Lecture Notes: QE 108: Electricity
Electric Circuits
Department of Electrical and Electronic Engineering
University of Peradeniya
Peradeniya
December 2005
Prepared by: Kithsiri M. Liyanage, Department of Electrical and Electronic Engineering, University of Peradeniya
© KML - DEEE, December 2005
QE 108: Electricity – Section 2 – Circuits
Contents
Course Structure...................................................................................................................................................... 1
Lectures and Tutorials ........................................................................................................................................ 1
Laboratory Classes ............................................................................................................................................. 1
Assignments......................................................................................................................................................... 1
1. Analysis of DC Circuits ...................................................................................................................................... 2
1.1. Elements and Laws..................................................................................................................................... 2
1.1.1. Current...................................................................................................................................................... 2
1.1.2. Voltage, Electromotive Force and Potential Difference ......................................................................... 3
1.1.3. Power in d.c. Circuits............................................................................................................................... 4
1.1.4. Circuit Elements....................................................................................................................................... 5
1.1.5. Resistance Conductance and Ohm’s law ................................................................................................ 5
1.1.6. Sources ..................................................................................................................................................... 7
1.1.7. Kirchchoff’s Laws ................................................................................................................................... 8
1.1.8. Practical (non-ideal) Sources................................................................................................................... 9
1.2. Circuit Analysis .......................................................................................................................................... 11
1.2.1. Definitions and terminology.................................................................................................................. 11
1.2.2. Mesh analysis......................................................................................................................................... 12
1.2.3. Nodal analysis........................................................................................................................................ 16
1.2.4. Loop Analysis ........................................................................................................................................ 18
TUTORIAL 1........................................................................................................................................................... 21
2. Network Theorems ....................................................................................................................................... 23
2.1. Thevenin’s Theorem ................................................................................................................................. 23
2.2. Norton’s Theorem..................................................................................................................................... 24
2.3. Maximum Power Transfer Theorem ........................................................................................................ 24
2.4. Y-∆ and ∆- Y Transformation ................................................................................................................. 25
2.5. Examples ................................................................................................................................................... 26
TUTORIAL 2........................................................................................................................................................... 27
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QE 108: Electricity – Section 2 – Circuits
Course Structure
Lectures and Tutorials
1. Analysis of DC Circuits ( ~5 Hrs.)
Circuit Elements and Laws
Mesh and Nodal Analysis of DC Circuits
Tutorial
2. Circuit Theorems ( ~4 Hrs.)
Thevenin’s Theorem and Applications
Norton’s Theorem and Applications
Star Delta Conversion
Tutorials
Laboratory Classes
1. Circuits (4 Hrs.)
Assignments
1. End Semester Examination
2. Quizzes (Tentative)
3. Laboratory class
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QE 108: Electricity – Section 2 – Circuits
1. Analysis of DC Circuits
1.1. Elements and Laws
In this section we discuss about Current, Voltage, Resistance, Ohm’s law, Electric Power,
Kirchhoff’s Laws and their applications
1.1.1. Current
We know about electrostatic charge. Electric charge in
motion results in an electric current.
e.g. Lightning discharge
Current is a measure of the rate at which electric
charge passes through the circuit
i=
dq
(instantaneous value)
dt
Fig 1. Ammeters Commonly
Charge motion is from a low potential to a higher
Used in the Laboratory
potential. However the more usual convention is that
current flow from a point of higher potential to a point of lower potential.
Fig 1. shows ammeters commonly used to measure current in our laboratory.
Instantaneous values are represented using lowercase characters (i). Steady state quantities are
represented using upper-case characters (I)
q(t) (coulombs)
t (s)
E.g. Find the current waveform at the point of a
circuit given that the charge passes through the
point is as shown in Fig. 2.
Fig. 2. Charge Flow Waveform
Current Direction Representation
Fig. 3 shows the conventions used to represent current directions in circuits.
5A
Supply
Source
5A
-5A
Circuit
i(t)
Supply
Source
-5A
Circuit
i(t)
5A
5A
(a)
(b)
Fig. 3. Current Direction Representation
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QE 108: Electricity – Section 2 – Circuits
1.1.2. Voltage, Electromotive Force and Potential Difference
The Voltage across (or between) a pair of terminals is a measure of the energy required to
move charge through the element or circuit connected between the terminals.
The voltage across an element is the energy required
to move a charge of 1 Coulomb from one terminal to
the other
We usually represent instantaneous voltage by v and
steady state voltage V. Units of voltage is volt
(Joules/ Coulombs!)
Fig. 4. shows voltmeters commonly used in our
laboratory.
The energy converted per unit charge in an electrical
Fig. 4. Voltmeters Commonly Used in
the Laboratory
source is known an the electromotive force (e.m.f.)
of the source, and the electrical potential difference
(p.d.) between two points in a circuit is a measure of the energy required to move charge
through the element
General name given to e.m.f and p.d. is “voltage”.
Representation of Voltage
Voltage may be represented on a circular diagram either by a ‘+’and ‘-‘ pairs of symbols or
by an arrow pointing from one terminal to another as shown in Fig. 5.
