 + 50 + 2 ( I 2 - I 1 ) = 0
These equations can be rearranged as
7 I 1 - 2 I 2 = 10
and - 2 I 1 + 12 I 2 = - 50
Solving these equations, we get
I 1 = 0.25 A and I 2 = - 4.125 A
This means in the second loop, current of 4.125 A flows in anticlockwise
direction.
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Example 9, Cont.
I 1 = 0.25 A and I 2 = - 4.125 A
5Ω
0.25A
10V
2Ω
0.25A
10Ω
4.125A
4.375A
50V
F
i g
.
1
. 1
9
C
i r
c
u
i t
o
f
e
x
a
m
p
l e
1
. 1
w
i t h
e
=
( 0
Power consumed by 5Ω resistor
l e
m
e
n
t
c
u
r
r
e
n
t s
m
a
r
k
e
d
0
. 3
1
2
5
3
8
. 2
8
1
2
5
W
6
2
5
W
2
. 2
5
)
x
5
=
x
2
=
x
1
W
2
P
o
w
e
r
c
o
n
s
u
m
e
d
b
y
2
Ω
P
o
w
e
r
c
o
n
s
u
m
e
d
b
y
1
0
T
o
t a
l
p
o
w
n
s
u
P
o
w
e
r
s
u
p
p
l i e
d
b
y
1
0
V
b
a
t t e
r
y
=
1
0
x
0
. 2
5
P
o
w
e
r
s
u
p
p
l i e
d
b
y
5
0
V
b
a
t t e
r
y
=
5
0
x
4
. 1
2
T
o
t a
l
p
o
w
p
l i e
r
e
s
i s
e
s
t o
r
=
( 4
. 3
7
5
)
=
( 4
. 1
2
5
)
2
e
e
r
r
c
s
o
u
p
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m
e
d
d
b
Ω
b
y
r
y
t h
t h
e
i s
e
v
r
o
e
t o
r
s
i s
l t a
g
t o
e
r
s
0
s
o
u
r
51
c
e
s
=
5
=
1
7
0
. 1
5
=
2
0
8
. 7
5
2
. 5
W
W
=
2
0
6
. 2
5
W
=
2
0
8
. 7
5
W
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Class Exercise
 For the circuit shown below, find the branch currents
I1, I2, and I3 using mesh analysis.
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What to do this Week?
 Reading assignment.





Sinusoids and Phasors.
Phasor Representation of AC Circuit Elements.
Impedance and Admittance.
Analysis of AC Circuits using Phasors.
AC Power Analysis: Instantaneous, Average,
Reactive and Complex Powers.
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