ENGINEERING UTILITIES 1
CHAPTER 1
BASIC CONCEPTS
ELECTRIC CIRCUIT

An interconnection of electrical elements.
SYSTEMS OF UNITS
CHARGE AND CURRENT

Charge
 most
basic quantity in an electric circuit
 is an electrical property of the atomic particles of
which matter consists, measured in coulombs (C).
 charge
e on an electron is negative and equal in
magnitude to 1.602×10−19 C, while a proton carries a
positive charge of the same magnitude as the electron.
The presence of equal numbers of protons and
electrons leaves an atom neutrally charged.
CHARGE AND CURRENT

Points should be noted about electric charge:
The coulomb is a large unit for charges. In 1 C of charge,
there are 1/(1.602 × 10−19) = 6.24 × 1018 electrons. Thus
realistic or laboratory values of charges are on the order of
pC, nC, or μC.
 According to experimental observations, the only charges
that occur in nature are integral multiples of the electronic
charge e = −1.602 × 10−19 C.
 The law of conservation of charge states that charge can
neither be created nor destroyed, only transferred. Thus the
algebraic sum of the electric charges in a system does not
change.

CHARGE AND CURRENT

Electric charge or electricity is mobile
• Positive charges move in one
direction while negative charges
move in the opposite direction
• Motion of charges creates
electric current
• Conventionally take the
current flow as the movement of
positive charges, that is,
opposite to the flow of negative
charges.
CHARGE AND CURRENT

Electric current is the time rate of change of charge,
measured in amperes (A).
1 ampere = 1 coulomb/second


Direct Current (DC) is a current that remains constant
with time.
Alternating Current (AC) is a current that varies
sinusoidally with time.
CHARGE AND CURRENT
1.
2.
3.
4.
5.
6.
How much charge is represented by 4,600 electrons?
Calculate the amount of charge represented by two
million protons.
The total charge entering a terminal is given by q =
5tsin4πt mC. Calculate the current at t = 0.5s.
If in Example 3, q = (10 − 10e−2t ) mC, find the current
at t = 0.5 s.
Determine the total charge entering a terminal between
t = 1 s and t = 2s if the current passing the terminal is i
= (3t2 − t) A.
The current flowing through an element is
Calculate the charge entering the element from t = 0 to
t = 2s.
VOLTAGE


Voltage (or potential difference) is the energy
required to move a unit charge through an element,
measured in volts (V).
Voltage vab between two points a and b in an electric
circuit is the energy (or work) needed to move a unit
charge from a to b; mathematically,
 where
w is energy in joules (J) and q is charge in
coulombs (C).
 Voltage vab or simply v is measured in volts (V)
1 volt = 1 joule/coulomb = 1 newton meter/coulomb
VOLTAGE

The plus (+) and minus (−) signs are used to define
reference direction or voltage polarity.
(1) point a is at a potential of vab volts higher than point b
(2) the potential at point a with respect to point b is vab
vab = −vba
(a), there is a 9-V voltage drop from a to b or
equivalently a 9-V voltage rise from b to a.
(b), point b is −9 V above point a.
A voltage drop from a to b is equivalent to a
voltage rise from b to a.
constant voltage is called a dc voltage and is represented by V, whereas a
sinusoidally time-varying voltage is called an ac voltage and is represented by v.
POWER AND ENERGY

Power is the time rate of expending or absorbing
energy, measured in watts (W).
where p is power in watts (W), w is energy in joules (J), and t
is time in seconds (s).
 power p is a time-varying quantity and is called the
instantaneous power.


If the power has a “+” sign, power is being delivered
to or absorbed by the element. If, on the other hand, the
power has a “−” sign, power is being supplied by the
element.
POWER AND ENERGY

Passive sign convention is satisfied when the
current enters through the positive terminal of an
element and p = +vi. If the current enters through
the negative terminal, p = −vi.
POWER AND ENERGY

law of conservation of energy must be obeyed:
 algebraic
sum of power in a circuit, at any instant of
time, must be zero.


