EE 287 CIRCUIT THEORY
DR ELVIS TWUMASI
LECTURER
ELVIS TWUMASI (PhD)
Department Electrical & Electronic Engineering
Email- etwumasi.coe@knust.edu.gh
Phone- 0248656932
Office – Room 408A Caesar Building
Teaching Assistant
Opoku Tawiah Bernard
0247720760
Oppong Albert
0242304639
Course Content
Network
Theorems in
DC networks
First and Second
Order Circuits- RL RC
and RLC
Network
Theorems
in AC
networks
Frequency Domain
Circuit Analysis –
Laplace Transform
Active Circuits Operational
Amplifiers
Two-port networks
Frequency
Response of
Circuits
4
Course Objectives
This course is designed to enable students develop problem solving skills and
understanding of circuit theory through the application of techniques and
principles of electrical circuit analysis to common circuit problems.
5
Grading System
•
Exams
70%
•
Quiz
10%
•
Assignments
5%
•
Mid-Semester Exams
10%
•
Attendance
5%
Software: Multisim
6
Recommended Textbooks
1. Engineering Circuit Analysis, William H. Hayt, and Jack E. Kemmerly,
International Student Edition
2. 2. Shaum’s Outline of Theory and Problems of Basic Circuit Analysis, John
O’malley, Ph.D
7
Classroom Norms
1. No phone usage
2. No eating
3. No noise-making
4. No lateness
5. No intimidation
6. No sleeping
7. Daily Word
8
1
NETWORK THEOREMS IN
DC NETWORKS
9
Electronic Components
Electronic components can be classified into:
1. Active Elements
2. Passive Components
These are elements that produce
energy in the form of current or
voltage.
These are elements that use energy.
Examples: Resistors, Inductors,
Capacitors
Examples: Generators, batteries,
Operational amplifiers, Transistors,
Solar Cells
10
Types of Active Elements
1. Independent Voltage Source
It is a two-terminal element that maintains a specified voltage between its
terminals regardless of the current through it.
A
A
v(t)
B
Figure 1: General Symbol
V
B
Figure 2: Constant Symbol
Figure 1 shows the symbol for time varying voltage (General symbol). The voltage
v(t) is referenced positive at A.
11
Types of Active Elements
2. Independent Current Source
It is a two-terminal element that maintains a specified current regardless of the
voltage across its terminals.
A
i(t)
B
Figure 3: General Symbol
Figure 3 shows the general symbol, and the arrow indicates the direction of the
current source when it is positive.
12
Types of Active Elements
3. Dependent or Controlled Voltage Source
A dependent or controlled voltage source is a
voltage source whose terminal voltage
depends on, or is controlled by a voltage or a
current at a specified location in the circuit.
Figure 4 VCVS
The two types are:
•
Voltage-Controlled Voltage Source (VCVS)
controlled by a voltage.
•
Current-Controlled Voltage Source (CCVS)
controlled by a current.
Figure 5 CCVS
13
Types of Active Elements
4. Dependent or Controlled Current Source
A dependent or controlled current source is a current
source whose current depends on, or is controlled by a
voltage or current at a specified location in the circuit.
The two types are:
•
Voltage-Controlled Current Source (VCCS)
controlled by voltage.
•
Current-Controlled Current Source (CCCS)
controlled by current.
The dependent sources are very important because they
form part of the mathematical models used to describe
the behaviour of many electronic circuit devices.
Figure 6 VCCS
Figure 7 CCCS
14
Example 1
Find Vo and Io.
(a) Vo = 20Vs = 20×2 = 40 V
(b) Io = 50Is = 50×1 = 50 mA
15
1. SUPERPOSITION THEOREM
“
Superposition theorem states that the current through, or the
voltage across an element in a linear network is equal to the
algebraic sum of the currents or voltages produced by each
source acting alone.
16
Steps to apply the Superposition Principle
1. Turn off all independent sources except one source. Find the output (voltage or
current) due to that active source using some of the basic techniques. (Ohm’s Law,
Kirchhoff's Laws, Current Divider rule)
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to the
independent sources.
Vo = Vo1 + Vo2 + Vo3
or
Io = Io1 + Io2 + Io3
N.B. Superposition can be used for linear circuits containing dependent sources. However,
it is not useful in this case because the dependent source is never made zero.
17
Two things we have to keep in mind:
1. We consider one independent source at a time while all other independent
sources are turned off.
