DET 101/3
Basic Electrical Circuit 1
DC CIRCUITS:
CHAPTER 3
Methods of Circuit Analysis and
Circuit Theorems:







Nodal Analysis (Node-Voltage Method)
Mesh Analysis (Mesh-Current Method)
Superposition Theorem
Source Transformation
Thevenin’s Theorem
Norton’s Theorem
Maximum Power Transfer
Introduction



Direct methods are not suitable to solve
complex and large circuits or as we
demand many unknowns.
To aid the analysis of complex circuit
structures-need to develop more powerful
techniques from the basic laws by
systematic approaches: Nodal Analysis
and Mesh Analysis.
These two techniques can be used to
solve almost any kind of circuit analysis
problems by means of a set of
simultaneous equations.
Introduction (Continued…)
Four ways of solving simultaneous
equations:
1. Cramer’s rule
2. Calculator (real numbers only)
3. Normal substitution and elimination
(not more than two equations)
4. Computer program packages:
matcad, maple, mathematica etc.

Introduction (Continued…)


Circuit analysis problems in this course
will be limited to three linear
simultaneous equations for conventional
hand solutions.
Circuit theorems merely developed to
simplify circuit analysis applicable to
linear circuits such as Thevenin’s and
Norton’s theorems, superposition
theorem, source transformation and
maximum power transfer.
Introduction (Continued…)


Circuit theorems are not analysis
techniques, rather they add up to the list
of simplifying/reduction techniques such
as the series-parallel reductions and -Y
transformations.
Although many computer aids facilitate us
as effective mathematics tools to solve
engineering problems they cannot replace
the compulsory needs to study the circuit
theories govern circuit behavior in
performing both circuit analysis and
design.
Circuit Analysis Method
NODAL
ANALYSIS
Concept


Developed based on the systematic
approach of Kirchhoff’s current law
(KCL) to find all circuit variables
without having to sacrifice any of
the elements.
General procedure which is making
use of node voltages in circuit
analysis as key solutions.
Importance terms



Node Voltage: Potential difference
between a marked node and the
selected reference node.
Element Voltage: Potential
difference across any element or
branch in the circuit.
When Node Voltage = Element
Voltage?
Why use Node Voltage?



Further reduce the number of
equations to be solved
simultaneously.
No of independent equations = No
of the marked nodes exclusive of
the reference node.
Element voltages and currents can
be obtained in few steps using the
solved node voltages.
Assumptions


KCL is performed with current going
out from a node as positive (+ve)
while current entering a node as
negative (-ve).
in – negative (subtract)
out – positive (add)
All unknown currents assumed to be
leaving a particular node.
Nodal Analysis Procedures:
1.
2.
3.
Mark all essential nodes and assign
proper voltage designations except for
the appointed reference node.
Apply KCL to each nonreference nodes.
Use Ohm’s law to formulate the equation
in terms of node voltages. Assume all
unknown currents are directing out of
the nodes.
Solve the resulting simultaneous
equations to obtain the unknown node
voltages.
Applying Nodal Analysis on Simple
Circuit
Example 1 (3 unknowns)
Is2
R1
Is1
R2
R3
R4
Solution
Step 1:
 Mark all essential
nodes
 Assign unknown
V1
node voltages
Is1
 Indicate the
reference node.
Is2
R1
V2
R3
R2
V3
R4
Solution (continued…)

Step 2: Perform KCL at each
marked nonreference nodes using
Ohm’s law to formulate the
equations in terms of the node
voltages.
Solution (continued…)

KCL V1: I s 1  I s 2
or
V1  V2

R1
(1)
I s 1  I s 2  G1 (V1  V2 )

KCL V2:
V2  0 V2  V3
V2  V1


0
R1
R3
R2
(2)

KCL V3:
I s2 
V3  V2
V 0
 3
0
R2
R4
(3)
Solution (continued…)

Step 3: Solve the
resulting
simultaneous
equations from
V1
step 2 above.
2 mA
3 mA
10 k
V2
5 k
4 k
V3
2 k
Solution (continued…)

KCL V1:
V1  V2
2m  3m 
10k
Simplify to

KCL V2:
Simplify to

V1 - V2 = 50
(1)
V2  V1 V2  0 V2  V3


0
10k
4k
5k
-2V1 + 11V2 - 4V3 = 0
KCL V3:
V3  V2 V3  0
3m 

0
5k
2k
Simplify to
-2V2 + 7V3 = -30
(2)
(3)
Solution (continued…)

Cramer’s rule: Put the equations in matrix forms.
Col-1
a1 b1 c1  V1  d1 
a b c  V   d 
 2 2 2  2  2
a3 b3 c3  V3  d 3 
Left
Col-1:
Col-2:
Col-3:



=
Col-2
Col-3
1  1 0  V1   50 
 2 11  4 V    0 

  2 

0  2 7  V3   30
Matrix:
coefficients of V1 i.e. a1, a2 and a3
coefficients of V2 i.e. b1, b2 and b3
coefficients of V3 i.e. c1, c2 and c3
Middle Matrix:
Right Matrix:
Unknown variables i.e. V1, V2 and V3
Constants i.e. d1, d2 and d3
Solution (continued…)


For third-order determinants, we use
shorthand methods of expansion
solution.
Shorthand method consists of
repeating the first two columns of the
determinant to the right of the
determinant and then summing the
products along the specific diagonals
as shown below.
Solution (continued…)

Use determinants to solve for each
variable:
  a2 b2
a3 b3
1 0
1
c2   2 11  4 -2
0 2 7 0
c3
a1 b1 c1
1
-1
11
-2
  (1)(11)(7)  (1)( 4)(0)  (0)( 2)( 2) 
(0)(11)(0)  (2)(4)(1)  (7)(2)(1)}
 77  (22)  55
Solution (continued…)

Determinant 1 when coefficients
for V1 is replaced by the constants.
d1 b1 c1
50  1
0
1  d 2 b2 c2  0 11  4
d3 b3 c3  30  2 7
50
0
-30
-1
11
-2
1  (50)(11)(7)  (1)( 4)( 30)  (0)(0)( 2) 
(30)(11)(0)  (2)(4)(50)  (7)(0)(1)}
 3730  (400)  3330
Solution (continued…)

Determinant 2 when coefficients
for V2 is replaced by the constants.
a1 d1 c1
1
50
0
 2  a2 d 2 c2   2 0  4
0  30 7
a 3 d 3 c3
1
50
-2
0
0 -30
 2  (1)(0)(7)  (50)( 4)(0)  (0)( 2)( 30) 
(0)(0)(0)  (30)(4)(1)  (7)(2)(50)}
  (580)  580
Solution (continued…)

Determinant 3 when coefficients
for V3 is replaced by the constants.
a1 b1 d1
1 1
50
 3  a 2 b2 d 2   2 11
0
0  2  30
a3 b3 d 3
1
-2
0
-1
11
-2
 3  (1)(11)( 30)  (1)(0)(0)  (50)( 2)( 2) 
(0)(11)(50)  (2)(0)(1)  (30)(2)(1)}
  130  (60)   70
Solution (continued…)




V1 = 1/ = 3330/55= 60.55 V
V2 = 2/ = 580/55 =10.55 V
V3 = 3/ = -70/55 =-1.27 V
You should verify your answers
using calculator for three unknowns.
Solution (continued…)

Knowing all the node voltages, we can
obtain the element voltages, element
currents and even power dissipated by the
passive elements such as:
VR1= V1 – V2
IR1 = (V1 – V2)/R1
PR1 = IR12R1 = VR12/R1
VR2= V2 – V3
IR1 = (V2– V3)/R2
PR2 = IR22R2 = VR22/R2
**VR3= V2
IR1 = V2/R3
PR3 = IR32R3 = VR32/R3
**VR4= V3
IR4 = V32/R4
PR4 = IR42R3 = VR42/R4
**In these two cases, the element voltages
identical to node voltages because one of its
terminals is at reference node.
Can you find the power dissipated by
the 10 k resistor?


Need to find the element voltage of
10-k resistor because power
calculation formula uses element
voltage.
P10k = (V1 – V2)2/10k
= (60.55 –10.55)2/10k
= 502/10k = 0.25 W
Applying Nodal Analysis on Simple
Circuit
Example 2 (2 unknowns)
Q: Find the power dissipated in the
20- resistor?
10 mA
50 
50 
50 
20 
9 mA
Solution
Step 1:
 Mark all
essential nodes
 Assign
V1
unknown node
voltages
50 
 Indicate the
reference node.
10 mA
Two nonreference nodes
50 
50 
V2
20 
9 mA
reference node
Solution (Continued…)

Step 2: Perform
V1
V1  V2
KCL at each
KCL V1:

 10m
100
50
marked
nonreference
(1)
3
V

2
V

1
Hence
1
2
nodes using
Ohm’s law to
formulate the
V
V

V
KCL
V
:
2
2
1
2
equations in

 9m  10m
20
50
terms of node
voltages.
Hence  2V1  7V2   0.1 (2)
Solution (Continued…)
 Step
3: Solve the resulting
simultaneous equations
which have been simplified
in step 2 above using
Cramer’s rule.
Solution (Continued…)