Energy absorbed or supplied by an element from
time t0 to time t is
Energy is the capacity to do work, measured in
joules ( J).
 electric
power utility companies measure energy in watthours (Wh)
POWER AND ENERGY
1.
2.
3.
4.
5.
6.
An energy source forces a constant current of 2A for 10s to flow
through a light bulb. If 2.3kJ is given off in the form of light and
heat energy, calculate the voltage drop across the bulb.
To move charge q from point a to point b requires−30 J. Find the
voltage drop vab if: (a) q = 2C, (b) q = −6C .
Find the power delivered to an element at t = 3 ms if the current
entering its positive terminal is i = 5cos60πt A and the voltage is: (a)
v = 3i, (b) v = 3 di/dt .
Find the power delivered to the element in Example 3 at t = 5 ms if
the current remains the same but the voltage is: (a) v = 2i V, (b) v =
How much energy does a 100-W electric bulb consume in two
hours?
A stove element draws 15 A when connected to a 120-V line. How
long does it take to consume 30 kJ?
CIRCUIT ELEMENTS

Two types of elements found in electric circuits:
Passive element – not capable of generating energy
(resistors, capacitors, and inductors)
 Active element - capable of generating energy (generators,
batteries, and operational amplifiers)


Two kinds of sources:
Independent Sources - an active element that provides a
specified voltage or current that is completely independent
of other circuit variables
 Dependent Sources - an active element in which the source
quantity is controlled by another voltage or current.
(transistors, operational amplifiers and integrated circuits)

CIRCUIT ELEMENTS

Ideal Independent Voltage Source delivers to the
circuit whatever current is necessary to maintain its
terminal voltage (batteries and generators).
 Symbols
for independent voltage sources: (a) used for
constant or time-varying voltage, (b) used for constant
voltage (dc).
CIRCUIT ELEMENTS

Ideal Independent Current Source is an active
element that provides a specified current
completely independent of the voltage across the
source.
 Symbol
for independent current source
CIRCUIT ELEMENTS

Four possible types of dependent sources:
 Voltage-Controlled
Voltage Source (VCVS).
 Current-Controlled Voltage Source (CCVS).
 Voltage-Controlled Current Source (VCCS).
 Current-Controlled Current Source (CCCS).
 Symbols
for: (a) dependent voltage source, (b) dependent
current source.
CIRCUIT ELEMENTS
1.
2.
Calculate the power supplied or absorbed by
each element
Compute the power absorbed or supplied by each
component of the circuit
CHAPTER 2
BASIC LAWS
OHM’S LAW


Resistance (R) –of an element denotes its ability to resist
the flow of electric current; it is measured in ohms (Ω).
The resistance of any material with a uniform crosssectional area (A) depends on A and its length (l)


where ρ is known as the resistivity of the material in ohmmeters.
Good conductors, such as copper and aluminum, have
low resistivities, while insulators, such as mica and
paper, have high resistivities.
OHM’S LAW
(a) Resistor, (b) Circuit symbol
for resistance.
OHM’S LAW



Ohm’s law states that the voltage “v” across a
resistor is directly proportional to the current “i”
flowing through the resistor.
Short circuit is a circuit element with resistance
approaching zero.
Open circuit is a circuit element with resistance
approaching infinity.
OHM’S LAW
◼ (a)
Short circuit (R = 0), (b) Open circuit (R =∞).
OHM’S LAW




Conductance is the ability of an element to conduct
electric current; it is measured in mhos ( ) or siemens (S).
Resistance can be expressed in ohms or siemens
The power dissipated by a resistor can be expressed in
terms of R.
The power dissipated by a resistor may also be
expressed in terms of G
OHM’S LAW
1.
2.
3.
An electric iron draws 2 A at 120 V. Find its
resistance.
In the circuit shown, calculate the current i, the
conductance G, and the power p.
A voltage source of 20 sin πt V is connected across a
5-kΩ resistor. Find the current through the resistor
and the power dissipated.
NODES, BRANCHES, AND LOOPS




A branch represents a single element such as a
voltage source or a resistor
A node is the point of connection between two or
more branches
A loop is any closed path in a circuit.
A network with b branches, n nodes, and l
independent loops will satisfy the fundamental
theorem of network topology:
b=l+n−1
NODES, BRANCHES, AND LOOPS