Turning off all other independent sources means:
•
Independent voltage sources are replaced by 0 V - Short Circuit
•
Independent current sources are replaced by 0 A - Open Circuit
2. Dependent sources are left intact because they are controlled by circuit
variables.
18
Example 1
Determine Vo in the circuit below using the Superposition principle
Let Vo = Vo1 + Vo2 + Vo3,
where Vo1, Vo2, and Vo3 are due to the 4-A, 24-V, and 38-V sources respectively.
19
For Vo1, we keep the independent 4-A current source and turn off all other
independent sources as shown in the circuit below
From the circuit above,
6||3 = 2 Ω, 4||12 = 3 Ω
Hence, Io =
5
5+5
4
×4= 2=2A
Vo1 = 5Io = 10 V
20
For Vo2 we keep the independent 24 V voltage source and turn off all other
independent sources as shown in the circuit below
From the circuit above,
12||4 = 3
3||8 =
V1 =
24
11
24
11
24
+6
11
× 24 = 6.4 V
21
Vo2 =
5
5+3
× V1 =
5
8
× 6.4 = 4 V
For Vo3, we keep the independent 38-V voltage source and turn off all other
independent sources as shown in the circuit below
6||3 = 2
7||12 =
84
19
Ω
22
V2 =
84
19
84
+4
19
Vo3 = −
5
2+5
× 38 = 19.95 V
× V2 = −
5
7
× 19.95 = −14.25V
Therefore, Vo = 10 + 4 – 14.25 = –0.25 V
23
Example 2
Use superposition to find Vo in the circuit
Current source acting alone:
Current in the 6 k: using the current divider rule,
(1+2)
3×2
2
×2=
= mA
1+ 2 + 6
9
3
2
Vꞌo = mA × 6k = 4 V
3
Iꞌo =
24
Voltage Source acting alone:
Io ꞌꞌ =
3
3
= mA and
1+2+ 6 9
Vo ꞌꞌ =
3
3
= mA × 6k = 2 V
1+2+ 6
9
Therefore Vo = Vo ꞌ + Vo ꞌꞌ = 4 + 2 = 6 V
25
Example 3
Using the Superposition principle, determine the current through the 4 Ω resistor
With the 54V source acting alone:
Total resistance = 24 + 12//4 = 24 +
Total Current =
12×4
= 27 Ω
16
54
=2A
27
26
Using current divider rule, current in the 4 Ω 𝐫𝐞𝐬𝐢𝐬𝐭𝐨𝐫:
Iꞌ =
12
12 × 2
×2=
= 1.5 A
12 +4
16
With the 48 V source acting alone:
Total resistance = 4 + 24//12 = 4 +
24 × 12
= 12 Ω
36
Total current = Current in the 4 Ω resistor,
Iꞌꞌ =
48
=4A
12
Since Iꞌ and Iꞌꞌ are not in the same direction through the 4 k resistor,
the actual current
I = Iꞌ - Iꞌꞌ = 4 – 1.5 = 2.5 A (Actual current is in the direction of Iꞌꞌ)
27
Example 4
Using the Superposition principle, find the current through the 12 k resistor
28
Solution
With the current source acting alone:
6 k and 12 k are in parallel
Iꞌ =
6
6×6
×6=
= 2 mA
12+ 6
18
With the 9 V source acting alone:
Iꞌꞌ =
9
9
= = 0.5 mA
6+12
18
I = Iꞌ + Iꞌꞌ = 2 + 0.5 = 2.5 mA
29
2. NODAL ANALYSIS
“
The nodal analysis is a method used to solve circuits containing
multiple nodes and loops.
This is based on Kirchhoff's Current Law (KCL) and Ohm’s Law.
30
Definition of terms used
•
A branch: It is a portion of a circuit containing a single element or more
elements in series.
•
A node: It is a point of connection of two or more branches or circuit
elements.
Note that all the connecting wire in unbroken contact with the point is part of
the node.
31
Case 1
“
Circuits that do not contain Voltage
Sources.
32
General Approach
Nodal Analysis Procedure
Given a circuit with n nodes without voltage sources
33
•
The reference node is commonly called the ground.
•
Assign voltage designations to the non-reference nodes.
•
Nodes 1 and 2 are assigned voltages v1 and v2 respectively.