3
2
2
7
1 
1
 0 .1
2
7
 (3)( 7)  ( 2)( 2)
 (1)(7)  (2)( 0.1)
 17
 6 .8
2 
3
2
1
 0 .1
 (3)( 0.1)  ( 2)(1)
Hence,
V1 = 1/ = 6.8/17 = 0.4 V
 1 .7
V2 = 2/ = 1.7 /17 = 0.1 V
P20 = V22/20 = 0.12(20) = 0.2 W #
Applying Nodal Analysis on Circuit with
Voltage Sources


Three different effects depending on
placement of voltage source in the
circuit.
Does the presence of a voltage
source complicate or simplify the
analysis?
Case 1: Voltage source between two
nonreference essential nodes.
R
E
P
U
S
Nonreference
essential node
V1
Supernode
Equation:
E
D
NO
Vs
V2
Nonreference
essential node
Vs  V1  V2
Case 2: Voltage source between a reference
essential node and a nonreference essential
node.
E
D
NO
N GE
W
O LTA
N
K O
V
Nonreference
essential node
V1
Known node voltage:
Vs
0V
V1  Vs
Reference
essential node
Case 3: Voltage source between an
essential node and a non-essential node.
Vs
Nonreference
essential node
V1
FIN
NO D N
N- OD
ES E
TE SE VO
RM NT LT
S IA A
VO OF L N GE A
LT NO OD T
Va AG D E IN
E E
Non-essential
node
Node voltage at
non-essential node:
Va  V1  Vs
Example 3
(Supernode or Known node voltage)

Q: Find the power of the 10-V voltage
source? Is it supplying energy to the
circuit or absorbing energy from the
circuit? Show your work according to the
nodal analysis procedure.
10 V
40 
5
8
50 
25 
3A
Solution 1 (Supernode)

Step 1: Mark essential nodes and assign
unknown node voltages and indicate the
reference node.
10 V
i
V1
DE
O
N
ER
P
40 
SU
V2 5 
8
50 
30 
Checklist:
3 essential nodes – 1
ref node
– 1 supernode
= 1 KCL Eqn. + 1
3A
Supernode Eqn.
Solution 1 (Continued…)

Step 2: Perform KCL at each marked
nonreference nodes using Ohm’s law to
formulate the equations in terms of node
voltages.
KCL supernode V1/V2:
Hence
Supernode Equation:
V1
V2

3
80
8
V1  10V2  240
(1)
V1  V2  10
(2)
Solution 1 (Continued…)




Step 3: Solve the resulting
simultaneous equations which have
been simplified in step 2 above.
Solving Eqn. (1) and (2)
simultaneously yields,
V1 = 30.91 V and V2 = 20.91 V
(You can check this answer by
calculator or Cramer’s rule).
Solution 1 (Continued…)




Finding current through the voltage
source,
V1 V1  V2

i0
KCL at V1:
80
40
 30.91 30.91  20.91
i

  0.636 A
80
40
Hence,
P10-V = Vi= (10)(-0.636) = -6.36 W.
(Delivering energy)
Solution 2 (Known node voltage)

Step 1: Mark essential nodes and
assign unknown node voltages and
indicate the reference node.
10 V
i
5
8
50 
30 
3A
V2
KNOWN NODE
VOLTAGE V1
40 
Checklist:
3 essential nodes –
1 ref node
– 1 known node
voltage
= 1 KCL Eqn.
Solution 2 (Continued…)

Step 2: Perform KCL at each marked
nonreference nodes using Ohm’s law to
formulate the equations in terms of node
voltages.
Immediately known node
voltage at V1:
KCL V2:
V1  10 V
V2  10 V2

 3
80
8
11V2   230
(1)
Solution 2 (Continued…)

Step 3: Solve the resulting simultaneous
equations which have been simplified in
step 2 above.
 230
V2 
  20.91 V
11
Solving Eqn. (1) yields,
Finding current through the voltage source,
KCL at V1:
V
V V
1
40

1
2
80
i0
 10 10  20.91
i

  0.636 A
40
80
Solution 2 (Continued…)


Hence,
P10-V = Vi= (10)(-0.636) = -6.36 W.
(Delivering energy)
Example 4 (One of the terminals not
an essential node)

Q: Find the current through the 10-k
resistor. Show your work according to the
nodal analysis procedure.
8 mA
4 k
15 V
4 k
10 k
1 k
5 k
25 V
Solution

