Chapter 4 Techniques of Circuit Analysis
So far we have analyzed relatively simple resistive circuits by applying KVL and KCL
in combination with Ohm’s law.
R1
Applying KVL and Ohm’s law, we have,
is
vs
R2
+
-
v - iR
s
+
1
iR
+
2
iR  0
3
Solving for i , we have,
R3
v
i 
R
+
1
R
s
+
2
R
3
However as circuit become more complicated and involve more elements, this direct method
become cumbersome ( ‫ مرهقة‬, ‫)غير سهل‬
However as circuit become more complicated and involve more elements, this direct method
become cumbersome ( ‫ مرهقة‬, ‫)غير سهل‬
R2
R1
R 7 R8
vs
+
-
R6
R3
R7
In this chapter we introduce two powerful techniques of circuit analysis that aid in the analysis
of complex circuit which they are known as
Node-Voltage Method or Nodal Analysis
Mesh-Current Method or Mesh Analysis
In addition to these method we will develop other method such as source transformations,
Thevenin and Norton equivalent circuit.
4.1 Terminology
Planar Circuit : a circuit that can be drawn on a plane with no crossing branches as shown
R1
R2
R 7 R8
vs
R6
+
-
R3
R4
R5
The circuit shown can be redrawn which is equivalent to the above circuit
R2
R1
Planar Circuit
R 7 R8
vs
+
-
R6
R3
R7
The circuit shown is non planar circuit because it can not be redrawn to make it planar the
circuit
R1
R2
R9
R 7 R8
vs
R6
+
-
R11
R3
R12
R5
R4
R10
Describing a Circuit – The vocabulary
Consider the following circuit
Node
Essential node
path
branch
Essential branch
loop
mesh
Planar Circuit
A point where two or more circuit elements join
a
A node where three or more circuit elements join
b
A trace of adjoining basic elements with no elements
included
R1
A path that connects two nodes
A path that connects two essential nodes without
passing through an essential node
A path whose last node is the same as the
starting node
A loop that does not enclose any other loops
A circuit that can be drawn on a plane with no
crossing branches
v 1 - R1 - R 5 - R 6
v 1 - R1
v 1 - R1 - R 5 - R 6 - R 4
-v
v 1 - R1 - R 5 - R 3 - R 2
2
Example 4.1 For the circuit identify the following
(a) All nodes ?
The nodes are a, b, c, d, e, f and g
(b) The essential nodes ?
A node where three or more circuit elements join
The essential nodes are b, c, e and
g
(c) All branches ?
A path that connects two nodes
The branches are
v 1 , v 2 , R1 , R 2 , R 3 , R 4 , R 5 , R 6 , R 7
and I
(d) All essential branches ?
A path that connects two essential nodes without passing through an essential node
The essential branches are
v 1 - R1
R 2 - R3
v 2 - R4 R5 R6 R7
I
(e) All meshes ?
The meshes are
v 1 - R1 - R 5 - R 3 - R 2 v 2 - R 2 - R 3 - R 6 - R 4 R 5 - R 7 - R 6 R 7 - I
(f) Two paths that are not loops or essential branches ?
R1 - R 5 - R 6
v 2 - R2
is a paths , but it is not a loop because it does not have the same
starting and ending nodes
Nor is it an essential branch because it does not connect two essential
nodes
Is a paths that are not loop or essential branches ?
(g) Two loops that are not meshes ?
v 1 - R1 - R 5 - R 6 - R 4 - v 2
is a loop , but it is not a mesh because there are two loops within it
I - R5 - R6
is a loop , but it is not a mesh because there are two loops within it
Simultaneous Equations- How many
Let the circuit as shown
There are nine branches in which the current is unknown
Note that the current I is known
i9
i1
i2
i6
i7
i3
i4
i8
i5
Since there are nine unknown current
Since there are 7 nodes
We still need 3 more equations
Therefore we need nine independent equations
There will be 6 independent KCL equations since the
7th one can be written in terms of the 6 equations
They will come from KVL around three meshes
or loops
Note the meshe or loop we apply KVL around should not contain the current source I
since we do not know the voltage across it
i9
i1
i2
i6
i7
i3
i4
i8
i5
Since KCL at the non essential (connecting two circuit elements only) like a , d , f
will have the following results
i 9  i1
i3  i7
i5  i8
Therefore we have 6 currents and 4 essential nodes (connecting three or more circuit
elements) like b, c, e, g
Therefore
i1
i2
i6
i3
i4
i5
Applying KCL to 3 of the 4 essential nodes since KCL on the 4th node will have an
equation that is not independent but rather can be derived from the 3 KCL equations
Applying KCL to the essential nodes b, c, and e ( we could have selected any three