34
Apply KCL to each non-reference node in the circuit
•
To be able to apply KCL, we need to know the direction of currents at each
node
•
Note that current flows from a HIGHER potential to a LOWER potential in a
resistor
35
At node 1, applying KCL gives
I1 = I 2 + i1 + i2
(1)
At node 2,
I2 + i 2 = i3
(2)
In nodal analysis if
•
We assign node voltages va and vb, the branch current i flowing from a to b
is then expressed as
i =
va− vb
R
36
We now apply Ohm’s law to express the unknown current i1, i2 and i3 in terms
of node voltages.
v −0
R1
i1 = 1
v − v2
R2
i2 = 1
v −0
R3
i3 = 3
(3)
Substituting (3) in (1) and (2) yields
I1 = I 2 +
v1
v −v
+ 1 2
R1
R2
v − v2
v
= 2
R2
R3
I2 + 1
Solve for the node voltages using the
•
Substitution method
•
Elimination method
•
Cramer’s rule
37
Example 1
Calculate the node voltages in the circuit shown below
38
Consider the circuit below prepared for nodal analysis:
•
We assigned voltages v1 and v2 to nodes 1 and 2
•
Current directions as shown on the circuit
39
Applying KCL at node 1 yields
i1 = i 2 + i 3
Expressing the currents in terms of voltages we’ve
v − v2 v1− 0
+
4
2
5= 1
Multiplying each term by 4, we obtain
20 = v1 − v2 + 2v1
or
3v1 − v2 = 20
(1)
40
Applying KCL at node 2 yields
i2 + i 4 = i 1 + i 5
Expressing the currents in terms of voltages we’ve
v1− v2
v −0
+ 10 = 5 + 2
4
6
Multiplying each term by 12 results in
3v1 − 3v2 + 120 = 60 + 2v2
or
−3v1 + 5v2 = 60
(2)
41
Now we have two simultaneous equations, equation (1) and (2)
Method 1
Using the elimination technique, we add equations (1) and (2)
4v2 = 80,
v2 = 20 V
Substituting v2 = 20 in equation (1) gives
3v1 − 20 = 20
v1 =
40
3
v1 = 13.33 V
42
Example 2
Apply nodal analysis to the network
Apply KCL at node 1:
V1
V −V
+ 1 2
12
6
1
1
V
1 = V1 ×
+ − 2
12
6
6
V
V
1= 1+ 2
(1)
4
6
1=
43
Apply KCL at node 2:
−4 =
−4 = −
V2
V −V
+ 2 1
6
6
V1
1
1
+ V2 × +
6
6
6
−4 = −
V1
V
+ 3
6
3
(2)
44
(1) × 2 + (2): − 2 =
From (1)
V1
, V1 = − 6 V
3
V2
V
6
5
5
= 1 − 1 = − − 1 = − , V2 = − × 6 = −15 V
6
4
4
2
2
Current I2 =
0 − V1
0−(−6)
6
=
= = 0.5 mA
12
12
12
I1 = 1 + I2 = 1 + 0.5 = 1.5 mA or use
V1 − V2
9
= = 1.5 mA
6
6
I3 = 4 − I1 = 4 − 1.5 = 2.5 mA or use
0 − V2
15
=
= 2.5 mA
6
6
45
Example 3 – Matrix Form
iA =
v1
v −v
v −v
1
1
1
+ 1 2+ 1 3=
+ +
R1
R2
R3
R1
R2
R3
0=
v2
v −v
v −v
1
1
1
1
+ 2 1 + 2 3 = − v1 +
+ +
R4
R2
R5
R2
R2
R4
R5
v1 −
1
1
v2 − v3
R2
R3
v − v2
v −v
1
1
1
1
+ 3 1 = − v1 − v2 +
+
R5
R3
R3
R5
R3
R5
−iB = 3
v2 −
v3
1
v
R5 3
46
In matrix form, the equations are as follows:
iA
0 =
−iB
1
1
1
+ +
R1
R2
R3
1
−
R2
1
−
R3
1
R2
1
1
1
+ +
R2
R4
R5
1
−
R5
−
1
R3
1
−
R5
−
1
1
+
R3
R5
v1
v2
v3
The equation is in the form
i = G v
The matrix G is called conductance or admittance matrix
47
The Addition of dependent sources
•
With the addition of dependent sources, a controlling equation is introduced.
•
Identify the equation governing the dependent source and substitute.
