EEE 2019 Principles of Electrical &
Electronic Engineering
Lecture 2: DC Circuit Theory:
Fundamental Laws
Instructor: Jerry MUWAMBA
Email: jerry.muwamba@unza.zm
jerry.muwamba@gmail.com
April 14, 2022
University of Zambia
School of Engineering,
Department of Electrical & Electronic Engineering
References
Our main reference text books in this course are:
[1] William H. Roadstrum and Dan H. Wolaver, Electrical Engineering for All Engineers,
(2008), John Wiley and Sons, ISBN :10:0471271780
[2] Jimmie Cathey and Sayed Nasar, Basic Electrical Engineering, Schaum’s Outline
Series, (1996), McGraw Hill 2nd edition, ISBN -10: 0070113556
[3] Charles I. Hubert , DC/AC Electric Circuits, (1982), McGraw Hill, ISBN-10:
0070308454; ISBN-13: 978-0070308459
[4] Charles K. Alexander and Matthew N. O. Sadiku, Fundamentals of Electric Circuits, 5th
Ed., 2012, McGraw-Hill, ISBN-13: 978-0077753603
[5] Theraja B.L., Theraja A.K., Tarnekar S.G., Electrical Technology-Basic Electrical
Engineering, vol. I, 1st Multicolor Ed., 2005, S. Chand, ISBN 81-219-24405.
However, feel free to use some additional text which you might find relevant to our
course.
Department of Electrical & Electronic Engineering, School of Engineering, University of Zambia
2
2.1 Introduction
Lecture 1 brought to the fore fundamental concepts such as current, voltage, and
power in an electric circuit.
To actually determine the values of these variables in a given circuit requires that
we understand fundamental laws that govern electric circuits.
These laws, namely Ohm’s law and Kirchhoff’s laws, form the bedrock upon which
electric circuit analysis is built.
In this lecture, in addition to these laws, we shall discuss some techniques
commonly applied in circuit design and analysis.
These techniques include combining resistors in series or parallel, voltage division,
current division, and delta-to-wye and wye-to-delta transformations.
We shall certainly restrict our application of these laws and techniques to resistive
circuits in this lecture.
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2.2 Ohm’s Law
Materials in general have a characteristic behaviour of resisting the flow of electric
charge.
This physical phenomenon, or ability to resist current, is known as resistance and is
represented by the symbol R.
The resistance of any material with uniform cross-sectional area A depends on A
and its length , as shown in Figure 1(a).
i
v
Cross-sectional
area A
R
Figure 1:
(a) Resistor, (b) Circuit
symbol for resistance.
Material with
resistivity 
(a)
(b)
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2.2 Ohm’s Law Cont’d
We can represent resistance (as measured in the laboratory), in mathematical form,
R=
A
(2.1)
where  is known as the resistivity of the material in ohm-meters.
Good conductors, such as copper and aluminum, have low resistivities, while
insulators, such as mica and paper, have high resistivities.
Table 2.1 presents the values of  for some common materials and shows which
materials are used for conductors, insulators, and semiconductors.
The circuit element used to model the current resisting behavior of a material is the
resistor.
Resistors are usually made from metallic alloys and carbon compounds.
The circuit symbol for the resistor is shown in Figure 1(a), where R stands for the
resistance of the resistor.
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5
2.2 Ohm’s Law Cont’d
Table 2.1
Material
Silver, Ag
Copper, Cu
Aluminum, Al
Gold, Au
Carbon, C
Germanium, Ge
Silicon, Si
Paper
Mica
Glass, SiO2
Teflon, (C2 F4 )n
Resistivity (  m)
1.64  10−8
1.72  10−8
2.8  10−8
2.45  10−8
4  10−5
47  10−2
6.4  102
1010
5  1011
1012
3  1012
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Usage
Conductor
Conductor
Conductor
Conductor
Semiconductor
Semiconductor
Semiconductor
Insulator
Insulator
Insulator
Insulator
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2.2 Ohm’s Law Cont’d
The resistor is the simplest passive element.
Georg Simon Ohm (1787-1854), a German physicist, is credited with finding the
