Circuits 1,2

1
1
Definitions and Laws
Of Electrical Circuits
1-1 Two Terminal- Electrical Elements
Resistors, capacitors, inductors, batteries, generators, etc… are examples of electrical two-
terminal-elements. These elements may be represented as shown in Fig 1-1, where A, and B are the
external terminals of the element.
Fig.1-1 Two Terminal – Electrical Element
1-2 Current
The current i is the rate of change of the electrical charge ( q), passing through the two-
terminal-element, with respect to the change of time. The arrow shows the direction of current:
The following Fig 1-2, shows an example for representing the same current

2
Fig.1-2
1-3 Voltage
Voltage (Potential-Difference) is a measure of the energy required for moving a charge across a
two terminal-element. We say that a voltage exists between terminal A and terminal B of the
element.
The sense of voltage is, usually, indicated by a (+) mark, at one terminal, and a (-) mark at the
other terminal. The terminal, with the (+) mark, is said to be positive with respect to the terminal
with the (-) mark, by the voltage indicated between the two marks.
Fig 1-3 shows an example for a voltage across a two terminal-element.
a b c d
Fig.1-3
In Fig.1-3-a,and Fig.1-3-b, terminal Ais 6V positive with respect to terminal B
In Fig.1-3-c,and Fig.1-3-d, terminal B is 6V positive with respect to terminal A
1-4 Power and Reference Directions
If the current ( i ) is assigned to flow into the terminal, indicated by a (+) mark, as shown in Fig.1-
4, then the current-direction is said to be associated with the voltage-direction. In this case, the
instantaneous power ( p) drawn by the element, can be determined from the following relationship:
Fig.1-4. Power and associated directions
If, at a certain moment, p is positive then the element is absorbing power, but if p is negative,
then the element is delivering power at that moment.
1-5 Resistive - Two-Terminal-Elements
An element is, referred to as, a resistive-two-terminal-element if it is possible to draw the direct
relation-ship between the element-voltage and element-current ( i
v  - characteristic). The
following are examples of resistive-two-terminal-elements:
1-5-1 Resistor

3
Fig 1-5-1 shows the symbol and ( i
v  ) - characteristic of a linear resistor.
Symbol Characteristic
Fig.1-5-1 The Linear Resistor
We can see the characteristic of the linear resistor is a straight line having a slope R and can be
expressed by the following relationship, known as "Ohm's Low:
i
*
R
v  (Ohm's Law)
where R is called "Resistance", having the basic unit: Ohm ( ).
Ohm's-Law can also be expressed also by
v
G
i *
 (Ohm's Law)
G is called "Conductance" having the unit: Mho ( ) or Siemens (S):
R
/
1
G 
Note that, in practice, Ohm’s Law is approximately valid as long as the power dissipated by the
resistor is less than a certain value called “Rating Power” of the resistor.
1-5-2 Ideal-Independent-Voltage-Source
The symbol and ( i
v  ) - characteristic of an ideal-independent-voltage-source is shown in Fig. 1-
5-2.
Symbol Characteristic
Fig.1-5-2 Ideal-independent-voltage-source
From its characteristic we can see that an ideal-independent-voltage-source produces a constant
voltage equal to s
v , to the circuit it is connected to, independent of the magnitude and direction of
the current i .
Note that a practical voltage source is approximately ideal for a limited amount of current called
“Rating Current”.
1-5-3 Ideal-Independent-Current -Source
Fig1-5-3 shows the symbol and ( i
v  )-characteristic of an ideal-independent-current -source.