Example 4
Determine the voltages at the nodes of the circuit shown below
48
We assign
•
voltages to the three nodes and
•
direction of currents at the nodes as shown in the circuit below
At node 1
3 = i1 + ix,
3=
v1− v3
v −v
+ 1 2
4
2
49
Multiplying by 4 and rearranging terms, we get
3v1 − 2v2 − v3 = 12
(1)
At node 2
ix = i2 + i3,
v1− v2
v −v
v −0
= 2 3+ 2
2
8
4
Multiplying by 8 and rearranging terms, we get
−4v1 + 7v2 − v3 = 0
(2)
At node 3
i1 + i2 = 2ix
v1 − v3 v2 − v3
+
= (v1 − v2)
4
8
50
Multiplying by 8, rearranging terms, and dividing by 3, we get:
2v1 − 3v2 + v3 = 0
(3)
Solving equations (1), (2) and (3)
v1 = 4.8 V
v2 = 2.4 V
v3 = − 2.4 V
51
Example 5
Obtain the nodal equations for the network given below
Apply KCL at node 1:
−βio =
v1
v −v
v
v
v −v
+ 1 2 or −β 2 = 1 + 1 2
R1
R2
R3
R1
R2
or
0=
1
1
+
R1
R2
v1 −
β
1
−
𝑅2
𝑅3
v2
52
Apply KCL at node 2:
iA =
v2
v −v
1
1
1
+ 2 1 = − v1 +
+
v
R3
R2
R2
R2
R3 2
0=
1
1
+
R1
R2
iA = −
v1 −
β
1
−
R2
R3
v2
1
1
1
v1 +
+
v
R2
R2
R3 2
In matrix form, we have:
0
=
iA
1
1
+
R1
R2
1
−
R2
−
β
1
−
R2
R3
1
1
+
R2
R3
v1
v2
53
Example 6
Obtain the nodal equations for the network given below
Apply KCL at node 1:
iA = G3v1 + G1(v1 − v2) or iA = (G1 + G3)v1 − G1v2
54
Apply KCL at node 2:
−iA − αvx = G1(v2 − v1) + G2(v2 −v3)
−iA = G1(v2 − v1) + G2(v2 −v3) + α(v2 − v3)
−iA = −G1v1 + (G1 + G2 + α) v2 − (G2 + α) v3
Apply KCL at node 3:
iB = G2(v3 − v2) + G4v3 = −G2v2 + (G2 + G4)v3
At node 1:
iA = (G1 + G3)v1 − G1v2
At node 2:
−iA = −G1v1 + (G1 + G2 + α)v2 − (G2 + α)v3
55
At node 3:
iB = −G2v2 + (G2 + G4)v3
In matrix form, we have:
v1
G1 + G3
−G1
0
iA
G1 + G2 + α G2 + α v2
−iA = −G1
iB
0
−G2
G2 + G4 v3
56
Case 2
“
Circuits containing Voltage Sources.
57
Circuits containing Voltage Sources
The presence of voltage sources reduces the number of equations and the
number of unknowns by one per voltage source.
One of the following two approaches may be used:
•
The concept of supernode.
•
Converting voltage sources to current sources.
NB: Converting voltage sources to current sources is applicable when the
independent voltage sources have resistances in series.
58
Method 1 - The Concept of Supernode
Voltage sources are enclosed in separate regions. Each closed region is called a
supernode.
A supernode contains two nodes which may consist of
•
a non-reference node and a reference node
or
•
two non–reference nodes.
59
The Concept of Supernode
To obtain the nodal equations, we apply KCL to
•
all supernodes not containing the reference node and
•
to all other reference nodes.
KCL is applied to a supernode using the generalized form of KCL which states
that “the algebraic sum of currents entering a closed region is zero.”
60
Example 1
Consider the circuit below
• The 10 V voltage source is
connected between v1 and the
reference node.
• Thus v1 is equal 10 V.
• The 5 V voltage source is
between two non-reference
nodes.
• v2 and v3 and v1 and the
reference node form two
supernodes.
61
If the voltage source is connected between two non-reference nodes:
•
the two non-reference nodes form a supernode.
•
we apply both KCL and KVL to determine the node voltages.