essential nodes) we have
KCL at node b we have
KCL at node c we have
KCL at node e we have
-i 1 + i 2 + i 6 - I  0
i1 - i 3 - i 5  0
i3 + i4 - i2  0
Note applying KCL on node g
i5 - i4 - i6 + I
0
Which is a linear combination
of the other KCL’s
i1
i2
i6
i3
i4
i5
Therefore the remaining 3 equations will be derived from KVL around 3 meshes
Since there are 4 meshes only, we will use the three meshes with out the current source
KVL around mesh 1
KVL around mesh 2
KVL around mesh 2
R 1i 1 + R 5i 2 + ( R 2
-( R 2
+ R 3 )i 3 - v 1  0
+ R 3 )i 3 + R 6i 4 + R 4i 5 - v 2
- R 5i 2 + R 7 i 6 - R 6 i 4  0
0
i1
i2
i6
i3
i4
i5
KCL Equations
-i 1 + i 2 + i 6 - I  0
i1 - i 3 - i 5  0
i3 + i4 - i2  0
6 equations and 6 unknown
which can be solved for the
currents
KVL Equations
R 1i 1 + R 5i 2 + ( R 2
-( R 2
+ R 3 )i 3 - v 1  0
+ R 3 )i 3 + R 6i 4 + R 4i 5 - v 2
- R 5i 2 + R 7 i 6 - R 6 i 4  0
0
i1, i 2 , i 3 , i 4 , i 6
4.2 Node-Voltage Method
Nodal analysis provide a general procedure for analyzing circuits using node voltages
as the circuit variables
Node-Voltage Method is applicable to both planar and nonplanar circuits
Using the circuit shown, we can summarize the node-voltage methods as shown
next :
1 W
10 V
+
-
2 W
5 W
10 W
2 A
1
1 W
10 V
+
-
2 W
5 W
2
10 W
3
Step 1 identify all essentials nodes
2 A
3
Do not select the non essentials nodes
Step 2 select one of the essentials nodes ( 1, 2, or 3) as a reference node
Although the choice is arbitrary the choice for the reference node is were most of
branches, example node 3
Selecting the reference node will become apparent as you gain experience using
this method (i,.e, solving problems)
v1
1 W
v2
2 W
+
10 V
+
-
5 W
v1
-
+
10 W
v2
2 A
-
A node voltage is defined as the voltage rise from the reference node to a nonreference node
Step 3 label all nonrefrence essentials nodes with alphabetical label as v1, v2…
Step 4 write KCL equation on all labels nonrefrence nodes as shown next
v1
1 W
+
i1
5 W
+
-
10 V
v1
i3
KCL at node 1
By applying KVL
i1
+
i2
+
i2
10 W
-
+
i3
v2
v 1 - 0  (5)i 3
2 A
-
=
0
Let us find
(1)i 1 + 10 -v 1  0
KVL on the middle mesh, we have
similarly
v2
2 W
i1
,
i2
i3
v 1 - 10
i1 
(1)
-v 1 + (2)i 2 + v 2  0
v1
i3 
(5)
,
i2 
v 1 -v 2
2
v1
1 W
10
+
i1
10 V
5 W
+
-
i3
Therefore
i1
+
i2
v2
2 W
v1
i2
+
10 W
-
+
i3
v2
2 A
-
=
0
We have
v 1 -v 2
v1
v 1 - 10
+
+
2
5
1
=
0
If we look at this KCL equation, we see that the current we notice that the potential at the
left side of the 1 W resistor which is tied to the + of the 10 V source is 10 V because
the – is tied to the reference
10
v1
1 W
2 W
i1
i2
5 W
v 1 - 10
i1 
1
v 1 -v 2
i2 
2
v
i3  1
5
v2
i3
Therefore we can write KCL at node 1 without doing KVL’s as we did previously
v 1 -v 2
v1
v 1 - 10
+
+
2
5
1
=
0
Similarly
v1
2 W
v2
i1
i3
10 W
2 A
i2
KCL at node 2
v 2 -v 1 + v 2
-2
10
2
=
0
10 V
v1
1 W
10
+
-
5 W
v2
2 W
10 W
2 A
v 1 -v 2
v1
v 1 - 10
+
+
2
5
1
v 2 -v 1 + v 2
-2 = 0
10
2
Two equations and two unknowns namely
100
v1 
 9.09 V
11
=
0
v1 , v2
we can solve and have
120
v2 
 10.91 V
11
4.3 The Node-Voltage Method and Dependent Sources
If the circuit contains dependent source, the node-voltage equations must be supplemented
with the constraint equations imposed by the dependent source as will be shown in the
example next
Example 4.3 Use the node voltage method to find the power dissipated in the 5 W resistor
i
1
2
i
3
3
The circuit has three essentials nodes 1, 2, and 3 and one of them will be the reference
We select node 3 as the reference node since it has the most branches (i.e, 4 branches)
We need two node-voltage equations to describe the circuit
20
v1
v2
8i 
i
The circuit has two non essentials nodes which are connected to voltage sources and will
impose the constrain imposed by the value of the voltage sources on the non essential
nodes voltage
Applying KCL at node 1
Applying KCL at node 2
v 1 - 20
v
v -v
+ 1 + 1 2 0
2
20
5
v 2 - 8i 
v 2 -v 1
v2
+
+
0
5
10
2
20
v1
8i 
v2
i
v 2 - 8i 
v 2 -v 1
v2
+
+
0
5
10
2
v 1 - 20
v
v -v
+ 1 + 1 2 0
2
20
5
The second node equation contain the current
v 1 -v 2
i 
5
which is related to the nodes voltages as
Substituting in the second node equation, we have the following
nodes equations
- 0.2v 2  10
-v 1 + 1.6v 2  0
0.75v 1
i
v 1  16
Solving we
havei  16-10 =1.2 A