4
symbol characteristic
Fig.1-5-3 Ideal-independent-current-source
From its characteristic, we can see that an ideal-independent-current-source produces a constant
current, to the circuit it is connected to, equal to s
i independent of the magnitude and direction of
the circuit voltagev .
Note that a practical current source can be considered as ideal only for a limited voltage, called
“Rating Voltage”.
1-6 Resistive Circuits and Definitions
If two, or more, passive electrical elements are connected to each other, using conducting wires
(in practice cupper wires), the connection is called an "Electrical Network", and if some of the
elements are active elements, the connection is called an "Electrical Circuit".
The circuit is called a "Resistive Circuit" if all its elements are of resistive type. Fig.1-6 shows an
example for a resistive circuit.
a Fig.1-6 b
Figs 1-6-a, -b represent, electrically, the same circuit. To understand this fact, we should be
familiar with the following circuit definitions:
1-6-1 Node
A point, where two or more elements are joined together, is called a "Node"
For example, in the circuit of Fig 1-6, there are 3 nodes; Node 1, Node2 , and Node 3
1-6-2 Independent Node
A node joining 3 or more elements may be called an "Independent Node".
For example, in the circuit of Fig 1-6., there are 2 independent nodes: Node 1 and Node2 .
1-6-3 Branch
Any number of elements, connected such that the same current is flowing
through each of them, is called a "Branch".
For example, in the circuit shown in Fig. 1-6, there are 4 branches.
1-6-4 Loop

5
If we start at any node, in a circuit, and move across some elements and end our movement at
the same node, then the path, we moved around, is called an "Electrical closed path" or "Loop".
For example, in the circuit shown in Fig 1-6., there are 6 loops.
1-6-5 Mesh
A loop, which does not contain other loops, is called a "Mesh".
For example, in the circuit shown in Fig 1-6, there are 3 meshes.
1-7 Electrical Circuit – Laws
The most important laws used in circuit analysis are the "Kirchhoff"s-Current Law" and
"Kirchhoff"s-Voltage Law", whose statements are as follows
1-7-1 Kirchhoff’s Current Law (KCL)
"The algebraic sum of all currents leaving a node, at the same moment, is zero "
Example 1-7-1 Apply KCL to the node 1 shown in Fig 1-7-1
Fig.1-7-1
Solution:
0
i
i
i
i D
C
B
A 




1-7-2 Kirchhoff’s Voltage Law (KVL)
"The algebraic sum of all voltages around a loop, at the same mome nt, is zero"
Example 1-7-2 Apply KVLon the loop shown in Fig 1-7-2
Fig.1-7-2
Solution:
0
v
v
v 3
2
1 



1-8 Circuit Analysis Using Kirchhoff's Laws
Circuit analysis means to find any unknown branch current or any unknown voltage between any
two terminals in a circuit.
To carry out the analysis, we have to consider the following notes:
-- An unknown branch current can be obtained, if the currents, in all

6
branches connected to this branch, are known.
-- An unknown voltage, between any two terminals, can be obtained if all
voltages, along a loop which includes this unknown voltage, are known.
1-8-1 Analysis Procedures
1 - Assign a symbol and direction for each unknown branch current.
2 - Apply Kirchhoff's laws as many times as the number of the unknowns.
Note: KCL can be applied at maximum ( 1
n  )-times, on independent
nodes, if the number of the independent nodes is n
Example 1-1 Find x
v and 1
R in the circuit shown in Fig 1-1
Fig.1-1
Solution:
KVL on the outer loop : 0
4
*
3
2
*
3
1
*
7
vx 



  25
vx 
KCL on node a : 0
7
3
i1 

  4
i1 
KVL on the right loop : 0
2
*
3
4
*
3
R
*
4 1 

  5
.
4
R1 
Example 1-2 Find x
v and the power absorbed by the element x in the
circuit of Fig.1-2
Fig 1-2
Solution:

7
KCL on node a: 0
6
1
8
i1 



  15
i1 
KVL on the outer loop: 0
3
*
15
v
37
2
*
6 x 


  20
vx 

KCL on node b: 0
6
1
i2 

  7
i2 
W
140
7
*
20
i
*
v
P 2
x
x 



 .Thus the element x is delivering W
140
Example 1-3 Find the voltage, across the 200  - resistor, in the circuit
of Fig.1-3
Fig.1-3
Solution:
KVL on the assigned loop:
0
10
i
200
40
i
100 


  1
.
0
i 
  20
i
*
200
v2 


Example 1-4 Find the current flowing through the 15 - conductance in
the circuit of Fig.1-4
Fig.1-4
Solution:
KCL on node a: 0
mA
120
mA
30
i
i 2
1 


  mA
90
i
i 2
1 
 ....(1)
KVL on the indicated loop: 0
i
30
1
i
15
1
2
1 
  0
i
i
2 2
1 
 ….(2)
mA
30
i1 