Two or more elements are in series if they are
cascaded or connected sequentially and
consequently carry the same current
Two or more elements are in parallel if they are
connected to the same two nodes and consequently
have the same voltage across them.
NODES, BRANCHES, AND LOOPS
1.
2.
Determine the number of branches and nodes in
the circuit shown. Identify which elements are in
series and which are in parallel.
How many branches and nodes does the circuit in
have? Identify the elements that are in series and
in parallel.
KIRCHHOFF’S LAWS


Kirchhoff’s current law (KCL) states that the
algebraic sum of currents entering a node (or a
closed boundary) is zero.
where N is the number of branches connected to the
node and in is the nth current entering (or leaving) the
node. By this law, currents entering a node may be
regarded as positive, while currents leaving the
node maybe taken as negative or vice versa.
KIRCHHOFF’S LAWS

The sum of the currents entering a node is equal to
the sum of the currents leaving the node.
KIRCHHOFF’S LAWS


Kirchhoff’s voltage law (KVL) states that the
algebraic sum of all voltages around a closed
path(or loop) is zero.
where M is the number of voltages in the loop (or the
number of branches in the loop) and vm is the mth
voltage.
KIRCHHOFF’S LAWS

By KVL:
−v1 + v2 + v3 − v4 + v5 = 0
v2 + v3 + v5 = v1 + v4

Sum of voltage drops = Sum of voltage rises
KIRCHHOFF’S LAWS
1.
For the circuit, find voltages v1 and v2.
2.
Determine vo and i in the circuit shown.
KIRCHHOFF’S LAWS
3.
Find current io and voltage vo in the circuit shown
4.
Find the currents and voltages in the circuit shown
SERIES RESISTORS AND VOLTAGE
DIVISION


Using Ohms Law
By KVL (clockwise direction)
The equivalent resistance
of any number of resistors
connected in series is the
sum of the individual
resistances.
SERIES RESISTORS AND VOLTAGE
DIVISION


Voltage across each resistor
Principle of Voltage Division - source voltage ‘v’ is
divided among the resistors in direct proportion to
their resistances; the larger the resistance, the
larger the voltage drop.
PARALLEL RESISTORS AND CURRENT
DIVISION

By Ohms Law

By KCL @ Node ‘a’

Then:
PARALLEL RESISTORS AND CURRENT
DIVISION

Where:
The equivalent resistance of two
parallel resistors is equal to the
product of their resistances divided
by their sum.
Req is always smaller than the resistance
of the smallest resistor in the parallel
combination. If R1 = R2 = … = RN = R,
then
PARALLEL RESISTORS AND CURRENT
DIVISION

Equivalent conductance for N resistors in parallel :
 The
equivalent conductance of resistors connected in
parallel is the sum of their individual conductance's.

Equivalent conductance Geq of N resistors in series:
PARALLEL RESISTORS AND CURRENT
DIVISION

Principle of Current Division - total current ‘i’ is
shared by the resistors in inverse proportion to their
resistances.
PARALLEL RESISTORS AND CURRENT
DIVISION


Divide both the numerator and denominator by R1R2
In general, if a current divider has N conductors
(G1,G2, . . . , GN) in parallel with the source current
i, the nth conductor (Gn) will have current:
Example
1.
2.
3.
Find Req for the circuit shown in Fig. A.
Calculate the equivalent resistance Rab in the circuit
shown in Fig. B
Find the equivalent conductance Geq for the circuit
in Fig. C.
Fig. A
Fig. B
Fig. C
Example
4.
5.
Find io and vo in the circuit shown in Fig. D. Calculate
the power dissipated in the 3Ω- resistor.
For the circuit shown in Fig. E, determine: (a) the
voltage vo, (b)the power supplied by the current
source, (c) the power absorbed by each resistor.
Fig. D
Fig. E
WYE-DELTA TRANSFORMATIONS
The bridge network
Two forms of the same network: (a) Y, (b) T.
Two forms of the same network: (a) Δ, (b) π.
WYE-DELTA TRANSFORMATIONS


Delta to Wye Conversion
Each resistor in the Y network is the product of the
resistors in the two adjacent Δ branches, divided by
the sum of the three Δ resistors.
WYE-DELTA TRANSFORMATIONS