5
V
v 2  10
V
p5W  (1.2) 2 (5)  7.2 W
4.4 The Node-Voltage Method : Some Special Cases
Let us consider the following circuit
1
100 V
10 W
2
25 W
+
-
3
50 W
5 A
3
The circuit has three essentials nodes 1, 2, and 3 which means that two simultaneous
equations are needed.
From the three essentials nodes, a reference node has been chosen and the other two
nodes have been labeled
v1
100
100 V
10 W
25 W
+
-
v2
50 W
5 A
However since the 100 V source constrains the voltage between node 1 and the reference
node to 100 V.
v 1  100
V
This mean that there is only one unknown node voltage namely (v2)
v 2 -v 1
v2
+
+ -5  0
10
50
Applying KCL at node 2
Since
Knowing
v1 and v2
v 1  100
V
v 2  125
V
we can calculate the current in every branch
Nodal Analysis with voltage sources Some Special Cases
This voltage
source is
connected
between non
reference and
reference
This voltage source is connected
between two non reference nodes
reference
Enclose the voltage source in a
region called supper node
Apply KCL to
that region
KCL on the Supper node
3v 2 + 3v 2
 36
KCL on the Supper node
3v 2 + 3v 2
 36
The second equation come from the constraint on the voltage source
v 2 -v 3
5
Two equations and two unknowns , we can solve
Two voltage sources are connected
between two non reference nodes
reference
Two supper nodes touching or
intersection
Dependent Voltage source
supper node apply as will to
dependent sources
Combine the two supper nodes
to supper supper node
4.5 Introduction to the Mesh-Current Method
Consider the following planar circuit
R
R
1
2
R
v
s
R
+
-
R
R
6
5
There are 4 essentials nodes
8
R
4
3
R
7
There are 7 essentials branches were the current is unknown
R
R
1
R
R
+
-
R
v
+
-
s
R
R
8
s
R
5
3
R vs
8
R
6
3
R
5
R
7
R
R
1
R
R
5
4
4
R
1
2
2
R vs
3
R
+
-
R
8
R
6
3
R vs
R
+
-
8
R
6
3
7
5
R
R
5
R
4
4
4
R
R
R
R
+
-
7
R
Therefore to solve via the mesh-current method we must
write 4 mesh -current
2
R
+
-
R
8
6
R
6
1
2
2
R
5
s
R
7
R
v
7
R
+
-
R
R
R
v
4
R
1
R
3
R
1
2
R
R
R
R
R
1
8
6
5
R
R
2
8
R
6
R
4
R
3
7
7
Number Exxential Branches
-(
4
Number Exxential Nodes
- 1)  4
R
7
A mesh current is the current that exists only in the perimeter of a mesh as shown in the
circuit below
R
R
1
2
R
i
v
s
+
-
R
1
i
8
2
R
6
R
3
i
i
R
5
R
7
4
3
4
The mesh current satisfy Kirchhoff’s Current Law (KCL)
At any node in the circuit, a given mesh current both enters and leaves the node
Mesh current is not always equal branch current
Mesh current is not always equal branch current
Ia
R
Ib
1
R
2
R
i
v
s
+
-
R
1
i
8
2
R
6
R
3
i
i
R
5
R
4
3
4
The branch current
Ia
Equal the mesh current
i1
The branch current
Ib
Doesn't equal any of the mesh currents
7
Consider the following circuit were we want to solve for the currents
R
v
1
+
-
R
1
i1
i3
R
i2
3
i1 , i 2 , i 3
2
+
-
v
2
The circuit has 3 essential branches and 2 essentials nodes
Therefore to solve via the mesh-current method we must write 2 mesh current
3
Number Exxential Branches
-(
2
Number Exxential Nodes
- 1)  2
R
v
1
+
-
R
1
i1
i3
R
2
i2
3
+
-
v
2
Applying KCL to the upper node (one KCL for two essentials nodes)
i1  i 2 + i 3
Applying KVL around the two meshes
v 1  R1i 1 + R 3i 3
-v 2  R 2i 2 - R 3i 3
Solving the KCL for i3 and substituting in the KVL equations we have
v 1  (R1 + R 3 )i 1 - R 3i 2
-v 2  -R 3i 1 + (R 2 + R 3 )i 2
R
v
i1
+
-
1
R
1
i3
R
2
v 1  (R1 + R 3 )i 1 - R 3i 2
i2
+
-
3
v
2
-v 2  -R 3i 1 + (R 2 + R 3 )i 2
Now consider the mesh currents ia, ib , and applying KVL around the two meshes we have
R1
R2
v 1  R1i a + R 3 (i a - i b )