-
+
8V
Circuit
-
8V
Circuit
-8V
Circuit
-8V
Circuit
+
(a)
(c)
(b)
Fig. 5. Voltage Direction Representation
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(d)
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QE 108: Electricity – Section 2 – Circuits
1.1.3. Power in d.c. Circuits
Power is the rate of transfer of energy. Energy requited to move
1 coulomb across an element is p.d. between two terminals of
the element. Therefore power (p) is equal to (energy required (v)
÷ time taken (t)) to move 1 coulomb across the element. For d.c.
circuits current (i) = (charge ÷ time). In the case where charge is
1 coulomb i = 1/t. Therefore p = v/t = v.i
Since we deal with instantaneous quantities here we get
instantaneous power. Average power is given by P = V.I. Power
could be measured using a wattmeter. Such wattmeter is shown
in Fig. 6.
Fig.6. A Wattmeter
Used in the
Laboratory
There are mainly two types of elements in electric circuits.
Power source - Power-supplying elements
Loads - Power absorbing elements
Fig. 7 shows several power supply units used in our laboratory for experiment.
Fig. 7. Power Supplies Commonly Used in the Laboratory
Current flows out of the positive terminal of a power source and current flows into the
positive terminal of a load as shown in Fig. 8.
+
5A
+
Source
Load
i(t)
-
5A
-
Fig. 8. Current Flow Directions in a
Source and a Load
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QE 108: Electricity – Section 2 – Circuits
1.1.4. Circuit Elements
Types of circuit elements
Linear
-
Nonlinear
Passive
-
Active
(a)
(b)
(c)
(d)
Linear elements - Voltage across the
element very linearly with the current
Passive elements - Circuit Components
that can only dissipate or store energy
Energy dissipating - Resistors
Energy Storing - Capacitors and
inductors
However most practical elements are
nonlinear
Fig. 9. Circuit Elements
(a) Resistors – passive, energy dissipating,
linear!
(b) Capacitors – passive, energy storing,
linear!
(c) Diodes – passive, nonlinear
(d) Transistors – active, nonlinear
1.1.5. Resistance Conductance and Ohm’s law
There is a resistance to flow of current in a material. This results is a loss of energy which is
usually converted to heat
Conductors
have low resistance
e.g. Copper, Brass, Aluminum
Insulators
have a high resistance
e.g. Wood, Plastic. Glass
Semi conductors
have a resistance between that of conductors and insulators
e.g. Silicon, Germanium
Super conductors
resistance is zero (at very low temperatures)
e.g. Tin, Lead, Thallium
Ohm’s Law
Ohm’s law as describes relationship between potential
difference (v) between and current (i) through a linear
element.
v= iR
or
R
V=IR
R is the constant of proportionality called resistance the
unit of resistance is Ohm (Ω)
Another from of expressing the relationships between v
(or V) and I (or I) is
(a)
R
(b)
Fig. 10. Representation of Resistors
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QE 108: Electricity – Section 2 – Circuits
i= G v
or
I=G V
where G is conductance. The unit of conductance is Siemens (S)
Resistance or conductance is represented by means of a rectangular box in circuit diagram as
shown in Fig. 10.
Power in a resistive circuit
We know that p = v.i and for a resistive circuit v = i .R = i/G
Therefore for a resistive circuit:
p = v.i = R.i2 = v2/R = v2.G = i2/G Watts(W)
Power is also measured in µW, mW, kW and MW depending upon the values involved.
E.g.
1. Given that v = 10 V and i = -15 mA, what is the power consumed
(a) 150 W
(b) – 150 W
(c) 0.15 W
(d) –0.15 W
Is this circuit a source or a load?
2. Given that P = 200 mW and I = 10 mA
(i) R = 20 Ω
(ii) R = 20 kΩ
(iii) G = 0.05 S
(iv) G = 50 µS
(a) Only (i) is true
(b) Only (i) and (iii) are true
(c) Only (ii) and (iv) are true
(d) Only (ii) and (iii) are true
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QE 108: Electricity – Section 2 – Circuits
1.1.6. Sources
Sources could be Voltage or Current and they could be Independent or Dependent
Ideal Sources: Can provide infinite amount of power
Independent Sources
Independent sources are represented using the symbols shown in Fig. 11.
1.5 mA
3A
+
+
4A
12 V
6V
Voltage Source
Current Source
Battery
Fig. 11. Representation of Independent Sources
Dependent Sources (Controlled Sources)
Dependent sources, also referred to as controlled sources as well, are represented using the
symbols shown in Fig. 12.
+
10 V1
10 I2
-
5 V3
20 I4
+
Voltage Sources
Current Sources
Fig. 12. Representation of Dependent Sources
Out put of a dependent Source is dependent on a voltage or current at some other point in the
circuit
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QE 108: Electricity – Section 2 – Circuits
1.1.7. Kirchchoff’s Laws
Kirchchoff’s Current law (KCL)
The algebraic sum of currents entering any node is zero
i1 – i2 – i3 + i1 = 0
i1
I2
Σ in = 0
i4
Another form:
The algebraic sum of currents entering any node is equal to the
algebraic sum of the currents leaving the node.