V1
1 k
Step 1: Mark essential nodes and assign
unknown node voltages and indicate the
reference node. For voltage sources, indicate the
node voltages at both ends with respect to the
assigned unknown node voltages at the essential
nodes
8 mA
4 k
15 V
V2
4 k
V3
Va = V2 + 15
10 k
i
Vb = -25 V
25 V
5 k
Checklist:
4 essential nodes – 1
ref node
= 3 KCL Eqns.
Solution (Continued…)

Step 2: Perform KCL at each marked
nonreference nodes using Ohm’s law to
formulate the equations in terms of node
voltages.
KCL V1:
V1
V1  (V2  15)

  8m
1k
4k
Hence
5V1  V2   17
KCL V2:
V2  25 (V2  15)  V1 V2  V3


0
10k
4k
4k
Hence
 10V1  18V2  10V3   250
(1)
(2)
Solution (Continued…)
KCL V3:
Hence
V3
V3  V2

 8m
5k
4k
 5V2  9V3  160
(3)
Solution (Continued…)
Step 3: Solve the resulting
simultaneous equations which have
been simplified in step 2 above.
 Solving Eqn. (1) till (3)
simultaneously yields,
V1 = -5.43 V, V2 = -10.17 V and
V3 = 12.13 V
(You can check this answer by
calculator and Cramer’s rule).

Solution (Continued…)


Finding current through the 10-k
resistor,
(V2  15)  V1 V2  V3

0
KCL at V2: i 
4k
4k
10.17  15  5.43 10.17  12.13
i

 3.01 mA
4k
4k
Applying Nodal Analysis on Circuit with
Dependent Sources



Circuits contain dependent sources
either VCVS, CCVS, VCCS or CCCS.
The presence of the dependent
sources require ‘Constraint Equation’
(CE).
CE describes the dependent term of
the dependent sources in relation to
the assigned unknown node voltages
or known values at the essential
nodes.
Example 5 (Circuit with dependent
sources)

Q: Use the node-voltage method to find
both dependent terms iO and Vx of the
dependent sources of the circuit in Figure
below.
3V
x
8
10 io
+ Vx -
10 
2
12 V
io
Solution

Step 1: Mark essential nodes and assign
unknown node voltages and indicate the
reference node.
3 Vx
DE
O
N
8  V3
V2
KNOWN
ER
P
10 io
SU
V1
+ Vx -
10 
2
12 V
io
Checklist:
4 essential nodes – 1
ref node – 1 s/node –
1 known
= 1 KCL Eqn. + 1
s/node Eqn. + 2
contraint Eqns.
Solution (Continued…)

Step 2: Perform KCL at each marked
nonreference nodes using Ohm’s law to
formulate the equations in terms of node
voltages.
Known node voltage:
KCL s/node V2:
Hence
S/node equation:
V3   12 V
V2 V2  12

  10io
2
8
5V2  80io   12
(1)
3V x  V1  12
(2)
Solution (Continued…)
Constraint equations: V x  V2
and
V2  12 V1
io 
  10i o
8
10
Hence
720io  8V1  10V2   120
(3)
(4)
Substituting Eqn. (3) into (2) yields
 V1  3V2  12
(2’)
Solution (Continued…)


Step 3: Solve the resulting simultaneous
equations which have been simplified in
step 2 above.
Solving Eqn. (1), (2’) and (4)
simultaneously yields,
V1 = -6.51 V, V2 = 1.83 V and io = -0.264 A

(You can check this answer by calculator
and Cramer’s rule).
Chapter 3, Problem 16.

Determine voltages v1 through v3 in the circuit of
Fig. 3.64 using nodal analysis. (Ans:V1=18.86V,
v2=6.29V, V3=13V)
Figure 3.64
Chapter 3, Problem 30.

Using nodal analysis, find vo and io in the circuit
of Fig. 3.78. (Ans: Vo=-1.344kV, io=-5.6A)
Figure 3.78
Chapter 3, Problem 31.

Find the node voltages for the circuit in Fig. 3.79.
(Ans: V1=4.97V, V2=4.85V, V3=-0.1212V)
Figure 3.79
Chapter 3, Problem 32.