In the circuit shown earlier
•
Nodes 2 and 3 form a supernode
•
KCL at the supernode is:
i1 + i4 = i2 + i3
or
v1 − v2 v1 − v3
v2 v3
+
=
+
2
4
8
6
62
To apply KVL to the supernode we redraw the circuit as shown below:
Going around the loop in the clockwise direction gives:
−v2 + 5 + v3 = 0
v2 − v3 = 5
63
Example 2
For the circuit shown below find the node voltages
64
•
The supernode contains the 2 V source, nodes 1 and 2 and the 10-Ω
resistor.
•
Applying KCL to the supernode as shown in the circuit below gives:
2 = i1 + i2 + 7
(1)
65
Expressing i1 and i2 in terms of node voltages
v −0
v −0
+ 2
+7
2
4
2= 1
8 = 2v1 + v2 + 28
or
2v1 + v2 = −20
(2)
66
•
To get the relationship between v1 and v2 ,we apply KVL to the circuit
shown below:
•
Going round the loop, we obtain:
−v1 − 2 + v2 = 0
v2 = v1 + 2
(3)
67
From equations (2) and (3), we write
v2 = v1 + 2 = −20 − 2v1
or
3v1 = −22,
v1 = −7.333 V
and
v2 = v1 + 2 = −7.333 + 2
v2 = −5.333 V
NB: Nodal analysis applies KCL to find unknown voltages in a given circuit
68
Example 3
Obtain the nodal equations for the network given below
69
Two supernodes A and B are shown
70
Voltage V1 at node 1 in supernode A (containing the reference node) may be
determined immediately as
V1 = 5 + 0 = 5 V
Voltage V5 at node 5 in supernode B can be expressed in terms of V4 at the
other non-reference node 4 in the supernode as
V5 = V 4 + 4
The number of nodal equations required = 5 – 2 = 3
71
Apply KCL to non-reference node 2:
6 = 1(V2 – 5) + 4(V2 – V3) + 2(V2 – V4 – 4)
19 = 7V2 – 4V3 – 2V4
Apply KCL to non-reference node 3:
0 = 3V3 + 4(V3 – V2) + 2(V3 – V4)
0 = –4V2 + 9V3 – 2V4
Apply KCL to supernode B:
0 = 2(V4 – V3) + 1(V4 + 4) + 2[(V4 + 4) – V2]
–12 = –2V2 – 2V3 + 5V4
72
At node 2:
19 = 7V2 – 4V3 – 2V4
At node 3:
0 = –4V2 + 9V3 – 2V4
At supernode B:
–12 = –2V2 – 2V3 + 5V4
In matrix form, we have:
7 –4 –2 𝑉2
19
0 = –4 9 –2 𝑉3
–2 –2 5 𝑉4
–12
73
Example 4
Find V and I in the circuit using nodal analysis
Network has two supernodes and we must apply KCL to the supernode which
does not contain reference node.
74
The supernodes are shown below
75
The voltage:
6=
V
V+3
+
+
4
2
– 20
V+3
6
6 × 12 = 3V + 6(V + 3) + 2[(V + 3) – 20]
72 = 11V – 16 or 11V = 88
Therefore: V = 8 V
The current:
I=
– 20
𝑉+3
I=–
6
=
8+3 – 20
6
9
6
I = 1.5 mA
76
Example 5
Find the current Io in the circuit below
77
78
Applying KCL to the non-reference node, we obtain:
0=
V–6
V
V–3
+ +
6
3
6
0 = (V – 6) + 2V + V – 3
0 = 4V – 9
V=
9
𝑉
4
The current:
Io =
V
9
3
=
=
3
4× 3
4
Io = 0.75 mA
79
Method 2 - Converting voltage sources to current sources
Find the voltage across the 3 Ω resistor by nodal analysis
80
Converting sources and choosing nodes, we obtain:
81
By inspection, we have
4=
1
1
1
+
+
2
4
6
1
𝑉1 – 6 V 2
1
6
1
1
1
+ +
6
3
10
–0.1 = – V1 +
𝑉2
or
11
1
4 = 12 𝑉1 – 6 V2
1
3
–0.1 = – 6 V1 + 5 𝑉2
Solving the two equations simultaneously, we obtain voltage across the 3 Ω
resistor, V2 = 1.101 V
82
The addition of dependent voltage sources
•
Treat them in the same manner as circuits containing independent voltage
sources.
•
When writing the circuit equations, first treat the dependent source as
though it were an independent source and then write the controlling
equations and substitute.