I3
Fig. 13. A Circuit Node
i1 + i1 = i2 + i3
Σ iin = Σ iout
Kirchchoff’s Voltage Law (KVL)
The algebraic sum of e.m.f.s and p.d.s around any close circuit is zero
v1 - v2 - v3 = 0
Σ vn = 0
In any closed path the algebraic sum of the e.m.f.s is equal to the algebraic sum of p.d.s.
v1 = v2 + v3
Σ vemf = Σ vpd
E.g. Consider the circuit shown in Fig. 14.
12 A
R1
IB
+
10 Ω
- VA
+
R2
120 V
10 Ω
8A
-
0.5 A
Fig. 14. Circuit for Example
(i) VA = ? (a) 120 V
(b) –120 V
(c) 40 V
(d) –40 V
(ii) IB = ? (a) 12 A
(b) 0.5 A
(c) –0.5 A
(d) 4.5 A
(iii) R1 = ? (a) 22.33 Ω
(b) 33.33 Ω (c) 43.33 Ω (d) 15 Ω
(iii) R2 = ? (a) 20 Ω
(b) 41.22 Ω
(c) 31.33 Ω (d) 21 Ω
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QE 108: Electricity – Section 2 – Circuits
VAN
The Double–Suffix voltage Notation for specifying
the voltage of one mode with respect to another
mode
VSA
A
VNA
S
N
The notation shown in Fig. 15 called double-suffix
voltage notation is also used to describe the voltage of
a node with respect to another node in the circuit.
VBS
B
C
VCB
VSN
Fig. 15. Double-Suffix Notation
E.g. Consider the circuit shown in Fig. 16.
6Ω
3Ω
+
-
5Ω
10 V
A
B
9Ω
40 V
+
4Ω
2Ω
5Ω
I
Fig. 16. Circuit for Example
VAB = ? (i) 28 V (ii) 26 V
(iii) –28 V (iv) –26 V
I=?
(iii) 0 A
(i) 2 A
(ii) 1 A
(iv) –1 A
1.1.8. Practical (non-ideal) Sources
The characteristics of practical
sources deviate from that of ideal
sources as shown in Fig.17. This
deviation is accounted for in the
model of:
ƒ
ƒ
v
v
Ideal Source
Es
practical voltages source by
introducing a resistance called
internal (or source or output)
resistance connected in series
with an ideal voltage source
practical
current
source
by
introducing a conductance called
internal (or source or output)
conductance connected in parallel
with an ideal current source as
shown in Fig. 18 (compare them with
the models shown in Fig11).
Practical Source
Practical
Source
Ideal
Source
i
Voltage Source
Current Source
Fig. 17. Source Characteristics
io
Rs
vo
Es
Voltage Source
i
Is
Gs
vo
Current Source
Fig. 18. Models of Practical Source
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QE 108: Electricity – Section 2 – Circuits
Transformation of Practical Sources
Rs
vL = Es – iL Rs (for voltage source)
iL
iL
Es Es
vL
vL = (Is – iL )/ Gs (for current source)
≡
Is
Gs
vL
Fig. 19. Equivalent Sources
i.e. Gs = 1/ Rs and Is = Es / Rs
E.g. Consider the circuit shown in Fig. 20.
Current I = ?
(a) 1/2 A
(b) 1/3 A
(c) 1/4 A
(d) 1/5 A
10 Ω
10 V Es
I
0.2 S
1A
Fig. 20. Circuit for Example
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QE 108: Electricity – Section 2 – Circuits
1.2. Circuit Analysis
Here we learn about mesh and nodal analysis.
Anyone of these methods can be applied to almost any circuit and, in many cases there is no
simple way of saying which is the 'best' method of solution; we cannot lay down simple rules
to determine the 'best' approach. Knowledge of each method can only be gained by acquiring
a sound understanding of the features of each type of solution.
1.2.1. Definitions and terminology
An electrical network is a system of
interconnected circuit elements containing, for
example, resistors, inductors, capacitors, voltage
and current sources, transformers, amplifiers, etc.
If the network contains at least one closed path or
mesh it is an electrical circuit
Every circuit is a network, but not all networks
are circuits! (Fig. 21)
Open Circuit
V
(a)
Short Circuit
V
(b)
Fig. 21. (a) Network (b) Circuit
An ideal circuit element is one, which does not,
strictly speaking, represent a practical element.
For example, a practical resistor has an ideal resistance element as part of its make-up, but it
also has some inductance because the current passing through it produces a magnetic field,
and it has some capacitance because it is insulated from, say, earth (the insulation acting as a
dielectric).
In many situations the major feature of an element (such as the resistance of a resistor, or the
capacitance of a capacitor) can be thought of as being at one point within the element. If this
is the case, then we say we are dealing with a lumped-constant (parameter) element. In other
cases we say that we are dealing with a distributed-constant
(parameter) element.
A node is a point in a circuit, which is common to two or more
circuit elements. A junction or principal node is a point in the
circuit where three or more elements are connected together;
nodes 0 and 2 in Fig. 22 are principal nodes.
1
Is
R1 2 R3
R2
3
R4
0
Fig. 22. A Circuit
In the majority of cases of circuit analysis, we choose one
node (usually a principal node) to be a reference node or datum node, so that the voltage of
other nodes can be defined with respect to it.
A branch in a circuit is a path containing one circuit element, and which connects one node to
another.
A path in a network is a set of connected elements that may be traversed without passing
through the same node twice. E.g. R1 and R2 are in the path that connects nodes A and C.