Obtain the node voltages v1, v2, and v3
in the circuit of Fig. 3.80. (Ans:V1=2V,
V2=12V, V3=-8V)
5 k
10 V
V2
20 V
V1
4 mA
Figure 3.80
V3
12 V
10 k
Circuit Analysis Method
MESH ANALYSIS
Concept



Similar to nodal analysis.
Developed based on the systematic
approach of Kirchhoff’s voltage law
(KVL) to find all circuit variables
without having to sacrifice any of
the elements.
General procedure which is making
use of mesh current in circuit
analysis as key solutions.
Importance terms



Mesh Current: Assigned unknown
current flows around the perimeter
of the particular mesh/loop.
Element Current: Actual current
thru any element or branch in the
circuit.
When Mesh Current = Element
Current?
Assumptions

KVL is performed in clockwise
direction.
Voltage rise – negative (subtract)
Voltage drop– positive (add)
Mesh Analysis Procedures:
1.
2.
3.
Label all independent meshes and
assign proper unknown mesh
currents in clockwise direction. Do the
checklist.
Formulate KVL/Supermesh/Constraint
Equation.
Solve the resulting simultaneous
equations to obtain the unknown
mesh current.
Applying Mesh Analysis on Simple
Circuit
PP 3.5 (2 unknowns)
Q: Find power dissipated in 12-resistor
and 3-resistor using mesh analysis.

2
9
12 
12 V
4
3
8V
Solution
1.
Label all independent meshes and assign
proper unknown mesh currents in clockwise
direction.
2
12 V
I1
12 
4
9
Checklist:
2 meshes = 2 KVL
Eqns.
I2
3
8V
Solution (Continued…)
2.


Formulate
KVL/Supermesh/Constraint Eq.
KVL I1: 18I1 – 12I2 = 12
KVL I2: -12I1 + 24I2 = -8
(1)
(2)
Solution (Continued…)
3.



18
 12
Solve the
 12 24
resulting
 (18)( 24)  ( 12)( 12)
simultaneous
 288
equations to
12  12
1 
obtain the
 8 24
unknown
 (12)( 24)  (12)( 8)
mesh current.
 192
18 12
I1 = 1/
2 
 12  8
I2 = 2/
 (18)( 8)  (12)(12)
0
Solution (Continued…)





Using calculator/Cramer’s rule we
obtain:
I1 = 0.667 A and I2 = 0 A
P12 = (I1 -I2)2(12) = 5.33 W
P3 = I22(3) = 0 W
Notice that the branch (3-resistor)
forming the outer most boundary of
the circuit will have mesh current =
element current.
Circuit with current sources and
dependent sources



Two different effects depending on
placement of voltage source in the circuit.
Does the presence of a current source
complicate or simplify the analysis?
The presence of dependent source in the
circuit need to impose constraint equation
to describe the r/ship btw. dependent
term of the dependent sources in relation
to the mesh currents.
Case 1: Current source located at the
outer most boundary



Connecting mesh current immediately known.
No need to apply KVL around that loop/mesh.
Mesh Current = Element Current = Current
Source Value
R1
R2
Is
I2
I1
R3
12 V
Immediately known
mesh current,
I3
R4
I3 = -Is
Case 2: Current source located at the
boundary between 2 meshes



Enclose the current source and combine
the two meshes to form a SUPERMESH.
KVL is performed around the supermesh –
do not consider voltage across cur.
source.
Formulate s/mesh equation – express the
r/ship btw mesh currents that form the
s/mesh and cur.source that it encloses.
SUPERMESH

KVL S/Mesh I2/I3:
-12 + I2R2 +I3R3 = I3R4
R1
R2
12 V
I1
S/Mesh Eq:
I3 – I2 = 3 mA
R3
I2

I3
3 mA
R4
Practice Problem 3.7 (S/Mesh)

Use the mesh analysis to determine i1,
i2 and i3.
Figure 3.25
Solution

Step 1: Checklist.
Checklist:
3 meshes – I s/m =
2 KVL Eqns. + 1 s/node
Eq.
Solution (Continued…)




Step 2: Formulate KVL/s.mesh
equation.
KVL i3: -2i1 – 4i2 + 8i3 = 0 (1)
KVL s/mesh i1/i2:
2i1 + 12i2 – 6i3 = 6
(2)
S/Mesh Eq: i1 – i2 = 3
(3)
Solution (Continued…)


Step 3: Solve the simultaneous
equations using Cramer rule or by
calculator.
We obtain, i1 = 3.474 A
i2 = 0.4737 A
i3 = 1.1053 A
Example 6 ( Known current &
dependent source)

Find the voltage of the dependent source
(CCCS).
9
5 ix
9
8V
ix
12 
Solution

Step 1: Assign mesh currents in CW
direction and perform checklist.
9
5 ix
I1
9
I
8V 2
ix
12 
Checklist:
2 meshes – I known =
1 KVL Eqn. + 1 CE
Solution (Continued…)
Step 2: Formulate
KVL/Constraint equations.
 Immediately known, I1 = 5ix
 KVL I2:
21I2 = -8
 CE:
ix = I2

(1)
(2)
(3)
Solution (Continued…)



Step 3: Solve the simultaneous
equations.
Substitute (3) into (1) and solve (1)
and (2) simultaneously, we obtain
I1 = 1.9048 A
I2 = -0.3810 A
Solution (Continued…)

To find voltage across the CCCS,
perform KVL around loop I1.
9
9
ix
KVL I1: -V + 9I1 – 8 = 0
+
V
-
I1
8 V 12 
V = 9(1.9048) – 8
= 9.1432 V
Chapter 3, Problem 55.

