M. Afendi M. Piah, PhD
Circuit Theory
y
(SKEE 1023)
Assoc. Prof. Dr. Mohamed Afendi Mohamed Piah
Institute of High Voltage & High Current
Faculty of Electrical Engineering, UTM
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Topics
™ Methods
M th d
off Circuit
Ci it Analysis:
A l i Nodal
N d l
Analysis; Node Voltages; Mesh
A l i
Analysis;
L
Loop
C
Currents;
t
Ci it
Circuit
Analysis with Voltage and Current
I d
Independent
d t
and
d
D
Dependent
d t
Sources.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
METHODS OF CIRCUIT ANALYSIS
Introduction
‰
‰
Two powerful techniques for circuit analysis, ie;
¾
Nodal analysis : based on application of KCL.
¾
Mesh analysis : based on application of KVL.
Analyze
ay e a
any
y linear
ea ccircuit
cu t by obta
obtaining
g a set o
of ssimultaneous
u ta eous equat
equations
o s
that are then solved to obtain the required values of current or voltage.
Nodal Analysis
‰
Provide a general procedure for analyzing circuits using node voltages
as the circuit variables.
‰
Interested in finding the node voltages.
voltages
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Steps
p to Determine Node Voltages
g ((Nodal Analysis)
y )
1. Select a node as the reference node. Assign voltages v1, v2, …. Vn-1 to
the remaining n – 1 nodes.
2 Apply KCL to each of the n – 1 nonreference nodes.
2.
nodes
3. Solve the resulting simultaneous equations to obtain the unknown node
voltages.
‰
The reference (datum) node is commonly called the ground (assumed to
have zero potential).
Symbols :
-Common ground
Ground
-Ground
-Chassis ground
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Method of Nodal Analysis
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
At node
d 1 and
d node
d 2,
2 apply
l KCL gives;
i
I1 = I 2 + i1 + i2 and
I 2 + i2 = i3
v1 − 0
i1 =
= G1v1
R1
v1 − v2
= G2 (v1 − v2 )
i2 =
R2
i3 =
v2 − 0
= G3v2
R3
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
At node 1
⎛1
1 ⎞
1
⎜
⎟
I1 = I 2 + ⎜ +
v1 − v2
R2
⎝ R1 R2 ⎠
⎛1
1 ⎞
1
⎜⎜ + ⎟⎟v1 − v2 = I1 − I 2
R2
⎝ R1 R2 ⎠
or
(G1 + G2 )v1 − G2v2 = I1 − I 2
⎛ 1
1
1 ⎞
⎜
I 2 = − v1 + ⎜ + ⎟⎟v2
R2
⎝ R2 R3 ⎠
At node 2
−
⎛ 1
1
1 ⎞
v1 + ⎜⎜ + ⎟⎟v2 = I 2
R2
⎝ R2 R3 ⎠
or
− G2 v1 + (G2 + G3 )v2 = I 2
INSPIRING CREATIVE AND INNOVATIVE MINDS
7
M. Afendi M. Piah, PhD
Circuit Theory
In matrix form;
1
⎡1
+
⎢R R
2
⎢ 1
⎢ − 1
⎢⎣ R2
or
1 ⎤
R2 ⎥ ⎡ v1 ⎤ ⎡ I1 − I 2 ⎤
⎥
=
1
1 ⎥ ⎢⎣v2 ⎥⎦ ⎢⎣ I 2 ⎥⎦
+
R2 R3 ⎥⎦
⎡G1 + G2
⎢ −G
2
⎣
− G2 ⎤ ⎡ v1 ⎤ ⎡ I1 − I 2 ⎤
=⎢
⎥
⎢
⎥
G2 + G3 ⎦ ⎣v2 ⎦ ⎣ I 2 ⎥⎦
−
Then, determine the values of v1 and v2.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Example 2.1
Calculate the node voltages in the circuit.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Problem 2.1
Obtain the node voltages in the circuit.
Ans: v1 = -2 V
V, v2 = -14 V
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Example 2
2.2
2
Determine the voltages at the nodes.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Problem 2.2
22
Determine the voltages at the nodes.
Ans : v1 = 80 V, v2 = -64 V, v3 = 156 V
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Nodal Analysis with Voltage Sources
‰
How voltage
H
lt
sources affect
ff t nodal
d l analysis.
l i Consider
C
id the
th following
f ll i
t
two
possibilities.