A loop in a circuit is a closed path within the network. E.g. Is – R1 – R3 – R4
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QE 108: Electricity – Section 2 – Circuits
An element that can be connected in a circuit in either direction without changing the
electrical performance of the circuit is known as a bilateral element.
E.g. resistors, inductors and capacitors.
The majority of networks are bilateral networks, that is, they contain only bilateral elements.
Certain network theorems, such as the reciprocity theorem (we’ll see soon!), are only
applicable to bilateral networks.
A planar network is one that may
be drawn on a flat surface, so that
none of the branches passes over or
under any other branch. When this
cannot be done, the network is
non-planar. The circuit drawn in
full line in Fig. 23 is a planar
network but, if the branch
containing R6 is introduced
(shown broken), it becomes a nonplanar circuit.
R5
R6
Vs
R1
Es
R2
R4
R3
Fig. 23. Planar and Non-planar Network
A mesh is a loop that does not contain any other loops within it. However in some cases it is
possible to re-draw the circuit so that a loop which is not a mesh in one version of the circuit
can become a mesh in another version.
1.2.2. Mesh analysis
Mesh analysis involves the concept of mesh current. Fig. 24 shows a circuit with mesh
currents, I1 and I2, circulating around the periphery of each mesh in a clockwise direction.
The relationship between the mesh current and the branch currents (IA, IB and IC) are
IA = I1, IB = I2, IC = I1 -I2
Applying KVL to each mesh we get
20 = 40I1 – 30I2
10 = -30I1 + 50I2
n
10 Ω IA
20 Ω
o IB
p
IC
20 V Es
I1
30 Ω
I2
10 V
Fig. 24. A Two Mesh Network
Solving the two simultaneous equations
we get the answers
I1 = 0.636 A and I2 = 0.182 A.
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QE 108: Electricity – Section 2 – Circuits
General rules for writing mesh equations
To understand how to write down mesh equations by just looking at a circuit, let’s consider a
more complex circuit shown in Fig. 5.
Applying KVL to mesh 1, mesh 2 and mesh 3
R1
V1 -I1R1 + I2R1 + V2 -I1R4 + I3R4 = 0
V1
I1
I1R4 -I3R4 + I2R3 -I3R3 -I3R5 = 0
E2 = R21I1 + R22I2 + R23I3
 E1   R11
E  = R
 2   21
 E 3   R31
R2
R3
R4
These equations can be rearranged to give the
following form
El = R11I1 + R12I2 + Rl3I3
I2
V2
-V 2 + I1R1 -I2R1 -I2R2 -I2R3 + I3R3 = 0
Mesh 2
Mesh 1
I3
R5
Mesh 3
Fig. 25. A Three- Mesh Network
R12
R22
R32
R13 
R23 
R33 
E3 = R31I1 + R32I2 + R33I3
 I1 
I 
 2
 I 3 
Resistance Matrix
Where
E1 = V1 + V2,
R11 = R1 + R4,
R22 = R1 + R2,
E2 = -V2,
R12 = R21 = - R1,
R23 = R32 = -R3,
E3 = 0,
R13 = R31 = -R4,
R33 = R3 + R4 + R5
In general
Rij = (-1) x The resistance in the branch mutual to meshes in which Ii and Ij flow.
Rii = The sum of resistance in the mesh in which Ii flows.
Ei = The sum of source voltages driving Ii
The resistance matrix is a square matrix that is it has as many rows as it has columns. For a
bilateral network it is symmetrical about the major diagonal, that is Rij = Rji. In the case of a
bilateral network, all elements on the major diagonal of the resistance matrix are positive;
elements not on the major diagonal are either zero or negative.
These comments do not always apply to a non-bilateral network or to networks containing
sources other than independent voltage sources, for example, current sources.
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QE 108: Electricity – Section 2 – Circuits
Depending on the circuit, there are either three or four general rules, which need to be
followed:
1. Draw a carefully labeled circuit diagram.
2. Assign mesh currents I1, I2, ..., Im to each mesh flowing in clockwise direction in the
circuit.
3. If the circuit contains only voltage sources, apply KVL to each mesh and solve the
resulting simultaneous equations for the unknown mesh currents (if there are m
meshes, there are m equations). If the circuit contains dependent voltage sources,
relate the dependent source volt- ages to the unknown mesh currents.
4. If the circuit contains one or more ideal current sources, replace each such source by
an open-circuit (note: the mesh currents assigned in step 2 must not be changed). Each
source current should then be related to the mesh currents assigned in step 2. The
resulting simultaneous equations should then be solved
Where a circuit contains practical current sources each can be converted into its equivalent
practical voltage source, and the problem solved as outlined in step 3 above.
Solution of simultaneous equations
Solution of simultaneous equations could be found using methods such as Inverse Matrix,
Cramer’s Rule, Gauss-Jordan Elimination (Reducing the augmented matrix to reduced row
echelon form) and Gauss Elimination (Reducing the augmented matrix to row echelon form)
methods.