83
Example 6
Find the current Io in the network
84
The supernodes are shown below
85
Apply KCL to the supernode A:
V +2Vx
V +2Vx – 6
V
V –6
0 = 2 12 + 2 6
+ 62 + 212
1
1
1
1
+
+
+
12
6
6
12
0=
3
3
V2 +
2
2
+
12
6
6
Vx – 1– 12
18
0 = 6 V2 + 6 Vx – 12
Controlling equation: Vx = V2
86
Substituting into the nodal equation, we obtain:
3
3
18
18
6
0 = 6 V2 + 6 V2 – 12 or 12 = 6 V2
V2 = 1.5 V
Current Io =
Io =
𝑉2+ 2𝑉𝑥
12
3V2
3
3
3
=
× = mA
12
12 2
8
87
Example 7
Find Vo in the circuit using nodal analysis
88
With currents in mA, the CCVS becomes 4Ix
89
Applying KCL to the non-reference node Vo, we obtain:
4=
Vo
V – (–4 I x )
+ o
10
10
V
or
4Ix
4 = 5o + 10 or 20 = Vo + 2Ix
Vo = 20 – 2Ix
The controlling equation is given by:
Ix =
Vo ––4 I x
V +4Ix
= o
10
10
90
𝑉
10Ix = Vo + 4Ix or 6Ix = Vo or 2Ix = 3𝑜
Substituting it in the nodal equation,
Vo = 20 – 2Ix
we obtain
V
4
Vo = 20 – 3o 𝑜𝑟 Vo × 3 = 20
Vo = 15V
91
3. MESH ANALYSIS
“
Mesh analysis uses KVL to determine currents in the circuit.
Once the currents are known, Ohm’s law
can be used to calculate voltages.
N independent equations are required if
the circuit contains N meshes.
92
Mesh Analysis
•
We assume that the circuits are planar. (Circuits that can be drawn on a plane
surface with no wires crossing each other).
•
In the case of non–planar circuits, we cannot define meshes and mesh analysis
cannot be performed.
•
Mesh analysis is thus not as general as nodal analysis, which has no topological
restrictions.
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Definition of terms used
•
A loop: It is a closed path through a circuit in which no node is encountered
more than once.
•
A mesh: A mesh is a special kind of loop that does not contain any loop
within it.
•
We note that as we traverse the path of a mesh, we do not encircle any
circuit element.
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Definition of terms used
•
A planar circuit: It is a circuit that can be drawn on a plane surface with no
crossovers, i.e. no element or connecting wire crosses another element or
connecting wire.
•
In planar circuits meshes appear as windows.
•
A non–planar circuit: It is a circuit that is not planar.
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General Approach to Mesh Analysis
We follow the systematic approach given below:
•
Place a loop current in the clockwise direction within each mesh or
window of the network.
•
Apply KVL around each mesh in the clockwise direction.
•
Solve the resulting simultaneous equations for the loop currents.
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Case 1
“
Circuits containing only Independent
Voltage Sources.
The General Approach or Format Approach can be used
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Independent Voltage Sources
To illustrate the steps, consider the circuit below
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Step 1
Mesh currents i1 and i2 are assigned to meshes 1 and 2
Step 2
Apply KVL to each mesh
Mesh 1: R1i1 + R3(i1 – i2) = V1
(R1 + R3)i1 – R3i2 = V1
(1)
Mesh 2: R2i2 + V2 + R3(i2 – i1) = 0
–R3i1 + (R2 + R3)i2 = –V2
(2)
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Step 3
Solve for the mesh currents, putting equations (1) and (2) in matrix form
yields:
R1 + R3
–R3
i1
V1
=
–R3
R2 + R3 i2
–V2
Note:
•
In equation (1) that the coefficient of i1 is the sum of the resistances in the
first mesh, while
•
The coefficient of i2 is the negative of the resistance common to meshes 1
and 2.