relationship between current and voltage for a resistor.
Ohm’s law states that the voltage v across a resistor is directly proportional to the
current i flowing through the resistor.
That is,
v i
(2.2)
Ohm defined the constant of proportionality for a resistor to be the resistance, R.
The resistance being a material property can change with respect to change internal
or external conditions of the element such as temperature.
Thus, Eqn (2.2) becomes
v = Ri
(2.3)
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2.2 Ohm’s Law Cont’d
Eqn (2.3) is the mathematical form of Ohm’s law and R therein is measured in the
unit of ohms, designated  .
The resistance R of an element denotes its ability to resist the flow of electric
current; it is measured in ohms (  ).
It follows from Eqn (2.3) that
v
R=
i
(2.4)
so that,
1 = 1V A
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2.2 Ohm’s Law Cont’d
For a passive element, by convention, current flows from a higher potential to a
lower potential which yields v = Ri .
If current flows from a lower potential to a higher potential, v = −Ri .
i=0
i
Linear
Circuit
v=0 R=0
Linear
Circuit
(a)
v R=
(b)
Figure 2: (a) Short circuit R = 0 , (b) Open circuit R =  .
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9
2.2 Ohm’s Law Cont’d
A resistor is either fixed or variable.
(a)
(a)
(b)
Figure 4: Variable (a) composition
type, (b) slider pot type.
(b)
Figure 3: Fixed (a) wirewound type,
(b) carbon film type.
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10
2.2 Ohm’s Law Cont’d
(a)
(b)
Figure 5: Circuit symbol for: (a) a
variable resistor in general, (b) a
potentiometer (pot).
A common variable resistor is known
as potentiometer or pot for short, with
the symbol shown in Figure 5(b).
The pot is a three-terminal element
with a sliding contact or wiper.
By sliding the wiper, the resistance
between the wiper terminal and the
fixed terminals vary.
It should be pointed out that not all
resistors obey Ohm’s law.
A resistor that obeys Ohm’s law is
said to be a linear resistor.
Its i-v graph is a straight line passing
through the origin, as depicted in
Figure 6(a).
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11
2.2 Ohm’s Law Cont’d
v
v
Slope = R
Slope = R
i
0
(a)
Figure 6: The i-v characteristic of: (a)
a linear resistor, (b) a nonlinear resistor.
i
0
(b)
A nonlinear resistor does not obey Ohm’s law. Its resistance varies with current and
its i-v characteristic is typically shown in Figure 6(b).
Examples of devices with nonlinear resistance are the light bulb and the diode.
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2.2 Ohm’s Law Cont’d
A useful quantity in circuit analysis is the reciprocal of resistance R, known as
conductance and denoted by G:
G =
1 i
=
R v
(2.5)
The conductance is a measure of how well an element will conduct electric current.
The unit of conductance is mho (ohm spelled backward) or reciprocal of ohm, with
symbol
, the inverted omega.
Although engineers often use the mho, in our discussion we prefer to use the
siemens (S), the SI unit of conductance.
1S = 1
= 1A V
(2.6)
Conductance is the ability of an element to conduct electric current; it is measured
in mhos ( ) or siemens (S).
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13
2.2 Ohm’s Law Cont’d
The same resistance can be expressed in ohms or siemens. For instance, 10  is the
same as 0.1 S.
From Eqn (2.5) it follows that,
i = Gv
(2.7)
Vividly, the power dissipated by a resistor in terms of R is of the form,
v2
p = vi = i R =
R
2
(2.8)
Thus, the power dissipated by a resistor in terms of G is of the form,
i2
p = vi = v G =
G
2
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(2.9)
14
2.2 Ohm’s Law Cont’d
Vividly, from Eqns (2.8) and (2.9) we should note two things:
 The power dissipated in a resistor is a nonlinear function of either current or
voltage.
 Since R and G are positive quantities, the power dissipated in a resistor is
always positive. Thus, a resistor always absorbs power from the circuit. This