Example 1: Consider the circuit shown in Fig. 26:
(i) I1 is
3Ω
2Ω
6Ω
(a) 0.75 A (b) 0.77 A (c) 0.79 A (d) 0.81 A
(ii) I2 is
(a) -0.12 A (b) -0.13 A (c) -0.14 A (d) -0.15
A
(iii) I3 is
8V
I2
10 V
I1
7Ω
5Ω
I3
4Ω
(a) 0.18 A (b) 0.20 A (c) 0.22 A (d) 0.24 A
Fig. 26. Circuit for Example 1
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QE 108: Electricity – Section 2 – Circuits
Example 2: Consider the circuit shown in Fig. 27:
(i) I1 is
6Ω
2Ω
(a) 0.45 A (b) 0.55 A (c) 0.65 A (d) 0.75 A
(ii) I2 is
(a) 0.05 A (b) 0.06 A (c) 0.07 A (d) 0.08 A
8V
0.5 A
10 V
3Ω
I1
I2
7Ω
5Ω
(iii) I3 is
4Ω
I3
(a) 0.18 A (b) 0.19 A (c) 0.20 A (d) 0.22 A
Fig. 27. Circuit for Example 2
Solution to Example 2
Applying KVL to the super-mesh in Fig. 28,
we get
10 - 8 = 2I1 + 3I2 + 7(I2 - I3) + 5(I1 -I3)
6Ω
2Ω
Open
Circuit
10 V
0 = -5I1 –7I2 + 16I3
I1
3Ω
8V
I2
7Ω
5Ω
0.5 = I1 – I2
4Ω
I3
Fig. 28. Circuit for Solving
Example 2
Example 3: Consider the circuit shown in Fig. 29:
(i) I1 is
(a) 4.00 A (b) 4.25 A (c) 4.58 A (d) 4.78 A
6Ω
2Ω
(ii) I2 is
10 V
(iii) I3 is
8V
3 VX
(a) -2.12 A (b) -2.53 A (c) -2.85 A (d) -2.97 A
3Ω
I1
(a) 0.25 A (b) 0.27 A (c) 0.29 A (d) 0.31 A
I2
7Ω
VX
5Ω
I3
4Ω
Solution to Example 3
Vx = 5(I1 - I3)
Fig. 29. Circuit for Example 3
10 + 3Vx = 13I1- 6I2- 5I3
-3Vx - 8 = -6I1 + 16I2- 7I3
0 = -5I1 –7I2 + 16I3
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QE 108: Electricity – Section 2 – Circuits
1.2.3. Nodal analysis
3S
n
o
Nodal analysis uses KCL to evaluate the
voltage at each principal node in the 2 A Es
3A
2S
4S
circuit, and is valid for all circuits, both
planar and non-planar. In this case we
m
write down and solve a set of simultaneous
Fig.
30.
A
Three-Node
Circuit
equations in terms of the unknown voltage
at each node. If the circuit has n principal
nodes, we need (n -1) simultaneous equations to solve the circuit. We can obtain (n-1)
simultaneous equations by applying KCL to each non-reference node in turn.
2 = 2Vl + 3Vl2 = 2Vl + 3(Vl –V2) = 5Vl -3V2
-3 = 3V21 + 4V2 = 3(V2 –V1) + 4V2 = -3V1 + 7V2
By solving two equations above we get V1 = 0.192 V and V2 = -0.346 V.
General rules for writing nodal equations
p
Consider the circuit shown in Fig. 31.
IA = (GA + GE)V1 -GAV2
IC
GC
IB - IA – IC = -GAV1 + (GA + GB + GC)V2 -GCV3
GB
m
o
IC = -GCV2 + (GC + GD)V3
These equations can be rewritten as
GD
IB
IA
I1 = G11 V1 + G12V2 + G13V3
I2 = G21 V1 + G22V2 + G23V3
GA
GE
n
Fig. 31. Four Node Network
I3 = G31V1 + G32V2 + G33V3
In general
Gij = (-1) x The conductance linking node i to node j.
Gii = Total conductance terminating on node i.
Ii = The sum of currents entering node i.
The conductance matrix is a square matrix, and is symmetrical about the major diagonal, that
is for all ij (i ≠ j), Gij = Gji that is, G12 = G21, G23 = G32 etc.
In the case of a bilateral network, all the elements on the major diagonal of the conductance
matrix are positive. The elements not on the major diagonal are negative or zero.
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QE 108: Electricity – Section 2 – Circuits
Depending on the circuit, there are either three or four steps to be carried out:
1. Draw a carefully labeled circuit diagram.
2. Mark the principal nodes on the circuit, and select a reference node. If there are n
principal nodes, (n -1) simultaneous equations are needed to solve the circuit.
3. If the circuit contains only independent current sources, apply KCL to each nonreference node. If the circuit contains dependent current sources, relate the source
current to the unknown node voltages.
4. If the circuit contains ideal voltage sources, we cannot deal with it in the normal way
because its internal resistance is zero. In that case, replace each voltage source by a
short-circuit (the voltages assigned in step 2 should not be changed). Each source
voltage should then be related to the unknown node voltages.