Wye to Delta Conversion
Each resistor in the Δ network is the sum of all
possible products of Y resistors taken two at a time,
divided by the opposite Y resistor.
WYE-DELTA TRANSFORMATIONS

The Y and Δ networks are said to be balanced when

Under these conditions, conversion formulas become
Example
1.
2.
Convert the Δ network in Fig. A to an equivalent Y
network.
Obtain the equivalent resistance Rab for the circuit
in Fig. B and use it to find current ‘i’.
Fig. A
Fig. B
CHAPTER 3
METHODS OF ANALYSIS
NODAL ANALYSIS


Nodal analysis provides a general procedure for
analyzing circuits using node voltages as the circuit
variables
Steps to Determine Node Voltages:
1.
2.
3.
Select a node as the reference node. Assign voltages v1,
v2, . . . , vn−1 to the remaining n − 1 nodes. The voltages are
referenced with respect to the reference node.
Apply KCL to each of the n − 1 nonreference nodes. Use
Ohm’s law to express the branch currents in terms of node
voltages.
Solve the resulting simultaneous equations to obtain the
unknown node voltages.
NODAL ANALYSIS

Current flows from a higher potential to a lower
potential in a resistor.
NODAL ANALYSIS

@ Node 1

@ Node 2

Then;
In Terms of Conductance:

So;
In Matrix Form:
NODAL ANALYSIS (Example)
1.
Calculate the node voltages in the circuits shown.
NODAL ANALYSIS WITH VOLTAGE
SOURCES


Case 1: If a voltage source is connected between
the reference node and a nonreference node, we
simply set the voltage at the nonreference node
equal to the voltage of the voltage source.
Case 2: If the voltage source (dependent or
independent)
is
connected
between
two
nonreference nodes, the two nonreference nodes
form a generalized node or supernode; we apply
both KCL and KVL to determine the node voltages.
NODAL ANALYSIS WITH VOLTAGE
SOURCES

A supernode is formed by
enclosing a (dependent or
independent)
voltage
source connected between
two nonreference nodes
and
any
elements
connected in parallel with
it.
Note the following properties of a supernode:
1. The voltage source inside the supernode provides a
constraint equation needed to solve for the node voltages.
2. A supernode has no voltage of its own.
3. A supernode requires the application of both KCL and KVL.
NODAL ANALYSIS WITH VOLTAGE
SOURCES (Example)
1.
For the circuits shown, find the node voltages.
NODAL ANALYSIS WITH VOLTAGE
SOURCES (Example)
2.
Find v and i in the circuit.
MESH ANALYSIS





Mesh analysis provides another general procedure for
analyzing circuits, using mesh currents as the circuit
variables
A mesh is a loop that does not contain any other loop
within it
Nodal analysis applies KCL to find unknown voltages in a
given circuit, while mesh analysis applies KVL to find
unknown currents
only applicable to a circuit that is planar
Planar circuit is one that can be drawn in a plane with no
branches crossing one another; otherwise it is nonplanar.
MESH ANALYSIS
(c)
(a)
(b)
(c)
A planar circuit with crossing branches
the same circuit redrawn with no crossing branches
A nonplanar circuit.
MESH ANALYSIS
MESH ANALYSIS (Example)
1.
For the circuit shown, find the branch currents I1, I2,
and I3 using mesh analysis.
MESH ANALYSIS (Example)
2.
3.
Calculate the mesh currents i1 and i2 in the circuit
shown.
Use mesh analysis to find the current io in the circuit
shown.
MESH ANALYSIS (Example)
4.
Using mesh analysis, find io in the circuit shown.
MESH ANALYSIS WITH CURRENT
SOURCES

Case 1: When a current source exists only in one
Set i2 = −5 A and write a mesh
mesh:
equation for the other mesh in the
usual way, that is,

Case 2: When a current source exists between two
meshes: We create a supermesh by excluding the
current source and any elements connected in series
with it.
MESH ANALYSIS WITH CURRENT
SOURCES

(a)
(b)
A supermesh results when two meshes have a
(dependent or independent) current source in
common.
Two meshes having a current source in common
a supermesh, created by excluding the current source.
MESH ANALYSIS WITH CURRENT
SOURCES