8
1-9 Problems
1-9-1 Find x
i and x
v , and the power absorbed by each element in the
circuit of Fig1-9-1.
Fig.1-9-1
1-9-2 Find x
i and x
v , and the power delivered by the 5V - voltage source
in the circuit of Fig.1-9-2
Fig.1-9-2
1-9-3 Which one of the elements ,
B
,
A and C can't be just a resistor, in
the circuit of Fig. 1-9-3
Fig.1-9-3
1-9-4 State the number of nodes, independent nodes, branches, loops, and
meshes, in the circuits of Fig.1-9-4 and then find x
i and x
v .
(a) Fig.1-9-4 (b)
1-9-5 Find i and v, and the power absorbed by the resistive load in the
circuit shown in Fig.1-9-5

9
Fig.1-9-5
1-9-6 Find x
i and x
v in each of the circuits, shown in Fig.1-9-6.
(a) Fig.1-9-6 (b
2
Analysis Techniques of
Electrical Circuits
There are useful analysis techniques , derived from Kirchhoff's laws , which offer a quicker and
simpler analysis of electrical circuits . In this chapter we shall present some of these analysis
techniques
2-1 Equivalent Resistance
If a number of resistors are connected together, and it is not required to find a current or a
voltage within this group of resistors, then this group of resisters may be replaced, between certain
two terminals, by an equivalent resistance such that the number of unknown branch currents in the
circuit is kept to a minimum, as we shall see in the following examples:
2-1-1 Series Resistors
If a number of resistors are connected as shown in Fig.2-1-1 the same current i will flow through
each of them. These resistors are said to be connected, electrically, in series, and may be replaced by
an equivalent resistance eq
R as follows:

10
Fig.2-1-1
0
v
i
R
......
i
R
i
R
i
R n
3
2
1 




  0
v
i
Req 

n
3
2
1
eq R
......
R
R
R
R 




2-1-2 Parallel Resistors
If a number of resistors are connected as shown in Fig.2-1-2 the same voltage will appear across
each of them. These resistors are said to be connected, electrically in parallel, and may be replaced by
an equivalent resistance eq
R as follows:
Fig.2-1-2
i
R
v
R
v
R
v
R
v
n
3
2
1







 , i
R
v
eq


eq
R
1
n
3
2
1 R
1
R
1
R
1
R
1







Example 2-1 Find the equivalent resistance eq
R for the circuit shown in
Fig. 2.1
Fig.2.1
Solution: 
eq
R
1
2
1 R
1
R
1
 =
2
1
2
1
R
R
R
R 

2
1
2
1
eq
R
R
R
R
R


Example 2-2 Repeat example 2-1 if 0
R2  (short circuit)

11
Fig.2-2
Solution : 0
0
R
0
*
R
R
1
1
eq 


Thus, the whole current i will flow through the short circuit.
2-1-3 Series - Parallel - Resistor - Combinations
The procedures for obtaining the equivalent resistance of series-parallel-resistor combinations are
as follows:
1- Replace any resistors connected in series by their equivalent resistance
2- Replace any resistors connected in parallel by their equivalent
resistance
3- Continue steps 1 and 2 until you reach the two terminals between which the equivalent
resistance is required.
Example 2-3 Find the equivalent resistance eq
R for the circuit of Fig. 2-3
Fig.2-3
Solution:
10
4
3
3
Req 



2-1-4 Y
to 

 - Transformation
If there are neither series- nor parallel resisters, we may simplify the given resistor- connection
using the following Y
to 

 - transformation method:

12
3
2
1
2
1
a
R
R
R
R
R
R


 ,
3
2
1
3
2
b
R
R
R
R
R
R


 ,
3
2
1
3
1
c
R
R
R
R
R
R



Example 2-4 Find eq
R for the circuit shown in Fig. 2-4, using Y
to 

 -
Transformation
Fig.2-4
Solution,
2-1-5 Star–to–Mesh – Transformation
For resistors which are neither series nor parallel, it is sometimes possible to find the equivalent
resistance using the Star–to–Mesh- transformation method, where a star- connection may be
replaced by a mesh-connection. The equivalent mesh-connection consists of
resistors connected between each two outer terminals of the star connection
An example for the transformation of a 4 - ray star to a mesh connection:

13
G
G
G
G b
a
ab


G
G
G
G c
a
ac


G
G
G
G d
a
ad


G
G
G
G c
b
bc


G
G
G
G d
b
bd


G
G
G
G d
c
cd


Example 2-5 Find eq
R for the circuit shown in Fig. 2-5. Using (Star–to–Mesh)
– Transformation
Fig. 2-5
Solution:

14
2
4
//
4
Req 

2-2 Problems
2-2-1 Find the equivalent resistance eq
R between terminals (a-b) for each
of the circuits shown in Fig, 2-2-1
(a) Fig 2-2-1 (b)
R between terminals (a-b) for each of
the circuits shown in Fig.2-2-2
(a) Fig.2-2-2 (b)
of the circuits shown in Fig.2-2-3 , if 2800
R1  .
Fig.2-2-3

15
of the circuits shown in Fig.2-2-4.
(a) (b) (c)
Fig.2-2-4
of the circuits shown in Fig 2-2-5.
(a) (b) (c)
Fig.2-2-5
2-3 Equivalent-Source
An equivalent source for a groupe of sources, connected in any way, should produce the same
current and same voltage to a certain circuit, produced by this group of sources when connected to
the same circuit. The following are examples for different source connections and their equivalent
sources
2-3-1 Series Voltage Sources
0
2
1 


 v
v
v s
s  2
1 s
s v
v
v 
 , 0
v
vseq 

  seq
v
v 

16
2
1 s
s
seq v
v
v 

2-3-2 Parallel Current Sources
0
2
1 


 i
i
i s
s  2
1 s
s i
i
i 
 , 0
i
iseq 

  seq
i
i 
2
s
1
s
seq i
i
i 

2-3-3 Parallel Voltage Sources
a) Assuming 2
s
1
s v
v 
0
v
v 2
s
1
s 

  2
s
1
s v
v   seq
v is undefined
b) Assuming 2
s
1
s v
v 
0
v
v 2
s
1
s 

  2
s
1
s v
v   2
s
1
s
seq v
v
v 

2-3-4 Series Current Sources
a) Assuming 2
s
1
s i
i  :
0
i
i 2
s
1
s 

  2
s
1
s i
i   seq
i is undefined
b) Assuming that 2
s
1
s i
i  :
0
i
i 2
s
1
s 

  2
s
1
s i
i   2
s
1
s
seq i
i
i 

2-3-5 Current and Voltage Sources in Parallel

17
0
v
vs 

  s
v
v  0
v
vseq 

  seq
v
v 
s
seq v
v 
2-3-6 Current and Voltage Sources in Series
0
i
is 

  s
i
i  , 0
i
i seq 
  seq
i
i 
s
seq i
i 
2-4 Voltage Divider
A series connection of resistors may be called " Voltage divider" since the voltage across them is
divided among them . The concept of voltage division is illustrated for two resistors as follows :
2
1
1
2
1
1
1
1
R
R
R
*
v
R
//
R
v
*
R
i
*
R
v




2
1
2
2
1
2
2
2
R
R
R
*
v
R
//
R
v
*
R
i
*
R
v




2-5 Current Divider
A parallel connection of resistors may be called " Current divider " since the total current flowing
through them is divided among them . The concept of current division is illustrated for two
resistors as follows :

18
2
1
2
1
2
1
1
1
R
R
R
*
i
R
)
R
//
R
(
*
i
R
v
i




2
1
1
2
2
1
2
2
R
R
R
*
i
R
)
R
//
R
(
*
i
R
v
i




Example 2-6 Find x
v in the circuit shown in Fig 2-6
Fig.2-6
Solution :
4
4
2
2
12
vx 


2-6 Source Transformation
A practical voltage source ( a voltage source in series with a resistor ) may be transformed to a
practical current source ( a current source in parallel with a resistor ) and vice versa as shown in Fig
2-6.
Fig.2-6
0
v
i
R
v sv
s 