If the circuit contains a practical voltage source, it can be converted to its equivalent practical
current source and dealt with as a normal current source
Example 4: Consider the circuit shown in Fig. 32:
o
(i) V1 is
(a) 0.089 V (b) 0.096 V (c) 0.150 V (d) 0.174 V
3S
2S
3A
(ii) V2 is
(a) 0.008 V (b) 0.009 V (c) 0.010 V (d) 0.011 V
2A
4S
n
p
(iii) V3 is
(a) 0.189 V (b) 0.284 V (c) 0.376 V (d) 0.432 V
6S
5S
Solution may be obtained by solving the following
simultaneous equations.
m
Fig. 32. Network for Example 4
0 = 12Vl - 2V2 - 4V3
-1 = -2Vl + 5V2 - 3V3
o
3 = -4Vl - 3V2 + 12V3
3S
2S
3A
Example 5: Consider the circuit shown in Fig. 33:
2A
(i) V1 is
n
4S
(a) 1.4 V (b) 2.0 V (c) 2.5 V (d) 3.2 V
6S
(ii) V2 is
(a) -1.4 V
(b) -1.0 V (c) –0.8 V (d) –0.6.2 V
Super
Node
5S
m
(iii) V3 is
(a) -1.4 V
4V
p
(b) -2.0 V (c) -2.5 V (d) -3.2 V
Fig. 33. Network for Example 5
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QE 108: Electricity – Section 2 – Circuits
Solution may be found by solving the following simultaneous equations.
Total current leaving super node = 2(V1 -V2) + 6(V1 -Vo) + 3(V3 -V2) + 5(V3 -Vo)
Total current entering super node = 3 A
3 = 2(V1 -V2) + 6V1 + 3(V3 -V2) + 5V3 = 8V1 – 5 V2 + 8V3
-1 = -2Vl + 5V2 -3V3
4 = V1 –V2
Example 6: Consider the circuit shown in Fig. 34:
3S
o
(i) V1 is
2S
(a) 0.071 V (b) 0.081 V (c) 0.091 V (d) 0.101 V
3A
(ii) V2 is
(a) 0.015 V (b) 0.016 V (c) 0.017 V (d) 0.018 V
2A
4S
n
(iii) V3 is
1.5 V23
6S
(a) 0.275 V (b) 0.315 V (c) 0.357 V (d) 0.387 V
p
5S
Solution may be found by solving the following
simultaneous equations.
m
Fig. 34. Network for Example 6
0 = 12V1 - 0.5V2 - 2.5V3
-1 = -2V1 + 5V2 - 3V3
3 = -4V1 - 3V2 + 12V3 – 1.5 (V2 - V3) = -4V1 – 4.5V2 + 13.5V3
1.2.4. Loop Analysis
Network topology
It was stated earlier that mesh current analysis is
applicable only to planar networks. However there
is a similar approach -known as loop analysis,
which allows us to solve non-planar networks.
m
2Ω
6Ω
8V
p
10 V
7Ω
n
4Ω
5Ω
Network topology is concerned with a
mathematical discipline known as graph theory
with special reference to electrical circuits. The
'graph' referred to here is not a conventional graph,
but is a collection of points (nodes) and
connecting lines (branches). When drawing the
'graph' of a network, the nature of the element in
the branch between a pair of nodes is suppressed,
and is replaced by a line or 'edge'.
Fig. 35 shows a four nodes and six branches
circuit and its graph.
3Ω
o
m
p
n
o
Fig. 35. A Circuit and its Graph
Given a graph, we define a tree ( or spanning tree)
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QE 108: Electricity – Section 2 – Circuits
as any set of branches in the graph which connect every node to all other nodes in the graph,
but not necessarily directly. Moreover, the tree does not contain a loop of any kind.
If the graph has N nodes, each tree has (N-l) branches in it. The four-node graph in Fig. 15
contains sixteen trees, one of which is shown in full line in Fig. 16. In loop current analysis,
we select a normal tree, that is one containing all the voltage sources in the network, together
with the maximum number of voltage-controlled dependent sources.
A cotree is a set of branches which do not belong to a tree; the cotree corresponding to the
tree in Fig. 16 is shown in broken lines. A branch in a cotree is known as a link. A cotree is
the complement of a tree, and a tree and its cotree form the complete graph of a network. An
N-node network contains a number of trees, each with (N-l) branches; if B is the number of
branches in the network, and L is the number of links in the cotree, the relationship between
them is
B = L + (N -1)
If we re-position anyone of the links from the cotree in the tree, we will form a loop in the
tree.
Since adding a link to the tree forms a loop, the number of links is equal to the number of
independent voltage equations we need to form the loop equations of the network.
Loop analysis
The following steps allow us to write a set of loop current equations for a circuit:
1. Draw a graph of the network and identify a normal tree.
2. Ensure that all voltage sources and, if possible, all control-voltage branches for voltagecontrolled dependent sources are in the tree.
3. Ensure that all-current sources and, if possible, all control-current branches for currentcontrolled dependent sources are in the cotree.
4. Reposition in the tree, one at a time, each link in the cotree. Using KVL, write down for
each loop the associated loop current equation; solve the equations.