In the circuit of Fig. 3.97, solve for i1,
i2, and i3. (Ans: i1=-1A, i2=0A, i3=2A)
Figure 3.97
Circuit Theorem
SUPERPOSITION
Advantages


Use of superposition theorem: to
find solution to circuits with multiple
independent sources which are
neither series nor parallel.
Advantage: no need to solve
simultaneous equations (tedious
computation for complex cct) in
order to find the circuit variables by
simplification techniques.
Concept

Concept: each independent source
is treated independently and the
algebraic sum is found to determine
a particular unknown quantity or
circuit variable of the circuit under
study.
Superposition Theorem
ST states that:
“ The current or voltage of any
element in a bilateral circuit is equal
to the algebraic sum of the currents
or voltages produced independently
by each source.”

Principle of Operation




To consider the effect of each source
independently requires that source to be
removed and replaced without affecting
the final results.
To remove voltage source – s.c the
terminals.
To remove current source – o.c the
terminals.
Any dependent source treated as though
they are passive element (must be left
intact during the process).
Example 6 (P3.5)

Find the voltage across the 12 resistor using
superposition hence the power dissipated by
this resistor.
2
9
+
V
-
12 V
4
12 
3
8V
Solution

i) Consider 12V/removed 8V.
2
9
+
V’
-
12 V
4
12 
3
s.c
12 // 12
V' 
x 12V
12 // 12  4  2
 6V
Solution (Continued…)

ii) Replace 8V/removed 12V
2
s.c
+
V’’
4
9
12 
3
8V
6 // 12
x 8V
6 // 12  9  3
 2V
V '' 
Solution (Continued…)


Hence, V = V’ + V’
= 6V + 2V = 8V.
P12 = V2/R = 82/12 = 5.33W
Example 7

Find the current in the 23  resistor using the
concept of superposistion.(Ans:11.23 A)
4
I
27 
47 
20 A
200 V
23 
Exercise 1

Using superposition, find the voltage V
in the circuit? (Ans: 40V)
6A
20 
10 
+
100 V
V
-
40 
I2
2R5
Circuit Theorem
SOURCE
TRANSFORMATON
What benefits from source
transformation?


Another tool to simplify circuit – the
simpler the cct, the easier will be
the solution.
How to simplify? – rearrange the
resistors/sources by S.Trans so that
they end up with series/parallel
connections.
Principle of Operation



The terminal v-i characteristics
must retain before and after
transformation as this concept is
based on equivalence.
S.Trans also applies to dependent
sources.
It does not affect the remaining
part of the circuit.
Definition

A Source Transformation is the
process of replacing a voltage
source Vs in series with resistor Rs
by a current source is in parallel
with the same resistor Rs or vice
versa.
Equivalent Circuits

The connections of each case should be between the
same terminals before and after transformation.
Rs
x
x
Is
Vs
Rs
y
In order for the
circuits in the left
and right sides to
be equivalent:
Vs  I s Rs
y
Vs
Is 
Rs
Example 8

Use series of source transformations to
find io in the circuit below.
10 A
4 
i
1 
40 
4 A
5 
2 
10 V
Solution

Transform 4A and 5 into voltage
source.
10 A
5 
20 V
4 
i
1 
40 
2 
10 V
Solution (cntd…)
9


Transform 10 A
and 1  into
voltage source.
Transform 10 V
and 40  into
current source.
1
10 V
i
20 V
0.25 A
40 
2
Solution (ctnd…)

Transform 10V and 10 into current
source.
i
1A
10 
0.25 A
40 
2
Solution (cntd…)



Combine the
current sources
2A and 0.25A.
Combine
resistors 10
and 40.
Solve for I
using CDR.
i
1.25 A
8
8
i
x 1.25 A  1 A
10
2
Practice Problem 4.6

Find io in the circuit of Figure 4.19 using
source transformation. (Ans:1.78A)
5V
5A
6
3
1
io
7
3A
4
Practice Problem 4.7

Use S.Trans to find ix in the cct shown in
Figure 4.22. (Ans:1.176A)
5
4A
ix
10 
2ix
Exercise 2