Note the following properties of a supermesh:
1.
2.
3.
The current source in the supermesh is not completely
ignored; it provides the constraint equation necessary
to solve for the mesh currents.
A supermesh has no current of its own.
A supermesh requires the application of both KVL
and KCL.
MESH ANALYSIS WITH CURRENT
SOURCES (Example)
1.
For the circuit shown, find i1 to i4 using mesh
analysis.
MESH ANALYSIS WITH CURRENT
SOURCES (Example)
2.
Use mesh analysis to determine i1, i2, and i3 in the
circuit shown.
CHAPTER 4
CIRCUIT THEOREMS
LINEARITY PROPERTY



Linearity is the property of an element describing a
linear relationship between cause and effect
The property is a combination of both the
homogeneity (scaling) property and the additivity
property
Homogeneity property requires that if the input
(also called the excitation) is multiplied by a
constant, then the output (also called the response) is
multiplied by the same constant.
LINEARITY PROPERTY




For example, Ohm’s law relates the input i to the output v,
If the current is increased by a constant k, then the voltage
increases correspondingly by k, that is,
The additivity property requires that the response to a
sum of inputs is the sum of the responses to each input
applied separately. Using the voltage-current relationship
of a resistor, if
then applying (i1 + i2) gives
LINEARITY PROPERTY



In general, a circuit is linear if it is both additive
and homogeneous. A linear circuit consists of only
linear elements, linear dependent sources, and
independent sources
A linear circuit is one whose output is linearly
related (or directly proportional) to its input
Relationship between power and voltage (or
current) is nonlinear – linearity theorems are not
applicable to power
LINEARITY PROPERTY






Consider the linear circuit shown
linear circuit has no independent sources inside it
Excited by a voltage source vs
Terminated by a load R
Current i through R as the output
Suppose vs = 10 V gives i = 2 A. According to the
linearity principle, vs = 1 V will give i = 0.2 A. By
the same token, i = 1 mA must be due to vs =5 mV.
LINEARITY PROPERTY
1.
2.
For the circuit, find io when vs = 12 V and vs = 24
V.
For the circuit , find vo when is = 15 and is = 30 A.
LINEARITY PROPERTY
3.
4.
Assume Io = 1 A and use linearity to find the actual
value of Io in the circuit.
Assume that Vo = 1 V and use linearity to calculate
the actual value of Vo in the circuit.
SUPERPOSITION


The superposition principle states that the voltage across
(or current through) element in a linear circuit is the
algebraic sum of the voltages across (or currents through)
that element due to each independent source acting
alone.
To apply the superposition principle, we must keep two
things in mind:
1.
2.
We consider one independent source at a time while all
other independent sources are turned off. This implies that
we replace every voltage source by 0 V (or a short circuit),
and every current source by 0 A (or an open circuit). This
way we obtain a simpler and more manageable circuit.
Dependent sources are left intact because they are
controlled by circuit variables.
SUPERPOSITION

Steps to Apply Superposition Principle :
 Turn
off all independent sources except one source.
Find the output (voltage or current) due to that active
source using nodal or mesh analysis.
 Repeat step 1 for each of the other independent
sources.
 Find the total contribution by adding algebraically all
the contributions due to the independent sources.
SUPERPOSITION
1.
2.
Use the superposition theorem to find v in the
circuit
Using the superposition theorem, find vo in the
circuit
SUPERPOSITION
3.
Find io in the circuit using superposition.
4.
Use superposition to find vx in the circuit
SUPERPOSITION
5.
6.
For the circuit, use the superposition theorem to
find i.
Find i in the circuit using the superposition principle.
SOURCE TRANSFORMATION

A source transformation is the process of replacing
a voltage source vs in series with a resistor R by a
current source is in parallel with a resistor R, or vice
versa.
SOURCE TRANSFORMATION

Points to be mind when dealing with source
transformation:
 The
arrow of the current source is directed toward the
positive terminal of the voltage source.
 Source transformation is not possible when R = 0, which is
the case with an ideal voltage source. However, for a
practical, nonideal voltage source, R = 0. Similarly, an
ideal current source with R =∞cannot be replaced by a
finite voltage source.
SOURCE TRANSFORMATION
1.
Use source transformation to find vo in the circuit
2.
Find io in the circuit using source transformation
SOURCE TRANSFORMATION
3.
Find vx using source transformation.
4.
Use source transformation to find ix in the circuit.
THEVENIN’S THEOREM