 0
i
R
v
i
si
s 



s
sv v
v
i
R 
 s
si
si i
R
v
i
R 

The above two practical sources are equivalent to each other if they produce the same current i
and same voltage v when they are connected to the same circuit as shown in Fig 6.4. Thus,
si
sv R
R  , s
si
s i
R
v  ,
sv
s
s
R
v
i 

19
Example 2-7 Use Source Transformation to find x
v in the circuit of Fig.2-7
Fig.2-7
Solution:
1
1
2
1
*
3
vx 


2-7 Superposition
If a circuit is containing n independent sources then any unknown current or unknown voltage
may be obtained as the sum of n parts , where each part is produced by one source acting alone
while all other independent sources are replaced by their internal resistances.
Note that the internal resistance of an ideal independent voltage source is zero (short circuit) and
the internal resistance of an ideal independent current source is infinity (open circuit).
Example 2-8 Use Superposition to find v in the circuit shown in Fig 2-8.
Fig. 2-8
8
)
3
/
1
(
)
6
/
1
(
)
2
/
1
(
)
2
/
1
(
16
v 









v 20
k
2
1
*
)
3
/
1
(
)
6
/
1
(
)
2
/
1
(
)
3
/
1
(
m
120 



20
v
v
v 



 12
20
8 



2-8 Problems
2-8-1 Find the current i in each of the circuits shown in Fig.2-8-1
(a) (b) (c)
Fig.2-8-1
2-8-2 Find the voltage v in each of the circuits shown in Fig.2-8-2
(a) (b) (c)
Fig.2-8-2
2-8-3 Find the currents 1
i and 2
i in each of the circuits shown in using the
voltage-divider- and or current divider- method.
(a) (b) (c)
Fig.2-8-3
2-8-4 Find the voltages 1
v and 2
v in each of the circuits shown in Fig.2-8-4
using the voltage divider- and, or current divider- method
(a) (b) (c)
Fig.2-8-4

21
2-8-5 Find the voltage o
v in each of the circuits shown in Fig.2-8-5, using
the source transformation-method.
(a) (b) (c)
Fig.2-8-5
2-8-6 Find the voltage v in each of the circuits shown in Fig.2-8-6, using the
super position method.
(a) (b) (c)
Fig.2-8-6
2-9 Thevenin's Equivalent
To simplify the analysis a resistive circuit may be replaced, between two certain points (a-b), by a
simple equivalent circuit called "Thevenin's Equivalent" which consists of a voltage source
th
v (Thevenin's voltage) in series with a resistance th
R (Thevenin's resistance) as shown in Fig.2-9
Fig.2-9 Thevenin’s Equivalent
The above two circuits are electrically equivalent if
-- The open circuit voltage of the first circuit
0
i
v 
is equal to the open circuit
voltage of the second circuit th
v :
0
i
v 
= th
v
-- The short circuit current of the first circuit (
0
v
i 
) is equal to the short
circuit current of the second circuit ( th
th R
/
v ):

22
0
v
i 
= th
th R
/
v
Thus, we can find th
v and th
R as follows:
0
i
th v
v 
 ,
0
v
0
i
th
i
v
R


 =
current
circuit
short
voltage
circuit
open




th
R is also equal to the equivalent resistance ab
R facing an external current ( d
i ), while all
independent sources in the circuit, are set to 0 as follows:
2-10 Norton's Equivalent
The Norton's equivalent , which consists of an independent current source (Norton's current N
i )
in parallel with a resistance (Norton’s resistance N
R ) can be obtained from the Thevenin's
equivalent using source transformation as shown in Fig. 2-10.
Thevenin's Equivalent Norton's Equivalent
Fig.2-10
th
N R
R  , th
th
N R
/
v
i 
Example 2-9 Find the Thevenin's equivalent and Norton's equivalent for the
circuit, to the left of terminals ( b
a  ) shown in Fig.2-9.
Fig.2-9
Solution:

23
18
3
6
6
*
27
vth 

 9
)
6
//
3
(
7
Rth 


Thevenin's Equivalent Norton's Equivalent
Example 2-10 Using Thevenin’s equivalent-theorem find v in the circuit of
Fig.2-10
Fig.2-10
Solution:
15
20
*
)
20
5
(
15
15
*
2
vth 


 10
)
5
15
//(
20
Rth 


2-11 Maximum Power - Transfer
A resistor L
R , connected between two terminals (a-b) in a resistive circuit, will absorb
maximum power, from this circuit, if L
R has a certain value calculated as follows:
1- Replace the circuit at terminals ( b
a  ) by its Thevenin’s equivalent as
shown in Fig.2-11

24
Fig.2-11
2- Calculate the power L
P absorbed by L
R , in general:
L
2
th
L
2
th
L
2
L
L R
*
)
R
R
(
v
R
*
i
P



3- Find the derivative ( L
L dR
/
dP ) and obtain the values of L
R , which
makes this derivative equal to 0:
0
)
R
R
(
R
R
*
v
)
R
R
(
)
R
R
(
R
2
)
R
R
(
*
v
dR
dP
3
th
L
L
th
2
th
4
th
L
th
L
L
2
th
L
2
th
L
L 








th
L R
R  , 

L
R
Thus, L
P will have a maximum value L
P (max) if we choose : th
L R
R 
In this case, the maximum power becomes:
th
2
th
L
R
4
v
(max)
P 
Example 2-11 For the circuit shown in Fig.2-11, find R , which would absorb
Maximum power from the circuit, and obtain this maximum
value of power.
Fig.2-11
Solution:

25
Node a: 0
6
3
i1 

  3
i1 

Node b: 0
6
4
i2 


  2
i2 
 9
6
3
Rth 


24
i
3
i
6
V 2
1
th 



Replacing the circuit, facing R , by its Thevenin’s equivalent, we obtain the following circuit:
Thus R would absorb maximum power L
P (max) if L
R = th
R = 9
and the maximum power is then:
W
16
9
*
4
)
24
(
R
4
v
(max)
P
2
th
2
th
L 



2-12 Problems
2-12-1 Find o
i in each of the circuits shown in Fig .2-12-1
(a) Fig 2-12-1 (b)
2-12-2. Find o
v in each of the circuits shown in Fig. 2-12-2
(a) Fig . 2-12-2 (b)
2-12-3 Find the Thevenin's Equivalent, at terminals ( b
a  ), for each of the
circuits of Fig. 2-12-3

26
(a) (b)
(c) (d)
Fig.2-12-3
2-12-4 Find o
i in each of the circuits shown in Fig .2-12-4, using the
Thevenin's Equivalent - Method.
(a) Fig .2-12-4 (b)
2-12-5 Find the value of R , which absorbs maximum power in each of the
circuits shown in Fig.2-12-5. Find also this maximum value of power
in each case
(a) Fig.2-12-5 (b)
2-13 Nodal Analysis
Using the nodal analysis method, the number of equations, required to completely analyze an
electrical circuit, may be considerably reduced. The steps of the Nodal analysis are:

27
1- Determine the number of independent nodes in the circuit.
2- Assign one of the independent nodes as the reference node, indicate it by the symbol (  ),
and consider the potential at the reference node equal to 0.
3- Assign a node-potential at each independent node, which is the voltage between that
independent node and the reference node.
4- If there is a voltage source in series with a resistor, connected between
two independent nodes, then use source transformation to change it to
a current source in parallel with the same resistor.
5- If there is a voltage source, connected between two independent nodes, these two nodes may
be combined to one node called “Super node” and then you can apply KCL on that Super
node.
6- Apply KCL on each independent node and on each super node, whose potential is
unknown, using the assigned node potentials.
7- Solving the obtained equations (Node Equations) you can find any unknown (current or
voltage) in the circuit.
Example 2-12 Obtain the node equations for the circuit shown in Fig.2-12
and then find v ..
Fig.2-12
Solution: We select, for example, the lower independent node as the
reference node, and assign the three unknown independent
node-potentials as follows:
Applying KCL on each of the three independent nodes, we get
0
)
v
v
(
4
)
v
v
(
3
3
8 3
1
2
1 