confirms that a resistor is a passive element incapable of generating energy.
[Example 2.1]
An electric iron draws 2 A at 120 V. Find its resistance.
[Solution]
From Ohm’s law,
R=
v 120
=
= 60 
i
2
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[Example 2.2]
In the circuit shown in Figure 7,
calculate the current i, the
conductance G, and the power p.
[Solution]
Vividly, Ohm’s law yields,
v  30 
i=
=
= 6 mA
3
R  5  10 
Thus, conductance is of the form,
1  1 
G =
=
= 0.2 mS
3
R  5  10 
i
30 V
5k
v
Figure 7:
Finally, power is of the form,
p = vi = 30(6  10−3 ) = 180 mW
or
p = i 2R = (6  10−3 )2 (5  103 ) = 180 mW
or
p = v 2G = (30)2 (0.2  10−3 ) = 180 mW
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[Example 2.3]
A voltage source of 20 sin  t V is connected across a 5 k resistor. Find the
current through the resistor and the power dissipated.
[Solution]
Applying Ohm’s law yields,
v 20 sin  t
i=
=
= 4 sin  t mA
3
R (5  10 )
Hence, power dissipated is of the form,
p = vi = 80 sin2  t mW
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2.3 Nodes, Branches, and Loops
A network can be regarded as an interconnection of elements or devices, whereas a
circuit is a network providing one or more closed paths.
In network topology, we study the properties relating to the placement of elements
in the network and the geometric configuration of the network.
Such elements include branches, nodes, and loops.
A branch represents a single element such as a voltage source or a resistor.
Simply put, a branch represents any two-terminal element.
A node is the point of connection between two or more branches.
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18
2.3 Nodes, Branches, and Loops Cont’d
a
5
10 V
Figure 8:
b
2
3
2A
c
Figure 8 shows a circuit having
five branches, namely 10 V
voltage source, 2 A current
source, and the three resistors.
The circuit in Figure 8 has three
nodes a, b, and c.
We demonstrate that the circuit
in Figure 8 has three nodes by
redrawing it in Figure 9.
The two circuits in Figures 8 and
9 are as a matter of fact identical.
Nevertheless, for the sake of
clarity, nodes b and c are spread
out with perfect conductors as
illustrated in Figure 8.
b
5
2
a
10 V
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3
c
2A
Figure 9:
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2.3 Nodes, Branches, and Loops Cont’d
A loop is any closed path in a circuit.
A loop is a closed path formed by starting at a node, passing through a set of nodes,
and returning to the starting node without passing through any node more that once.
A loop is independent if it contains at least one branch which is not part of any
other independent loop, and results in an independent equation.
A network with b branches, n nodes, and l independent loops will satisfy the
fundamental theorem of network topology of the form,
b =l +n −1
(2.10)
It is worthy noting that circuit topology is of great value to the study of voltages
and currents in an electric circuit..
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20
2.3 Nodes, Branches, and Loops Cont’d
Two or more elements are in series if they exclusively share a single node 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 voltages across them.
[Example 2.4]
Determine the number of branches
and nodes in circuit shown in
Figure 10.
Identify which elements are in
series and which are in parallel.
5
10 V
6
2A
Figure 10:
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21
[Example 2.4] [Solution]
1
5
10 V
Figure 11:
2
6
2A
3
Since there are four elements in the circuit, the circuit has four branches: 10 V,
5  , 6  , and 2 A .
The circuit has three nodes as identified in Figure 11.
The 5  resistor is in series with the 10 V voltage source.
The 6  resistor is in parallel with the 2 A current source because both are
connected to the same nodes 2 and 3.
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2.3 Kirchhoff’s Laws
Ohm’s law coupled with Kirchhoff’s laws form a sufficient powerful set of tools for