Use STrans to find Vo. (Ans:-135V)
340 V
8
80 
5A
5
+
Vo
20 
10 
45 
Circuit Theorem
THEVENIN’S
THEOREM
Purpose



Used when we are interested ONLY in the
terminal behavior of the circuit
particularly where a variable load is
connected to.
Provides a technique to replaced the fixed
part of the circuit by a simple equivalent
circuit.
Avoid the re-do on the analysis of the
entire circuit except for the changed load.
Thevenin’s Theorem

States that “ A linear two-terminal
circuit can be replaced by an
equivalent circuit consisting of an
open-circuit voltage source at the
terminals, VTh in series with a
resistor RTh where RTh is the input or
equivalent resistance at the
terminals when all independent
source are turned off.”
Replacing linear two-terminal (a-b)
circuit by its Thevenin equivalent

Original circuit
Replacing linear two-terminal (a-b)
circuit by its Thevenin equivalent (Cntd)

Thevenin equivalent circuit
VTh - Thevenin voltage
RTh - Thevenin resistance
Procedures to obtain VTh and RTh



Step 1: Priliminary – Omitting load resistor
RL (Not applicable if no load resistor)
Step 2: Find RTh – setting all independent
sources to zero. Find the resultant
resistance between the marked terminals.
Voltage source – s.c
Current sorce – o.c
Step 3: Find VTh – calculate VTh by returning
all sources back to their original positions.
Find the o.c voltage btw the marked
terminals using the method which takes
least effort.
Example 9

Find the Thevenin equivalent between
terminal a-b. (Ans: VTh=32V, RTh=8)
5
4
a
25 V
20 
3A
b
Example 10

Find the Thevenin equivalent circuit at the terminal ab of the circuit below. (Ans: VTh=-4.8V, RTh=2.4)
4
a
6
3
b
8V
2
Practice Problem 4.8

Use the Thevenin’s theorem to find the equivalent
circuit to the left of the terminals a-b in the circuit
below. Then find i. (Ans: VTh=6V, RTh=3, i=1.5A)
6
6
a
i
12 V
2A
4
b
1
Circuit Theorem
NORTON’s
THEOREM
Norton’s Theorem


The purpose of its use is similar to the
Thevenin’s theorem.
States that “A linear two-terminal circuit
can be replaced by an equivalent circuit
consisting of a short circuit current source
through the terminals, IN in parallel with a
resistor RN where RN is the input or
equivalent resistance at the terminals
when all independent source are turned
off.”
Replacing linear two-terminal (a-b)
circuit by its Norton equivalent


(a) Original Circuit
(b) Norton Equivalent Circuit
Finding Norton current, IN
Procedures to obtain VTh and RTh



Step 1: Priliminary – Omitting load resistor
RL (Not applicable if no load resistor)
Step 2: Find RTh – setting all independent
sources to zero. Find the resultant
resistance between the marked terminals.
Voltage source – s.c
Current sorce – o.c
Step 3: Find VTh – calculate VTh by returning
all sources back to their original positions.
Find the o.c voltage btw the marked
terminals using the method which takes
least effort.
Procedures to obtain IN and RN



Step 1: Priliminary – Omitting load resistor
RL (Not applicable if no load resistor)
Step 2: Find RN – setting all independent
sources to zero. Find the resultant
resistance between the marked terminals.
Voltage source – s.c
Current sorce – o.c
Step 3: Find IN – calculate IN by returning
all sources back to their original positions.
Find the o.c s.c current btw the marked
terminals using the method which takes
least effort.
NORTON EQUIVALENT CIRCUIT




A Norton equivalent circuit consists of an
independent current source in parallel
with the Norton equivalent resistance.
Can be derive from a Thevenin
equivalent circuit simply by making a
source transformation.
Norton current, IN = the short-circuit
current at the terminal of interest.
Norton resistance, RN = Thevenin
resistance, RTh
Series resistors combined, producing the
Thevenin equivalent circuit
8
a
32V
THEVENIN
EQUIVALENT
CIRCUIT


b
Producing the Norton equivalent circuit
a
4A
NORTON
EQUIVALENT
CIRCUIT
8
b
Example 10

Find the Norton Equivalent circuit with respect to
the terminals a-b.
2
a
9A
12 
9
b
4
Practice Problem 4.11

Find the Norton equivalent circuit in
Figure below. (Ans: IN=4.5A, RN=3)
3
15 V
4A
3
a
6
RL
b
Circuit Theorem
THEVENIN & NORTON’s
THEOREMS WITH
DEPENDENT SOURCE
Procedures to obtain VTh/IN and RTh/RN