  11
v
4
v
3
v
7 3
2
1 



0
v
)
v
v
(
2
)
v
v
(
3
3 2
3
2
1
2 





  3
v
2
v
6
v
3 3
2
1 



0
v
5
)
v
v
(
4
)
v
v
(
2
25 3
1
3
2
3 





  25
v
11
v
2
v
4 3
2
1 



The above node-equations, may be written, in a matrix form as follows:

28
















11
2
4
2
6
3
4
3
7










3
2
1
v
v
v
=










25
3
11
Solving the above node equations, we have
2
v2  , 3
v3  , 3
2 v
v
v 
 1


Fig.2-13
Solution:
Let us select the lower independent node as the reference node, and assign the unknown
independent node-potentials as follows:
Applying KCL on each of the three independent nodes we get
0
)
v
v
(
4
)
v
v
(
3
8 3
1
2
1 



  8
v
4
v
3
v
7 3
2
1 



0
v
)
2
v
v
(
6
)
v
v
(
3 2
3
2
1
2 




  12
v
6
v
10
v
3 3
2
1 




0
)
v
2
v
(
6
)
v
v
(
4
v
5
25 2
3
1
3
3 






  37
v
15
v
6
v
4 3
2
1 



The above node equations can be written, in a matrix form, as follows:
















15
6
4
6
10
3
4
3
7










3
2
1
v
v
v
=












37
12
8

29
Fig. 2-14
Let us select the lower independent node as the reference node and assign the two unknown node
potentials as follows:
Applying KCL on node 1, we get
0
)
22
v
v
(
4
)
v
v
(
3
3
8 2
1
2
1 





 
77
v
7
v
7 2
1 
 ….(1)
and applying KCL on the Super node , we have
0
)
22
v
(
5
)
v
(
1
)
v
v
(
3
3
)
v
22
v
(
4
25 2
2
1
2
1
2 









 
170
v
13
v
7 2
1 


 ….(2)
The above node equations can be written, in matrix form, as follows:








13
7
7
7






2
1
v
v
= 





 170
77
2-13-1 Deriving Node Equations by Inspection
The general form of node equations, in matrix form, is the following:

30
Conductance-matrix
The terms of the Conductance Matrix can be obtained as follows:
1) Each diagonal term ( 11
G , 22
G , 33
G nn
G
...... ) is equal to the sum of
the conductance's, connected to a particular node, for example, kk
G is
the sum of all conductance's connected to node k .
2) Each non-diagonal term is equal to : – (Sum of the conductance's
connected between two particular nodes), for example ( 53
35 G
G  )
is the sum of all conductance's connected between node 3 and node 5.
3 ) ( n
3
2
1 v
,.......,
v
,
v
,
v ) are the unknown node potentials.
4) Each term in the right column ( sn
3
s
2
s
1
s i
,.....,
i
,
i
,
i ) is equal to the
algebraic sum of the source currents flowing into a particular node, for
example : sk
i is the algebraic sum of the source-currents flowing into
node k .
by inspection.
Fig. 2-15
Solution: Let us select the lower independent node as reference node, and
assign the three unknown node potentials as follows:

31
The node equations can be written, by inspection, in a matrix form, as follows:
















11
2
4
2
6
3
4
3
7










3
2
1
v
v
v
=










25
3
11
Example 2-16 Obtain the node equations for the circuit shown in Fig.6.16
by inspection.
Fig.2-16
Solution: Let us select the lower independent node as the reference node,
use source transformation, and assign the three unknown
independent node-potentials as follows:
The node equations can be written by inspection, in a matrix form, as follows:
















15
6
4
6
10
3
4
3
7










3
2
1
v
v
v
=












37
12
8
Example 2-17 Obtain the node equations for the circuit shown in Fig.2.17
by inspection.