analyzing a huge variety of electric circuits.
Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a
node (or a closed boundary) is zero.
Mathematically, KCL which is based on the law of conservation of charge is of the
form,
N
i
n =1
n
=0
(2.11)
where N is the number of branches connected to the node and in is the n th
current entering (or leaving) the node.
By KCL currents entering the node may be regarded as positive, while currents
leaving the node may be taken as negative or vice versa.
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2.3 Kirchhoff’s Laws Cont’d
Kirchhoff’s second law is based on the law of conservation of energy.
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a
closed path (or loop) is zero.
Mathematically, KVL is of the form,
M
v
m =1
m
=0
(2.12)
where M is the number of voltages in a loop (or the number of branches in the loop)
and vm is the m th voltage.
When applying KVL the sign on each voltage is the polarity of the terminal
encountered first as we travel around the loop.
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[Example 2.5]
For the circuit of Figure 12(a), find
voltages v1 and v2 .
2
v1
20 V
2
i
v2
3
v1
v2
20 V
(b)
3
Figure 12:
(a)
[Solution]
We apply Ohm’s law and KVL
assuming current i flows through the
loop as illustrated in Figure 12(b).
From Ohm’s law,
v1 = 2i ; v2 = −3i
(2.5.1)
Applying KVL to the loop yields,
− 20 + v1 − v2 = 0
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(2.5.2)
25
[Example 2.5] [Solution]
Substituting Eqn (2.5.1) into Eqn (2.5.2) yields,
− 20 + 2i + 3i = 0; i.e., 5i = 20;
 i = 4A
Substituting i into Eqn (2.5.1) finally gives,
v1 = 2(4) = 8 V ;
v2 = −3(4) = −12 V
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[Example 2.6]
Determine v 0 and i in the circuit
shown in Figure 13(a).
i
12 V
2v 0
4
4V
v0
(b)
6
Figure 13:
v0
(a)
[Solution]
i
6
4V
12 V
2v 0
4
Applying Ohm’s law,
v 0 = −6i
(2.6.2)
Substituting Eqn (2.6.2) into Eqn
(2.6.1) yields,
− 16 + 10i − 12i = 0;  i = −8 A
(2.6.1)
Applying KVL around the loop yields,
− 12 + 4i + 2v 0 − 4 + 6i = 0
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[Example 2.6] [Solution]
Thus,
v 0 = −6(−8) = 48 V
[Example 2.7]
Find current i0 and voltage v 0 in
the circuit shown in Figure 14.
a
v0
4
Applying KCL to node a yields,
3 + 0.5i0 = i0 ;  i0 = 6A
i0
0.5i0
[Solution]
3A
For the 4  resistor, Ohm’s law
gives,
v 0 = 4i0 = 24 V
Figure 14:
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[Example 2.8]
Find currents and voltages in the circuit
shown in Fig. 15.
8
i1
a
i2
v1
8
i1
a
30 V
loop 1 v2
3 loop 2 v 3
6
i2
v1
30 V
i3
i3
v2
3
v3
6
(b)
Figure 15:
(a)
From Ohm’s law,
[Solution]
We apply Ohm’s law and Kirchhoff’s
laws to Fig. 15(b).
v1 = 8i 1 ; v2 = 3i2 ; v 3 = 6i3 (2.8.1)
At node a, KCL yields,
i1 − i2 − i3 = 0
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(2.8.2)
29
[Example 2.8] [Solution]
Applying KVL to loop 1 yields,
− 30 + v1 + v2 = 0
It follows that,
− 30 + 8i1 + 3i2 = 0;
i.e.,
(30 − 3i2 )
i1 =
;
8
(2.8.3)
Applying KVL to loop 2 gives,
− v2 + v 3 = 0;
i.e., v 3 =v2
(2.8.4)
as expected since the two resistors are in parallel.
Substituting Eqn (2.8.1) into Eqn (2.8.4) yields,
6i3 = 3i2 ;
i.e., i3 =
i2
2
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(2.8.5)
30
[Example 2.8] [Solution]
Substituting Eqns (2.8.3) and Eqn (2.8.5) into (2.8.2) yields,
30 − 3i2
8
That is,
− i2 −
i2
2
=0
i2 = 2A
Substituting the value of i2
into Eqns (2.8.1) to (2.8.5) gives
i1 = 3A; i3 = 1A; v1 = 24 V; v2 = 6 V; v 3 = 6 V;
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31
2.5 Series Resistors and Voltage Division
The equivalent resistance of any
number of resistors connected in
series is the sum of the individual
resistances.
For N resistors in series it follows
that,
Req = R1 + R2 +
+ RN =
a
v
R1
R2
v1
v2
i
b
vN
RN
Figure 16:
N
R
n =1
n
(2.13)
In general, if a voltage divider has N resistors (R1, R2 , , RN ) in series with the
th
source voltage v , the n resistor (Rn ) will have a voltage drop of the form