Step 1: Priliminary – Omitting load resistor
RL (Not applicable if no load resistor)
Step 2: Find VTh = Vo.c or IN = Is.c using the
method that takes the least effort.
Step 3: Find RTh/RN
Method 1:If circuit contains independent
source.
Rth = RN = Vo.c/Is.c = VTh/IN
Method 2: If circuit contains independent
source and without independent source.
Using Method 2 to find RTh/RN






Turn off all independent sources but dependent
sources left intact because they are controlled by
circuit variables.
Because of the presence of the dependent source,
we excite the circuit with a voltage source or
current source between the terminals.
Set Vo=1V to ease calculation since the circuit is
linear.
Goal? To find io so that RTh=RN=1/io
Alternatively, we may set io=1A.
Goal? To find Vo so that RTh=RN=Vo/1
Illustration of Method 2 to find RTh/RN
Example 11

Find the Thevenin and Norton equivalent circuit
for the circuit containing dependent sources below
between terminals a-b. (Ans: VTh=-5V, RTh=100,
IN=-50mA)
2 k
a
i
+
5V
3Vo
20io
25 
Vo
-
b
Example 12

Find the Thevenin and Norton equivalent circuit
for the circuit containing dependent source
below between terminals a-b. (Ans: VTh=20V,
RTh=0.625, IN=32A)
2ix
5 
a
ix
40 V
8 A
1 
b
Practice Problem 4.9

Find the Thevenin equivalent circuit of the circuit
in Figure 4.34 the left of terminals a-b. (Ans:
VTh=5.33V, RTh=0.44)
5
3
a
6V
1.5Ix
4
b
Practice Problem 4.10

Obtain the Thevenin equivalent of the circuit in
Figure 4.36. (Ans: VTh=0V, RTh=-7.5)
10 
4Vx
a
+
Vx
-
5
15 
b
Circuit Theorem
MAXIMUM
POWER
TRANSFER
Introduction


Power transfer from source to the load can be
analyzed and discussed from two basic types of
systems:
1. Efficiency – eg: power utility systems
concerned with generation, transmission and
distribution of large quantities of electric power.
2. Amount – eg. Comm. & instrumentation sys
because in the transmission of info or data via
electric signals, the power available at the
transmitter or detector is limited or small.
At this moment our concern is on the 2nd type
of system that is the amount of maximum
power transfer in purely resistive circuit.
Thevenin Equivalent Circuit


The Thevenin equivalent circuit is
useful in finding the max. power a
linear cct. can deliver to a load.
The entire cct can be replaced by its
Thevenin eq. except for the
adjustable load.
Thevenin Equivalent Circuit used for
maximum power Transfer
VTh and RTh
are fixed
Power delivered
to the load,
 VTh
p  i RL 
 RTh  RL
2
2

 RL

(1)
Power delivered to the load as a
function of RL.
Maximum Power Theorem

Maximum Power is transferred to
the load when the load resistance,
equals to the Thevenin resistance as
seen from the load (RL = RTh)
Proving Maximum Power Transfer
Theorem

Differentiate p in Eq.(1) with
respect to RL and set the result
equal to zero,
2

(
R

R
)
 2 RL ( RTh  RL ) 
dp
2
Th
L
 VTh 

4
dRL
( RTh  RL )


 VTh
2
 ( RTh  RL  2 RL 

0
3
 ( RTh  RL ) 
(2)
Proving Maximum Power Transfer
Theorem (Cntd…)





Implies that,
0 = (RTh + RL -2RL) = (RTh – RL)
Yields,
RL = Rth (3)
Eq (3) gives the maximum power
by showing that d2p/dRL2 < 0.
Maximum Power Formula

Substituting Eq.(3) into (1) to
obtain the maximum power transfer,
2
p max

VTh

4 RT h
(4)
Eq.(4) applies only when RL = RTh.
When RL ≠ RTh, compute power
from Eq.(1)
Practice Problem 4.13


Determine the value of RL that will draw the
maximum power from the rest of the circuit.
Calculate the maximum power.
(Ans: 4.22, 2.901W)
2
Vx
4
1
9V
RL
3Vx
Example 13

The variable resistor in the circuit below
is adjusted for maximum power. Find
the value of RL and the maximum
power. (Ans: 5k, 45mW)
6 k
18 V
5 mA
1k25
a
RL
10 k
b
Exercise 3

The load resistance in both circuits below are
adjusted until maximum power is delivered. Find
the power delivered to the loads and the value of
both RL. (Ans: 600, 38.4mW)
(a)
1 k
500 
1 k
I2
10 V
2 mA
RL
Exercise 3
(b)
(Ans: 21.7, 0.8W)
10 
50 V

RL
a
b
20 
30 