Case 1:
¾
If a voltage source is connected
between the reference node and a
nonreference
f
node;
d ⇒ v1 = 10 V.
V
Case 2:
¾
If the voltage source (dependent or
i d
independent)
d t) is
i connected
t d between
b t
two
nonreference
nodes.
The
nonreference
nodes
form
a
generalized node or supernode.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
¾
Nodes 2 and 3 form a supernode.
¾
There is no way knowing the
current through a voltage source.
source
¾
KCL must be satisfied at a
supernode, i.e.
i1 + i4 = i2 + i3
or
v1 − v2 v1 − v3 v2 v3
+
= +
2
4
8 6
and v2 − v3 = 5
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Example 2.3
For the circuit shown below, find the node voltages.
Applying
A
l i KCL to the
h supernode;
d
2 = i1 + i2 + 7
where,
v
v
i1 = 1 ; i2 = 2 and v2 − v1 = 2
2
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INSPIRING CREATIVE AND INNOVATIVE MINDS
15
M. Afendi M. Piah, PhD
Circuit Theory
Problem 2.3
Find v and i.
Ans: -0.2 V, 1.4 A
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Example 2.4
Find the node voltages in the circuit.
Solution
KCL at supernode (1 - 2)
10 + i3 = i1 + i2
where, i3 =
v3 − v2
v −v
v
; i1 = 1 4 ; i2 = 1
6
3
2
KCL at supernode (3 - 4)
i1 = i3 + i4 + i5
where, i1 , and i3 same as above and;
i4 =
v
v4
; i5 = 3
1
4
Supportive equations :
v1 − v2 = 20 and v3 − v4 = 3v x = 3(v1 − v4 )
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Problem 2.4
24
Find v1, v2 and v3 in the circuit using nodal analysis.
Ans: v1 = 3.043 V, v2 = -6.956 V, v3 = 0.6522
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Mesh Analysis
‰
Another general procedure for analyzing circuits, using
mesh current as the circuit variables.
‰
Definations: Loop is a closed path with no node passed
more than once.
once Mesh is a loop that does not contain any
other loop within it.
‰
Mesh analysis applies KVL to find unknown currents.
‰
Only applicable to a circuit that is planar (two-dimensional).
Nonplanar circuits can be handled using nodal analysis.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Types of circuit
A planar circuit
The same
ccircuit
cu t redrawn
ed a
A nonplanar circuit
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
¾
Paths abefa and bcdeb are meshes. i1
and i2 are mesh current.
¾
In mesh analysis, we are interested in
applying KVL to find the mesh current.
Steps to Determine Mesh Currents
‰
Assign mesh currents i1, i2, …. in to the n meshes. Assume mesh
currents flow clockwise.
‰
Apply KVL to each of the n meshes. Use Ohm’s law to express
the voltages in terms of the mesh current.
‰
Solve the resulting n simultaneous equations to get the mesh
currents.
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
For mesh 1
− V1 + R1i1 + R3 (i1 − i2 ) = 0
or
(R1 + R3 )i1 − R3i2 = V1
For mesh 2
R2i2 + V2 + R3 (i2 − i1 ) = 0
In matrix form;
⎡ R1 + R3
⎢ −R
3
⎣
¾
− R3 ⎤ ⎡ i1 ⎤ ⎡ V1 ⎤
=
R2 + R3 ⎦⎥ ⎢⎣i2 ⎥⎦ ⎣⎢− V2 ⎥⎦
or
− R3i1 + (R2 + R3 )i2 = −V2
Notice that the branch currents are different from the mesh currents unless the
mesh is isolated. i for a mesh current and I for a branch current.
I1 = i1 ; I 2 = i2 ; I 3 = i1 − i2
INSPIRING CREATIVE AND INNOVATIVE MINDS
22
M. Afendi M. Piah, PhD
Circuit Theory
Example 2.5
Find the branch currents I1, I2 and I3 using mesh analysis.