Developed in 1883 by M. Leon Thevenin (1857–1926), a
French telegraph engineer
Thevenin’s theorem states that a linear two-terminal circuit
can be replaced by an equivalent circuit consisting of a
voltage source VTh in series with a resistor RTh, where VTh is
the open-circuit voltage at the terminals and RTh is the input
or equivalent resistance at the terminals when the
independent sources are turned off.
G
THEVENIN’S THEOREM

CASE 1: If the network has no dependent sources,
we turn off all independent sources. RTh is the input
resistance of the network looking between terminals
a and b, as shown
THEVENIN’S THEOREM

CASE 2: If the network has dependent sources, we turn off all
independent sources. As with superposition, dependent sources
are not to be turned off because they are controlled by circuit
variables. We apply a voltage source vo at terminals a and b
and determine the resulting current io. Then RTh = vo/io, as shown.
Alternatively, we may insert a current source io at terminals a-b
as shown in and find the terminal voltage vo. Again RTh = vo/io.
Either of the two approaches will give the same result.
THEVENIN’S THEOREM

It often occurs that RTh takes a negative value. In this
case, the negative resistance (v = −iR) implies that
the circuit is supplying power.
THEVENIN’S THEOREM
1.
2.
Find the Thevenin equivalent circuit of the circuit
shown, to the left of the terminals a-b. Then find the
current through RL = 6Ω, 16Ω, and 36Ω .
Using Thevenin’s theorem, find the equivalent circuit
to the left of the terminals in the circuit. Then find i.
THEVENIN’S THEOREM
3.
4.
Find the Thevenin equivalent of the circuit shown.
Find the Thevenin equivalent circuit of the circuit
shown to the left of the terminals.
THEVENIN’S THEOREM
5.
6.
Determine the Thevenin equivalent of the circuit
shown.
Obtain the Thevenin equivalent of the circuit.
NORTON’S THEOREM


Developed by E. L. Norton, an American engineer at Bell
Telephone Laboratories in 1926, about 43 years after
Thevenin published his theorem.
Norton’s theorem states that a linear two-terminal circuit
can be replaced by an equivalent circuit consisting of a
current source IN in parallel with a resistor RN, where IN is
the short-circuit current through the terminals and RN is the
input or equivalent resistance at the terminals when the
independent sources are turned off.
NORTON’S THEOREM



By Source transformation, the Thevenin and Norton
resistances are equal
To find the Norton current IN,we determine the shortcircuit current flowing from terminal a to b in both circuits
.
Observe the close relationship between Norton’s and
Thevenin’s theorems: RN = RTh
NORTON’S THEOREM

Since VTh, IN, and RTh are related, to determine the
Thevenin or Norton equivalent circuit requires that
we find:
 The
open-circuit voltage voc across terminals a and b.
 The short-circuit current isc at terminals a and b.
 The equivalent or input resistance Rin at terminals a and
b when all independent sources are turned off.

Thus we find;
NORTON’S THEOREM
1.
Find the Norton equivalent circuit of the circuits
NORTON’S THEOREM
2. Find the Norton equivalent circuit of the circuit.
MAXIMUM POWER TRANSFER



Circuit is designed to provide power to a load
Electric utilities, minimizing power losses in the
process of transmission and distribution is critical for
efficiency and economic reasons, there are other
applications in areas such as communications where
it is desirable to maximize the power delivered to a
load
Address problem of delivering the maximum power
to a load when given a system with known internal
losses.
MAXIMUM POWER TRANSFER

Thevenin equivalent is useful in finding the maximum
power a linear circuit can deliver to a load
Maximum
power
is
transferred to the load
when the load resistance
equals
the
Thevenin Maximum power transferred is obtained by:
resistance as seen from
the load (RL = RTh).
MAXIMUM POWER TRANSFER
1.
Find the value of RL for maximum power transfer in
the circuits. Find the maximum power.