v1
+
-
+
-
R3
ia
ib
v2

v 1  (R1 + R 3 )i a - R 3i b
-v 2  R 3 (i b - i a ) + R 2i b
-v 2  -R 3i a + (R 2 + R 3 )i b
The two equations are identical if we replace the currents i1, i2 with the mesh currents ia, ib
R
v
1
+
-
R
1
i1
i3
R
i2
3
R1
v1
2
+
-
v
2
R2
+
-
+
-
R3
ia
v 1  (R1 + R 3 )i a - R 3i b
ib
v2
-v 2  -R 3i a + (R 2 + R 3 )i b
the branches currents i1, i2, i3 can be written in terms of the mesh currents ia, ib
i1  i a
i 2  ib
i 3  ia - ib
Once we solve for the mesh current we can solve for any branch current
Once we know the branch current we can solve for any voltage or power of interest current
Example 4.4
6W
2W
4W
+
40 V
8W
+
-
6W
vo
+
-
20 V
2W
40 V
+
-
6W
8W
i
a
4W
6W
i
b
i
+
c
20 V
4.6 The mesh-Current Method and dependent sources
Example 4.5
Use the mesh-current method to
determine the power dissipated in the
4 W resistor
i
i
i
1
2
i
i
3
KVL on mesh 1
i
50  5(i 1
2
- i 2 ) + 20(i 1 - i 3 )
KVL on mesh 2
i
1
i
i
0  5(i 2
- i 1 ) + 1i 2 + 4(i 2 - i 3 )
3
KVL on mesh 3
0  20(i 3
Since i   i 1
- i3