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QE 108: Electricity – Section 2 – Circuits
IA + IB + IC
1™
2
š
IB + IC 4 Ω
IB
IC
2Ω
›
4
6

›4
6

ID
(b)
8Ω
œ
5
5V
š
2
IB
3ž
›
4
(b)
(a)
1
™
IA – ID
IA
›4
1
™
œ5
œ
5
ID
6Ω
2
š
IA
ž
3
IA + IB - ID
7Ω
ž
3
1
™
2
š
5Ω
3Ω
3A
1
™
3ž
š2
IC
6
(a)
š2
›4
6

(c)
5
ID œ
(d)
IA = 3 A
0 = 7IB + 5(IA + IB - ID) + 4(IA + IB+ Ic) + 3(IB + IC)
0 = 5 + 5(IA + IB + IC) + 3(IB + IC) + 2IC
0 = 5 - 5(IA + IB – ID) - 6(IA - ID) + 8ID
Solving the above equations gives
IB = -0.524 A, IC = -1.482 A, ID = 1.336 A
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QE 108: Electricity – Section 2 – Circuits
TUTORIAL 1
1. The repetitive waveform of current entering a circuit is
0.5 A
shown in Figure 1.
(a) What is the value of the current at
t = 0.01
s?
t (ms)
0
(b) What charge enters the circuit between t = 80
100
ms and t =150 ms?
200
Figure 1
(c) What total charge has entered the circuit at t =
210 ms?
10Ω
B
2. The electrical charge entering a terminal in a circuit is 10 sin
100πt mC
-30V
(a) What charge has entered between –3 ms and 3 ms ?
(b) Calculate the current at t =2ms
10V
D
A
3. In Figure 2 calculate VBC , VCA,VBD,VEB and VDC
-100V
20Ω
25Ω
E
4. Calculate the power consumed by each resistor in Figure 2.
B
5. Calculate the power absorbed by each of the circuit elements in
40Ω
Figure 3.
Figure 2
6. The resistance, R, of a conductor in ohms is given by the equation
R=ρL/A where ρ is the resistivity (or volume
length of the conductor in m,and A is the cross–
sectional area of the conductor in m2 .if a
0.2 A
conductor of resistivity 0.027 µΩ m,which is 100
km long and of diameter 1 cm carries a current of
9e-5t
6V
2A
+
20 A, calculate the p.d.across the length of the
wire ,the power consumed in the wire and the
20 V
12 V
resistivity) of the conductor in Ω m, L is the
-10 V
5A
-
energy consumed in 20 mins.
7. For the circuit in Figure 4. Calculate
-4.25 A
(a) I,
5Ω
Figure 3
10 Ω
(b) V and
(c) the power absorbed by each
20 V
5A
element. The calculation should
verify
the
principle
V
of
conservation of energy.
Department of Electrical and Electronic Engineering – University of Peradeniya – 2005
30 V
R
Figure 4
Page 21
QE 108: Electricity – Section 2 – Circuits
8. In Figure 6, calculate V and the power supplied by the 50 mA source.
9. Calculate the resistance between A and B in Figure 7
2Ω
I
10. If 10 V is applied between A and B in Figure 7, calculate the
10 VX
10 V
voltage across and the current in each element.
3Ω
VX
Figure 5
A
20 mA
50 mA
2 kΩ
2Ω
B
8Ω
3Ω
1 kΩ
V
4Ω
Figure 6
7Ω
6Ω
5Ω
Figure 7
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QE 108: Electricity – Section 2 – Circuits
2. Network Theorems
2.1. Thevenin’s Theorem
A linear bilateral two-terminal network can be replaced by a Thevenin equivalent circuit
consisting of a voltage source and a resistor connected in series.
RTh
a
a
A
B
b
B
VTh
b
VTH is the open circuit voltage between terminals “a” and “b” with part B disconnected. RTH
is the resistance of part A between terminals “a” and “b” with all idea voltage sources short
circuited ideal current sources open circuited and partial sources represented by their internal
resistance (deactivation of independent sources).
a
A
Is
Vs
b
RTh = Vs/ Is
With independent
Source deactivated
a
A
Isc
RTh = VTh/ Isc
b
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QE 108: Electricity – Section 2 – Circuits
2.2. Norton’s Theorem
A linear bilateral two-terminal network can be replaced by a Norton Equivalent Circuit
consisting of a current source and a conductance connected in parallel.
The Thevenin Equivalent Circuit can be replaced by an equivalent current source in parallel
with a conductance.
IN =VTh / RTh = ISC
GN = 1 / RTh
a
GN
IN
b
2.3. Maximum Power Transfer Theorem
The maximum power is received by a load from a linear bilateral dc circuit when the load
resistance is equal to the RTH of the circuit.
RTh
VTh
PRL =
RL
VTH2
R
(RTH + R)2
R
dPRL = VTH2 d
dR
dR (RTH + R)2
= VTH (RTH + R)2 – 2 R (RTH + R)
(RTH + R)4
=
(RTH + R) (RTH + R-2R) VTH
(RTH + R)4
i.e R = RTH
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QE 108: Electricity – Section 2 – Circuits
2.4. Y-∆ and ∆- Y Transformation
B
A
C
RA
RCA
RB
RBC
RC
A
C
B
RAB
∆- Y Transformation
RA =
RAB RAC
RAB + RAC + RBC
RB =
RBA RBC
RAB + RAC + RBC
RC =
RCA RCB
RAB + RAC + RBC
Y-∆ Transformation
RAB =
RARB + RARC + RBRC
RC
RAC =
RARB + RARC + RBRC
RB
RBC =
RARB + RARC + RBRC
RA
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QE 108: Electricity – Section 2 – Circuits
2.5. Examples
2.5.1. Consider the circuit shown below. Find the value of the resistor which when connected
across terminal a and b draws a current of 5A.