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Example 1
For the circuit below find the branch currents I1, I2 and I3 using mesh analysis
101
We first obtain the mesh currents using KVL
Mesh 1:
–15 + 5i1 + 10(i1 – i2) + 10 = 0
or
3i1 – 2i2 = 1
(1)
Mesh 2:
6i2 + 4i2 + 10(i2 – i1) – 10 = 0
or
i1 – 2i2 = –1
(2)
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Substituting (2) into (1)
6i2 – 3 – 2i2 = 1
i2 = 1 A
From (2):
i1 = 2i2 – 1 = 2 – 1 = 1 A
Thus
I1 = i1 = 1 A
I2 = i2 = 1 A
I 3 = i1 – i2 = 0
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Example 2
Find the current through each branch of the network
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The mesh currents are shown below:
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Apply KVL to mesh 1:
12 = 6I1 + 6(I1 – I2) = 12I1 – 6I2 or
12 = 12I1 – 6I2
(1)
Apply KVL to mesh 2:
–3 = 3I2 + 6(I2 – I1) or
–3 =–6I1 + 9I2
(2)
(1) + 2 × (2) gives:
6 = 12I2
6
I2 = 12 = 0.5 mA and from (2)
6I1 = 9I2 + 3 = 9 × 0.5 + 3
I1 = 1.25 mA
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I1 = 1.25 mA
I2 = 0.5 mA
Current in the left outer branch = I1 = 1.25 mA
Current in the middle branch = I1 – I2 = 1.25 – 0.5 = 0.75 mA
Current in the right outer branch = I2 = 0.5 mA
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Example 3
Find the current through each branch of the network
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Loop 1
–6 = 4I1 + 6(I1 – I3) or
–6 = 10I1 – (0)I2 – 6I3
Loop 2
Therefore in matrix form:
–6
10
6 = 0
0
–6
0
12
–3
–6 I1
–3 I2
21 I3
6 = 3(I2 – I3) + 9I2 or
6 = (0)I1 – 12I2 – 3I3
Loop 3
0 = 6(I3 – I1) + 12I3 + 3(I3 – I2) or
0 = –6I1 – 3I2 + 21I3
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Case 2
“
Circuits containing Current Sources.
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As in the case of nodal analysis with voltage sources, the presence of current
sources reduces the number of unknowns in mesh analysis by one per current
source.
Method 1
Converting current sources to voltage sources.
NB: The sources must have resistances connected in parallel
Method 2
Concept of Supermesh
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Converting current sources to voltage sources
Example 1
Use mesh analysis to determine the currents for the network
112
Convert current sources to voltage sources
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12 + 64 = (2 + 6 + 8)I1
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I1 = 16 = 4.75 A
The currents in the various branches of the circuit:
Current in 6 Ω resistor = 4.75 A
Current in 2 Ω resistor = 6 – 4.75 = 1.25 A
Current in 8 Ω resistor = 8 – 4.75 = 3.25 A
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Example 2
For the network, determine the current I2 using mesh analysis and then find
the voltage Vab
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Convert current source to voltage sources:
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By inspection:
–4 = 14I1– 6I2
15 = –6I1 + 13I2
Solving the two simultaneous equations, we obtain:
I1 = 0.26 A
I2 = 1.274 A
Voltage Vab = V4 – 6
= 4I2 – 6
= 4 × 1.274 – 6
= 0.904 V
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The concept of supermesh
Scenario 1:
When a current source exists only in one mesh:
•
set the mesh current to the value of the current source
i2 = –5A
•
The mesh equation for the other mesh is:
–10 + 4i1 + 6(i1 – i2) = 0, i1 = –2 A
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Scenario 2:
When a current source exists between two meshes
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We create a supermesh by:
•
excluding the current source and any element connected in series with it
•
A supermesh results when two meshes have a (dependent or
independent) current source in common
120
We create a supermesh as the periphery of the two meshes and treat it
differently.
Applying KVL to the supermesh gives
–20 + 6i1 + 10i2 + 4i2 = 0
or
6i1 + 14i2 = 20
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We apply KCL to a node in the branch where the two meshes intersect.
Applying KCL to node 0 and obtain:
i2 = i 1 + 6
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Example 1
Find the current “I” in the circuit below
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KVL for loop 1 gives:
–30 + 2(i1 – i3) + 3(i1 – i2) = 0
or
5i1 – 3i2 – 2i3 = 30
(1)
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KVL for loop 2 gives:
i2 + 3(i2 – i1) + 6(i2 – i4) = 0
or
–3i1 + 10i2 – 6i4 = 0
(2)
KVL for the supermesh yields:
2(i3 – i1) + 4i3 + 8i4 + 6(i4 – i2) = 0
or
–2i1 – 6i2 + 6i3 + 14i4 = 0
(3)
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Applying KCL to a node in the branch where the two meshes intersect
i4 = i 3 + 4
(4)
Solving (1) to (4) by elimination gives
i = i1 = 8.561 A
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4. THEVENIN’S THEOREM
“
Thevenin’s theorem states that any two-terminal linear dc network can
be replaced by an equivalent circuit consisting of a voltage source, VTH
and a series resistance RTH.