50  25i 1 - 5i 2 - 20i 3
0  -5i 1 + 10i 2 - 4i 3
0  -5i 1 - 4i 2 + 9i 3
 25
  -5
 -5
- i 1 ) + 4(i 3 - i 2 ) + 15i 
-5 -20   i 1 
 
10 -4  i 2
 

-4 9  i
50
0
 
 3   0 
i
i
1
 25
 -5
 -5
2
i
i
we need only i2 and i3
25 50 -20
-5 0 -4
-5 0
9
3250
i2 =
=
 26 A
25 -5 -20
125
-5 10 -4
-5 -4 9
P4 W  4(i 3
50
0
 
 3   0 
3
To solve, we use Cramer method as follows,

-5 -20   i 1 
 
10 -4  i 2
 

-4 9  i
25 -5 50
-5
-5
i3 =
25
-5
-5
- i 2 )2 =4(28 - 26) 2 =16 W
10 0
-4 0
3500
=
 28 A
-5 -20
125
10 -4
-4 9
4.7 The Mesh-Current Method : Some Special Cases
i
i
b
+
a
v
-
i
c
i
b
Supermesh
i
a
i
c
The Mesh-Current Analysis of the Amplifier Circuit
i
a
i
i
i
b
B
c
Supermesh
i
a
i
i
i
b
B
c
4.8 The Node-Voltage Method Versus the Mesh-Current Method
iΔ
i
i
a
i
b
i
c
d
i
e
iΔ
Supernode
v1
v3
v2
iΔ
Supernode
va
vb
vc
2.5 W
4W
2W
+v
193 V
+
-
i
+
a
vo
-
6W
i
0.4 v
b
0.5 A
Δ
+v

-
7.5 W
8W
-
Δ
i
c
+
-
0.8 v

Supermesh
2.5 W
4W
2W
+v
+
193 V
+
-
i
a
vo
-
6W
0.4 v
i
Δ
+v

b
0.5 A
-
7.5 W
8W
Δ
i
c
+
-
0.8 v

By inspection
193
vo
4W
va
2.5 W
2W
+v
+
193 V
+
-
vo
-
0.4 v
0.5 A
Δ
+v

-
vb
Δ
+
-
4.9 Source Transformations
The Node-Voltage Method and the Mesh-Current Method are powerful techniques for
solving circuits.
We are still interested in in methods that can be used to simplify circuits like to what we did
in parallel and series resistors and D to Y transformations
A method called Source Transformations will allow the transformations of a voltage
source in series with a resistor to a current source in parallel with resistor.
R
a
a
vs
is
+
-
R
b
b
The double arrow indicate that the transformation is bilateral , that we can start with either
configuration and drive the other
R
a
a
vs
+
-
iL
is
RL
R
RL
iL
b
b
R
iL 
is
R + RL
vs
iL 
R + RL
Equating we have
,
vs
R