32
Fig.2.17
Solution: Let us select the lower independent node as the reference node and
assign the three unknown node potentials as follows:
The node equations can be written, by inspection, in a matrix form as follows:








13
7
7
7






2
1
v
v
= 





 170
77
2-14 Mesh Analysis
Using the mesh analysis method, the number of equations required to analyze an electrical circuit,
may be, considerably, reduced. The analysis steps are:
1- Determine the number of meshes in the circuit (to be analyzed).
2- Assign a mesh current to each mesh in the circuit, where a mesh current
is defined as the current flowing in each element in a particular mesh. If
an element (excluding a current source) is common between two meshes
then the actual current flowing through that element is equal to the
algebraic sum of the mesh currents.
3- Apply KVL around each mesh, which does not contain a current source.
If a current source is common between two meshes then combine
these two meshes to one mesh (Super Mesh) and apply KVL around the
Super mesh.
4- Solve the obtained equations (Mesh Equations) to find any unknown
current or unknown voltage in the circuit.
Example 2-18 Obtain the mesh equations for the circuit shown in Fig.2-18

33
Fig.2.18
Solution: Let us assign the three mesh currents and apply KVL around
each of them as follows:
Mesh 1: 0
)
i
i
(
1
)
i
i
(
2
i
2 2
1
3
1
1 



  0
i
2
i
i
5 3
2
1 


Mesh 2: 0
7
)
i
i
(
2
6
)
i
i
(
1 3
2
1
2 




  1
i
2
i
3
i 3
2
1 



Mesh3: 0
)
i
i
(
2
6
)
i
i
(
2
i
1 1
3
2
3
3 




  6
i
5
i
2
i
2 3
2
1 


The above mesh equations can be written, in a matrix form, as follows:
















5
2
2
2
3
1
2
1
5










3
2
1
i
i
i
=










6
1
0
Fig.2-19
Solution: Let us assign the mesh currents and the super mesh as follows:

34
Mesh 1: 0
)
i
i
(
1
)
2
i
i
(
3
i
2 2
1
2
1
1 




 
6
i
4
i
6 2
1 
 .... (1)
SuperMesh: 0
7
)
2
i
(
1
)
i
2
i
(
3
)
i
i
(
1 2
1
2
1
2 






 
1
i
5
i
4 2
1 


 ....(2)
The above mesh equations can be written in a matrix form as follows :








5
4
4
6






2
1
i
i
= 





 1
6
2-14-1 Deriving Mesh Equations by Inspection
The general form of mesh equations in matrix form is the following:
Resistance matrix
The terms of the Resistance Matrix can be obtained as follows:
1- Each diagonal term ( nn
33
22
11 R
,....
R
,
R
,
R ) is equal to the sum of all
Resistances in a particular mesh which does not contain current sources
or in a super mesh, for example kk
R is the sum of all Resistances in mesh
k or super mesh k .
2- Each non-diagonal term is equal to the sum of all resistances which are
common between two particular meshes, for example ( 53
35 R
R  ) is
the sum of all resistances which are common between mesh 3 and mesh 5
Note that 35
R and 53
R should be given a negative sign if the mesh
currents, through it, are assigned in opposite directions.
3- ( n
3
2
1 i
,.......,
i
,
i
,
i ) are the assigned unknown mesh currents.

35
4- Each term in the right column ( sn
3
s
2
s
1
s v
,.....,
v
,
v
,
v ) is equal to the
algebraic sum of all source-voltages in a particular mesh which would
produce a current in the same direction of this mesh current. For example
sk
v is the algebraic sum of all source-voltages in mesh k .
by inspection.
Fig.2-20
Solution: Let us assign the three meshes and mesh currents and obtain the
mesh equations by inspection as follows:
by inspection
Fig.2-21
Solution: Let us assign the mesh currents and obtain the mesh equations by
inspection as follows:

36
2-15 Problems
2-15-1 Using the Nodal Analysis –Method, find the current o
i , in each of
the circuits of Fig.2-15-1
(a) (b) (c)
Fig.2-15-1
2-15-2 Find the Nodal Equations, in matrix form, for each of the circuits of
Fig. 2-15-2
(a) (b) (c)
Fig.2-15-2
2-15-3 Using the Mesh Analysis –Method, find the voltage o
v in each of the
circuits shown in Fig.2-15-3

37
(a) (b) (c)
Fig.2-15-3
2-15-4 Find the Mesh Equations, in matrix form, for each of the circuits of
Fig.2-15-4
(a) (b) (c)
Fig.2-15-4

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