Rn
vn = 
 R1 + R2 +

v
+ RN 
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(2.14)
32
2.6 Parallel Resistors and Current Division
The equivalent resistance of two
parallel resistors is equal to the
product of their resistances
divided by their sum.
Thus,
Req =
Node a
i
R1
v
b
R1R2
(2.15)
R1 + R2
Figure 17:
i1
R2
i2
RN
iN
Node b
In general, the equivalent resistance of a circuit with N resistors in parallel is of the
form,
1
1
1
=
+
+
Req R1 R2
1
+
RN
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(2.16)
33
2.6 Parallel Resistors and Current Division
It is worth noting that Req is always smaller than the resistance of the smallest
resistance in the parallel combination.
= RN = R , it follows that,
Thus, if R1 = R2 =
Req =
R
N
(2.17)
It is often more convenient to use conductance rather than resistance when dealing
with resistors in parallel.
From Eqn (2.16), the equivalent conductance for N resistors in parallel is of the
form
Geq = G1 + G2 + G 3 +
+ GN
where Geq = 1 Req , G1 = 1 R1 , G2 = 1 R2 , G 3 = 1 R3 ,
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(2.18)
, GN = 1 RN .
34
2.6 Parallel Resistors and Current Division
Vividly, equation (2.18) states that:
The equivalent conductance of resistors connected in parallel is the sum of their
individual conductance.
It follows that the equivalent conductance Geq of N resistors in series is of the
form,
1
1
1
1
=
+
+
+
Geq G1 G2 G 3
+
1
GN
(2.19)
In general, if a current divider has N conductors (G1, G2 , ,GN ) in parallel with
the source current i , as illustrated in Fig. 17, the n th conductor (Gn ) will have
current of the form


Gn
in = 
i
(2.20)
G1 + G2 + + GN 
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35
[Example 2.9]
Find Req for the circuit shown in Figure 18.
4
1
2
Req
5
8
6
3
Figure 18:
[Solution]
First and foremost we ought to combine resistors in series and parallel as follows:
(6)(3)
6 3 =
= 2 ;
6+3
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36
[Example 2.9] [Solution] Cont’d
Here the symbol
is used to indicate a
parallel combination.
For the series resistors 1 and 5  ,
their equivalent is of the form
4
Req
1 + 5  = 6 ;
2.4 
8
Thus, the circuit in Fig. 18 reduces to that
in Fig. 19(a) and eventually Fig. 19(b).
(b)
4
Figure 19:
8
It follows from Fig. 19(a) that,
(4)(6)
(2  + 2 ) 6  =
= 2.4  ;
4+6
Thus, Fig. 19(b) yields,
2
Req
6
2
(a)
Req = [4  + 2.4  + 8 ] = 14.4 
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37
[Example 2.10]
Calculate the equivalent resistance Rab in the circuit in Fig. 20.
a
Rab
10 
c
1
1
d
6
3
4
5
12 
b
b
b
Figure 20:
[Solution]
The 3  and 6  resistors are in parallel as they share nodes c and b, i.e.,
(3)(6)
3 6 =
= 2 ;
(2.10.1)
3+6
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38
[Example 2.10] [Solution] Cont’d
Similarly, the 12  and 4  resistors
are in parallel as they share nodes d and
b, i.e.,
(12)(4)
12  4  =
= 3 ;
(2.10.2)
12 + 4
Also the 1 and 5  resistors are in
series, i.e.,
(2.10.3)
1 + 5  = 6 ;
10 
a
c
Rab
b
1
d
6
3
b
(a)
a
c
Rab
b
2
b
(b)
3
b
Figure 21:
2
b
10 
b
It follows from Fig. 21(b) that,
(2)(3)
2 3  =
= 1.2  ;
2+3
Thus, Fig. 21(b) yields,
Rab = [10  + 1.2  ] = 11.2 
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39
[Example 2.11]
Find the equivalent conductance Geq for
the circuit in Fig. 22(a).
5S
Geq
6S
5S
Geq
8S
12S
20S
(b)
(a)
Figure 22:
1