is
R + RL R + RL
 is 
vs
R
OR v s  Ri s
Example 4.8 (a) find the power associated with the 6 V source
(b) State whether the 6 V source is absorbing or
delivering power
We are going to use source transformation to reduce the circuit, however note that we
will not alter or transfer the 6 V source because it is the objective.
40
 8A
5
(20W || 5W)  4Ω
(8A )(4Ω)  32 V
(6 + 4)  10Ω
(10 + 10)  20Ω
32
 1.6A
20
(30W || 20W)  12Ω
(4 + 12)  16Ω
(1.6A )(12Ω)  19.2 V
i
i=
1.2-6
 0.825 A
4+12
 P6V  (0.825)(6)  4.95 W
It should be clear if we transfer the 6V during these steps you will not be able to find
the power associated with it
Example 4.9 (a) use source transformations to find the voltage vo ?
+
vo
Since the 125 W resistor is connected across or in parallel to the 250 V source then we can
remove it without altering any voltage or current on the circuit except the 250 V current
which is not an objective any how
Our objective is
vo
Therefore we remove the 125 W
Similarly the 10 W resistor is connected in series with the 8 A source then we can remove
it without altering any voltage or current on the circuit
Now the circuit become
+
vo
We now use the source transformation to replace the 250 V with the 25 W resistor with a
current source and parallel resistor
+
vo
250
25
 10 A
We now combine the parallel
resistors
(25W ||100W || 20W)  10Ω
+
vo
We now combine the parallel resistors
(25W ||100W || 20W)  10Ω
The circuit now become
+
vo
-
 v o  (2A )(10W)  20 V
Let us find the Thevenin Equivalent of the following circuit
By inspection
25
v1
v1
v 1 - 25
v1
+
-3  0
5
20
Therefore the Thevenin voltage for the circuit is
v 1  32
32 V
To find the RTh we deactivate the sources by shortening independent voltage sources and
open independent current source
5W
4W
a
Rab  RTh
20 W
b
R ab  RTh  (4 + (5 || 20))  (4 + 4)  8W
4
Let us find the Thevenin Equivalent using source transformations
25
 5A
5
4W
8A
4W
4W
+
-
32 V
(5W || 20W)  4Ω
8W
4W
+
-
32 V
The Thevenin Equivalent is represented as an independent voltage source VTh and a series
resistor RTh
RTh
a
VTh
+
-
b
Now let us put a short across the terminal and calculate the resulting current
RTh
VTh
+
-
a
i SC
i SC V Th
RTh
b
And this another method of finding RTh as
RTh V Th
i SC
Let us find the
iSC for the circuit shown before
By inspection
25
v2
iSC
can be found easily once
we know the voltage v2
v2
i sc 
4
To find
v2
select the reference as the lower node and write KCL
v 2 - 25
v
v
+ 2 -3 + 2  0
5
20
4
16
i sc   4 A
4
v 2  16 V
RTh V Th  32 8 Ω
i SC 4
Norton Equivalent
i
RTh
4.64 Find Thevenini equivalent with respect to the terminals
a
b
A Using short circuit current
B Deactivating the independent sources
a
and b
A Using short circuit current
17.4
Since RTh V Th V OC
i SC i SC
B Deactivating the independent sources
4.12 Maximum Power
Transfer
Question : what is the value of the load resistant RL that will absorb the maximum power
i
0
0
i
Substituting
in
4.13 Superposition
A linear system obeys the principle of superposition which state that :
Whenever a linear system is excited or driven by more than one independent source of
energy , the total response is the sum of the individual responses by the independent
sources
Example Consider the following linear circuit
i3
i1
i2
i4
We will use the principle of superposition to find the branch currents
i1 , i 2 , i 3 , i4
1- Voltage Source is active and deactivate the current source by opening it
v1
i 3'
i 1'
i '2
KCL at
node
v1
i 4'
v 1  30 V
i 3'
i 1'
i '2
i 4'
2- Current Source is active and deactivate the voltage source by shorting it
v3
i 1''
i 3''
i ''2
Using Nodal analysis
Solving we get
v4
i 4''
i3
i1
i2
i 1''
i 3'
i 1'
i '2
Now to find the branch
currents
i4
i 3''
i ''2
i 4'
i1 , i 2 , i 3 , i4
We sum the currents
i 4''
P488 Use the principle of superposition to find the voltage
v
in the circuit
ib
4W
2 ib
20 W
+ +
70 V
+
-
v
2W
-
10 W
+
50 V
70-V Source acting
alone
70
1
2
3
Therefore from 2 and 3 we have
4
Since
From 4
Therefore from 1 and 5 we have
5
OR
OR