Answer
5Ω
6Ω
a
VTH = 80V
∴Rtotal = 80/10 = 16
20 Ω
100V
RTH = 10Ω
∴RL = 6Ω
b
2.5.2. Find the Thevenin Equivalent Circuit of the following circuit
Answer
VTH = 10/1000 x 30 x 500
= -150V
1 kΩ
10V
30 IB
IB
ISC = 30 x IB = 30 x 10/1000 A
= 3/10 A
500 Ω
∴RTH = 150/3 * 10
= 500 Ω
2.5.2. Find the NOrton Equivalent Circuit of the following circuit
Answer:
2 kΩ
1V
25 IB x 40 = -VC
((1-0.0004VC)/2000) x 25 x 40 = -VC
VC = -625
ISC = 25 x IB = 25x (1/2000) = 1/80
∴ IN = 12.5 mA
∴GN = ISC / VC = 0.02 mS
IB
0.0004VC
40
VC
25IB
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QE 108: Electricity – Section 2 – Circuits
TUTORIAL 2
1. If, in Figure 1 ,V1 = 10 V, V2 = 20 V, R1 = 1 Ω , R2 = 2 Ω, R3= 3
R1
Ω, R4= 4 Ω and R5 = 5 Ω, calculate the mesh currents.
R2
2. Calculate voltages across 2 S and 4 S elements in Figure 2 and
V2
compute the total power consumed.
V1
3. Using mesh analysis, calculate I1 and I2 in Figure 3.
R3
4. Using nodal analysis, calculate V1 and V2 in Figure 3.
R5
R4
5. The mesh equations of a network are as follows:
Figure 1
− 10.7
 11 − 4 − 2 0   I 1 
 19.9 

 

 = − 4 14 0 − 3  I 2 
 − 5.3 
− 2 0 17 − 8  I 3 



 
 − 1.1 
 0 − 3 − 8 17   I 4 
Draw the corresponding circuit diagram.
V1
2A
6Ω
V2
3S
n
2 A Es
o
2S
3A
4S
m
4Ω
Figure 2
5Ω
7Ω
I1
I2
6Ω
10 V
6V
4Ω
Figure 3
7. Using mesh analysis, calculate v in Figure 4.
4V
3Ω 5Ω
v
8. Use nodal analysis to calculate v in Figure 5.
Figure 4
9. In Figure 4,the 3 Ω, 5 Ω and 6 Ω resistors from a tree. Use loop
v
analysis with respect to this tree to calculate v.
10. Use mesh analysis to calculate v in Figure 5.
6Ω
2A
2Ω
11. For the circuit in Figure 6, use mesh analysis to calculate
(a) the voltage gain V2 / V1
(b) the input resistance (V1/ I1) of the circuit.
12. For the simplified emitter follower equivalent circuit in Figure 7,
3A
3Ω
5Ω
use mesh analysis to calculate
(a) the voltage gain of the circuit (V2 / V1)and
(b) the input resistance (V1/ IB).
Figure 5
13. Use nodal analysis to solve problem 11.
14. Solve problem 12 using nodal analysis.
15. Use nodal analysis to calculate I1, I2 and I3 in the circuit in Figure 8.
16. Use nodal analysis to determine the node voltages V1and V2 in Figure 9.
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QE 108: Electricity – Section 2 – Circuits
3Ω
100 kΩ
I1
V1
1500 Ω
5Ω
4Ω
6Ω
10 VX
2Ω
VX
100×10-6 V2
V2
IB
V1 = 0.1 V
Figure 6
50 µS
70IB
Figure 7
3Ω
I1
V2
5Ω
-10 A
I2
2Ω
I3
4Ω
2A
3V1
5Ω
V2
2Ω
V1
10 V
V1
3Ω
15 A
4Ω
Figure 8
Figure 9
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Page 28
10 kΩ
QE 108: Electricity – Section 2 – Circuits
Annex-1
Material
Silver
Copper
Aluminum
Tungsten
Iron
Platinum
Manganin
Lead
Mercury
Nichrome
(Ni,Fe,Cr alloy)
Constantan
Carbon*
(graphite)
Germanium*
Silicon*
Glass
Quartz
(fused)
Hard rubber
Temperature
Conductivity σ x 107
Resistivity ρ (ohm m) coefficient per degree
/Ωm
C
1.59
1.68
2.65
5.6
9.71
10.6
48.2
22
98
x10^-8
x10^-8
x10^-8
x10^-8
x10^-8
x10^-8
x10^-8
x10^-8
x10^-8
0.0061
0.0068
0.00429
0.0045
0.00651
0.003927
0.000002
...
0.0009
6.29
5.95
3.77
1.79
1.03
0.943
0.207
0.45
0.1
100
x10^-8
0.0004
0.1
49
x10^-8
...
0.2
Mar-60
x10^-5
-0.0005
...
1-500
0.1-60
1-10000
x10^-3
...
x10^9
-0.05
-0.07
...
...
...
...
7.5
x10^17
...
...
1-100
x10^13
...
...
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Page 29