5
[Solution]
The 8 S and 12 S resistors
are in parallel, i.e.,
6S
Req
1

6
1

8
1

12
8S + 12S = 20 S;
(c)
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40
[Example 2.11] [Solution] Cont’d
This 20 S resistor is now in series with 5 S as depicted in Fig. 21(b), i.e.,
20S 5S =
(20)(5)
= 4 S;
20 + 5
This is in parallel with the 6 S resistor, i.e.,
Geq = 6S + 4S = 10 S
It is worth noting that the circuit in Fig. 21(a) is same as that in Fig. 21(c).
In order to prove this, we obtain Req from Fig. 21(c) and hence, Geq .
Vividly, from Fig. 21(c) we obtain,
 1  1 
  
6  4 


1
1 1 1 1
1 1 1 
1 1

=
;
Req =
; i.e., Req =
 +
 =
 +
=
1 1
10
6  5 8 12  6  5 20 
6 4
+
6 4
1
Geq =
= 10 S ; Q.E.D
Thus,
Req
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41
[Example 2.12]
For the circuit in Fig. 23(a), determine:
(a) the voltage v 0 , (b) the power
supplied by current source, (c) the power
absorbed by each resistor.
30mA
6k
30mA
v0
9k
i0
i2
i1
v0
9k
18 k
(b)
12k
(a)
[Solution]
(a)
Combining the series resistors in Fig.
23(a) yields Fig. 23(b).
Figure 23:
Applying the current divider rule
yields,
 18 k 
 (30mA) = 20 mA;
i1 = 
 9k + 18 k 


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42
[Example 2.12] [Solution] Cont’d
Also,