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•
VTH is the open-circuit voltage at the terminals
•
RTH is the Thevenin (equivalent) resistance when all the independent
sources are turned off.
•
It often occurs in practice that a particular element in a circuit is
variable (load) while other elements are fixed as shown in the circuit
below
•
Each time the variable element, RL, changed, the entire circuit has to
be analyzed again.
128
•
Thevenin’s theorem helps in the development of a technique by
which the fixed part of the circuit is replaced by an equivalent
circuit.
129
Thévenin Equivalent Resistance
Case 1
If the network has no dependent sources:
•
Turn off all independent voltage and current sources.
•
Compute the total resistance between the load terminal with
the load removed.
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Thévenin Equivalent Resistance
Case 2
If the network has dependent sources:
•
Turn off all independent voltage and current sources.
•
Dependent sources are not turned off.
•
We apply a voltage source vo (1V) at terminals a and b and determine the
resulting current io
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Example 1
Compute the equivalent resistance the load RL ‘sees’ at port a-b for circuit the
below
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We find RTH by short circuiting the 32-V voltage source and open circuiting
the 2-A current source. The circuit becomes what is shown below:
The Thevenin Equivalent Resistance is
RTH = (4||12) + 1 =
4 × 12
+1= 4Ω
16
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Example 2
Compute the Thevenin equivalent resistance of circuit the below
134
•
This circuit contains a dependent source:
•
To find RTH, we set the independent source equal to zero but leave the
dependent source alone.
•
We excite the network with a voltage source vo connected to the terminals
as shown below:
•
We set vo = 1 V to ease calculation
•
We find current io through the terminals and then obtain RTH =
1
io
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Applying KVL to mesh 1 yields:
−2vx + 2(i1 − i2) = 0
or
vx = (i1 − i2)
But
Vx = −4i2 = i1 − i2
Hence
i1 = −3i2
(1)
Applying KVL to meshes 2 and 3 produces
4i2 + 2(i2 − i1) + 6(i2 − i3) = 0
(2)
and
6(i3 − i2) + 2i3 + 1 = 0
(3)
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Solving equations 1, 2 and 3 gives:
1
i3 = − 6 A
But
io = −i3 =
1
6
Hence
1V
RTH = i = 6 Ω
o
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Computing the Thevenin Voltage
The Thevenin equivalent voltage is equal to the open-circuit voltage present at
the load terminal
Procedure
•
Remove the load, leaving the load terminals open-circuited
•
Define the open-circuit voltage voc across the given load terminal
•
Apply any preferred method (eg: node analysis) to solve for voc
•
The Thevenin voltage is vTH = voc
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Example 3
Compute the Thevenin voltage in the circuit the below
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Procedure
•
Disconnect the load as shown below
•
Applying mesh analysis to the two loops we’ve
−32 + 4i1 + 12(i1 − i2) = 0,
i2 = −2 A
•
Solving for i1, we get i1 = 0.5 A
•
Thus, VTH = 12(i1 − i2) = 12(0.5 + 2) = 30 V
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The Thevenin Equivalent circuit is as shown below:
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5. NORTON’S 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
142
•
Where iN is the short-circuit current through the terminals
•
RN is the equivalent resistance at the terminals when the independent
sources are turned off.
RN = RTH
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Computing Norton Current
Procedure
•
Replace the load with a short-circuit
•
Define the short-circuit current isc to be the Norton equivalent current
•
Apply any preferred method (eg: mesh analysis) to solve for isc
•
The Norton current iN = isc
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Example 1
•
Find the Norton equivalent circuit of the circuit below at terminals a-b
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We find
•
RN in the same way we find RTH in the Thevenin equivalent circuit
•
Set the independent sources equal to zero and this leads to
RN = 5 || (8 + 4 + 8) = 5 || 20 =
5×20
=4Ω
25
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To find:
•
IN we short-circuit terminals a and b as shown below:
•
We ignore the 5-Ω resistor because it has been short-circuited
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•
Applying mesh analysis, we obtain:
i1 = 2 A,
•
20i2 − 4i1 − 12 = 0
From these equations, we obtain:
i2 = 1 A = isc = IN
•
Thus the Norton equivalent circuit is as shown below:
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Thanks!
Any questions?
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