Solution
Apply KVL for mesh 1
− 15 + 5i1 + 10(i1 − i2 ) + 10 = 0
or
3i1 − 2i2 = 1
Apply KVL for mesh 2
6i2 + 4i2 + 10(i2 − i1 ) − 10 = 0
In matrix form;
⎡ 3 − 2⎤ ⎡ i1 ⎤ ⎡1⎤
⎢− 1 2 ⎥ ⎢i ⎥ = ⎢1⎥
⎣
⎦⎣ 2 ⎦ ⎣ ⎦
Ans : i1 = 1 A ; i2 = 1 A
or
− i1 + 2i2 = 1
Therefore, the branch currents are;
I1 = i1 = 1 A; I 2 = i2 = 1 A; I 3 = i1 − i2 = 0 A
INSPIRING CREATIVE AND INNOVATIVE MINDS
23
M. Afendi M. Piah, PhD
Circuit Theory
Problem 2.5
Calculate the mesh current i1 and i2.
Ans: i1 = 2/3
/ A, i2 = 0 A
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Example 2.6
Use mesh analysis to find the current Io in the circuit.
Solution
Apply KVL for mesh 1
− 24 + 10(i1 − i2 ) + 12(i1 − i3 ) = 0
or
11i1 − 5i2 − 6i3 = 12
Apply KVL for mesh 2
24i2 + 4(i2 − i3 ) + 10(i2 − i1 ) = 0
Apply KVL for mesh 3
4 I 0 + 12(i3 − i1 ) + 4(i3 − i2 ) = 0
or
− 5i1 + 19i2 − 2i3 = 0
and I 0 = i1 − i2 ; so that
4(i1 − i2 ) + 12(i3 − i1 ) + 4(i3 − i2 ) = 0
Ans : I 0 = i1 − i2 = 2.25 − 0.75 = 1.5 A
or − i1 − i2 + 2i3 = 0
INSPIRING CREATIVE AND INNOVATIVE MINDS
25
M. Afendi M. Piah, PhD
Circuit Theory
P bl
Problem
26
2.6
Using mesh analysis, find I0 in the circuit.
Ans: -5 A
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Mesh Analysis
y
with Current Sources
‰
The presence of the current sources reduces the number of
equations. The number of equations (based on KVL) that need to be
write is n – m, where n: number of meshes and m: number of current
sources.
‰
Consider the following 2 possible cases:
Case 1: When current source exists only in one mesh.
In the figure, n − m = 2 − 1 = 1 ; therefore,
KVL at mesh 1: − 10 + 4i1 + 6(i1 − i2 ) = 0
and
d i2 = −5 (supportiv
(
ti e equation)
ti )
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INSPIRING CREATIVE AND INNOVATIVE MINDS
M. Afendi M. Piah, PhD
Circuit Theory
Case 1: When current source exists between two meshes. We create a
supermesh by excluding the current source and any elements connected in
series
i with
i h it.
i
supermesh
KVL at supermesh gives : − 20 + 6i1 + 10i2 + 4i2 = 0
or
6i1 + 14i2 = 20
Supportive equation : i2 − i1 = 6
⇒ (a)
⇒ (b)
Solving eqns (a) and (b), we get,
i1 = −3.2A ; i2 = 2.8 A
INSPIRING CREATIVE AND INNOVATIVE MINDS
28
M. Afendi M. Piah, PhD
Circuit Theory
Example 2.7
Find i1 to i4 using mesh analysis in the circuit.
Solution
Number of meshes, n = 4 ; Number of Current Sources, m = 2
⇒ Number of equations need to be applied, n − m = 4 − 2 = 2
Meshes 1 & 2 form a supermesh. Also
meshes 2 & 3 form another supermesh.
p
The
two supermeshes intersect and form a larger
supermesh.
⇒ Applying KVL to the larger supermesh;
2i1 + 4i3 + 8(i3 − i4 ) + 6i2 = 0
Supportive equations;
⇒ (c)
or i1 + 3i2 + 6i3 − 4i4 = 0
⇒ i2 − i1 = 5
⇒ i2 − i3 = 3I 0 = 3(−i4 )
⇒ (d)
⇒ (a)
⇒ Applying KVL in mesh 4;
2i4 + 8(i4 − i3 ) + 10 = 0
or 5i4 − 4i3 = −5
⇒ (b)
From eqns (a) to (d), we get,
i1 = −7.5 A, i2 = −2.5 A, i3 = 3.93 A, i4 = 2.143 A
INSPIRING CREATIVE AND INNOVATIVE MINDS
29
M. Afendi M. Piah, PhD
Circuit Theory
Problem 2.7
27
Use mesh analysis to determine i1, i2 and i3.
Ans: i1 = 3.474 A; i2 = 0.4737 A; i3 = 1.1052 A
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INSPIRING CREATIVE AND INNOVATIVE MINDS