9k
 (30mA) = 10 mA;
i1 = 
 9k + 18 k 


Notice that the voltage across the 9k and 18 k resistors is the same, and
v 0 = 9000i1 = 18000i2 = 180 V , as expected.
(b) Power supplied by source is
p0 = v 0i0 = (180 V)(30mA) = 5.4 W
(c) Power absorbed by the 12k resistor is
p = vi = i2 (i2R) = i22R = (10−2 )2 (12  103 ) = 1.2 W
Power absorbed by the 6k and 9k resistors are respectively,
p = i22R = (10−2 )2 (6  103 ) = 0.6 W ; and p = v 0i1 = (180)(20  10 −3 ) = 3.6 W
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2.7 Wye-Delta Transformations
In circuit analysis situations arise when resistors are neither in parallel nor in series.
Many circuits of the nature can be simplified by using three-terminal networks
called wye (Y) or tee (T) and the delta () or pi () networks.
Delta to Wye Conversion
Delta () to Wye (Y) conversion is
achieved using equations of the form,
Rb Rc
R1 =
(2.21)
Ra + Rb + Rc
R2 =
R3 =
Rc Ra
Ra + Rb + Rc
Ra Rb
Ra + Rb + Rc
a
b
R2
R1
n
Rb
(2.22)
(2.23)
Rc
Figure 24:
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R3
Ra
c
44
2.7 Wye-Delta Transformations Cont’d
Each resistor in the Y network is the product of the resistors in the two adjacent 
branches, divided by the sum of the three  resistors.
Wye to Delta Conversion
Wye (Y) to Delta () conversion is achieved using equations of the form,
Rc
a
R1
b
R2
Ra =
R1R2 + R2R3 + R3R1
R1
n
Rb
Figure 25:
R3
Ra
Rb =
Rc =
R1R2 + R2R3 + R3R1
R2
R1R2 + R2R3 + R3R1
c
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R3
(2.24)
(2.25)
(2.26)
45
2.7 Wye-Delta Transformations Cont’d
Each resistor in the  network is the sum of the all possible products of the Y
resistors taken two at a time, divided by the opposite Y resistor.
The Y and  networks are said to be a balanced when,
R1 = R2 = R3 = RY ;
Ra = Rb = Rc = R
(2.27)
Under these conditions, conversion formulas are of the form,
RY =
R
3
;
or
R = 3RY
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(2.28)
46
[Example 2.13]
Convert the  network in Fig. 26(a)
to an equivalent Y network.
Rc
a
Rc
a
5
b
b
7.5
R1
R2
25
Rb
Rb
10
15
R3
3
Ra
Ra
c
(b)
c
(a)
[Solution]
Using Eqns (2.20) to (2.22) we obtain,
Figure 26:
Rb Rc
(10)(25)
R1 =
=
;
Ra + Rb + Rc 15 + 10 + 25
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47
[Example 2.13] [Solution] Cont’d
That is,
R1 =
R2 =
250
= 5
50
Rc Ra
Ra + Rb + Rc
=
(25)(15)
= 7.5 
50
Ra Rb
(15)(10)
R3 =
=
= 3
Ra + Rb + Rc
50
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48
[Example 2.14]
Obtain the equivalent resistance Rab for the circuit in Fig. 27 and use it to find the
current i .
i
a
a
10
12.5
5
c
120 V
n
20
15
b
30
b
Figure 27:
[Solution]
One approach to this problem is to use wye to delta transformation.
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49
[Example 2.14] [Solution] Cont’d
Let the Y network comprising 5 , 10 , and 20 resistors, be
R1 = 10 ,
R2 = 20 ,
R3 = 5 
It follows that,
R1R2 + R2R3 + R3R1 (10)(20) + (20)(5) + (5)(10)
Ra =
=
; i.e.,
R1
10
350
Ra =
= 35 
10
Furthermore,
Rb =
Rc =
R1R2 + R2R3 + R3R1
R2
R1R2 + R2R3 + R3R1
R3
=
350
= 17.5 
20
=
350
= 70 
5
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50
[Example 2.14] [Solution] Cont’d
a
a
12.5
c
15
b
4.545
17.5
35
70
d
30
2.273
1.8182
c
b
a
7.292
b
(b)
20
(c)
Figure 28:
21
10.5
n
15
(a)
30
With the Y converted to  , the
equivalent circuit (with the voltage
source removed for now) is of the
form of Fig. 28(a).
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51
[Example 2.14] [Solution] Cont’d
Combining the three pairs, we obtain
(70)(30)
70 30 =
= 21  ;
70 + 30
12.5 17.5 =
(12.5)(17.5)
= 7.292  ;
12.5 + 17.5
(15)(35)
15 35 =
= 10.5  ;
15 + 35
Thus, the equivalent circuit is depicted in Fig. 28(b), i.e.,
(17.792)(21)
Rab = (7.292 + 10.5) 21 =
= 9.632 
17.792 + 21
Therefore,
vs
120
i=
=
= 12.458A
Rab 9.632
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52
[Example 2.14] [Solution] Cont’d
Proof. We evaluate if the answer is correct by solving the same problem using
delta-wye transformation.
To transform the delta, can, into wye, we let
Rc = 10 ,
Ra = 5 ,
Rn = 12.5 
This yields.,
Rc Rn
(10)(12.5)
Rad =
=
= 4.545  ;
Ra + Rc + Rn 5 + 10 + 12.5
Ra Rn
(5)(12.5)
Rcd =
=
= 2.273  ;
Ra + Rc + Rn
27.5
Rnd =
Ra Rc
Ra + Rc + Rn
=
(5)(10)
= 1.8182  ;
27.5
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53
[Example 2.14] [Solution] Cont’d
This now leads to the circuit shown in Fig. 28(c).
Looking at the resistance between d and b, we have two series combination in
parallel, giving us
(17.273)(21.8182)
Rdb = (2.273 + 15) (1.8182 + 20) =
= 9.642 
17.273 + 21.8182
This is in series with the 4.545  resistor, both of which are in parallel with the
30  resistor. This yields
Rab = (9.642 + 4.545) 30 =
(14.187)(30)
= 9.632 
14.187 + 30
Therefore,
vs
120
i=
=
= 12.458A ; Q.E.D
Rab 9.632
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54
End of Lecture 2
Thank you for your attention!
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55