CIRCUIT
ANALYSIS
METHODS
Topic 3
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
INTRODUCTION OF NODEVOLTAGE METHOD
• Use KCL.
• Important step: select one of the
node as reference node
• Then define the node voltage in
the circuit diagram.
Node-voltage example
• In the diagram, node 3 is define as
reference node and node 1 and 2 as
node voltage V1 and V2.
• The node-voltage equation for node 1
is,
V1  10 V1 V1  V2
0
 
1
5
2
• Node-voltage equation of node 2,
V2  V1 V2
0
 2
2
10
• Solving for V1 and V2 yields
100
V1 
 9.09V
11
120
V2 
 10.91V
11
THE NODE-VOLTAGE METHOD
AND DEPENDENT SOURCES
• If the circuit contains dependent
sources, the node-voltage
equations must be supplemented
with the constraint equation
imposed by the presence of the
dependent sources.
example…
Use the node-voltage method to find the
power dissipated in the 5Ω resistor.
• The circuit has 3 node.
• Thus there must be 2 node-voltage
equation.
• Summing the currents away from node 1
generates the equation,
V1  20 V1 V1  V2


0
2
20
5
• Summing the current away from
node 2 yields,
V2  V1 V2 V2  8 i
 
0
5
10
2
• As written, these two equations contain
three unknowns namely V1, V2 and iØ.
• To eliminate iØ, express the current in
terms of node-voltage,
V1  V2
i 
5
• Substituting this relationship into
the node 2 equation,
0.75V1  0.2V2  10
 V1  1.6V2  0
• Solving for V1 and V2 gives,
V1  16V V2  10V
• Then,
16  10
i 
 1.2 A
5
p  i  R  1.445
2
 7.2W
SPECIAL CASE
• When a voltage source is the
only element between two
essential nodes, the nodevoltage method is simplified.
Example…
• There is three essential nodes, so
two simultaneous equation are
needed.
• Only one unknown node voltage, V2
where as V1=100V.
• Therefore, only a single nodevoltage equation is needed which is
at node 2.
V2  V1 V2

5  0
10
50
Using V1 =100V, thus V2=125V.
SUPERNODE
• When a voltage source is
between two essential
nodes, those nodes can be
combine to form a
supernode (voltage sourse
is assume as open circuit).
Supernode example…
• Nodes chosen,
• Node-voltage equation for node
2 and 3,
V2  V1 V2

i  0
5
50
V3
i 4  0
100
• Summing both equation,
V2  V1 V2 V 3


4  0
5
50 100
Above equation can be generates
directly using supernode approach
Supernode
• Starting with resistor 5Ω branch
and moving counterclockwise
around the supernode,
V2  V1 V2 V3


4  0
5
50 100
• Using V1 =50V and V3 as a function
of V2,
V3  V2  10 i
V2  50
i 
5
• Substitute into the node-voltage
equation,
1
10 
 1 1
V2   

  10  4  1
 50 5 100 500 
V2 (0.25)  15
V2  60V
• Using V2 value, gives
60  50
i 
 2A
5
V3  60  20  80V
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
INTRODUCTION OF MESHCURRENT METHOD
• A mesh is a loop with no loop
inside it.
• A mesh current is the current that
exist only in the perimeter of a
mesh.
• Mesh-current method use KVL to
generates equation for each mesh.
Mesh-current example…
• Mesh-current circuit with mesh
current ia and ib.
• Use KVL on both mesh,
V1  ia R1  ia  ib R3
 V2  ib  ia R3  ib R2
• Solving for ia and ib, and you
can compute any voltages
or powers of interest.
THE MESH-CURRENT METHOD
AND DEPENDENT SOURCES
• If the circuit contains dependent
sources, the mesh-current
equations must be
supplemented by the appropriate
constraint equations.
Example…
• Use the mesh-current method to determine
the power dissipated in the 4Ω resistor.
• Using KVL,
50  5i1  i2   20i1  i3 
0  5i2  i1   1i2  4i2  i3 
0  20i3  i1   4i3  i2   15i
i  i i
• But

1
3
• Substituting into the mesh-current
equation,
50  25i1  5i2  20i3
0  5i1  10i2  4i3
0  5i1  4i2  9i3
• Using Cramer rule, the values of i2 and i3
can be determine,
 25
 5

 5
i2 
 25
 5

 5
50  20

0 4 
0
9 
 5  20

10  4 
4
9 
  5  4
50

5 9 

i2 
  5  20
 25  20  25  5
 5
 10
 4



9 
9    5  4
 4
 5
50(65)
i2 
 5(125)  10(125)  4(125)
 3250

625  1250  500
 3250
i2 
 26 A
 125
 25  5 50
 5 10 0 


 5  4 0 
i3 
125
 5 10 
50

 5  4


125
3500

 28 A
125
• Power dissipated by 4Ω resistor
is
2
pi R
 (28  26) 4
2
 16W
SPECIAL CASE (SUPERMESH)
• When a branch includes a
current source, the mesh-current
method can be simplified.
• To create a supermesh, remove
the current source from the
circuit by simply avoiding the
branch when writing the meshcurrent equations.
• Supermesh equation,
 100  3ia  ib   2ic  ib 
 50  4ic  6ia  0
50  9ia  5ib  6ic
• Mesh 2 equation,
0  3ib  ia   10ib
 2ib  ic 
• From the circuit,
ic –ia= 5A
• Using Cramer rule, the three
mesh current can be obtain.
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
SOURCE TRANSFORMATION
• Source transformation allows a
voltage source in series with a
resistor to be replaced by a
current source in parallel with the
same resistor or vice versa.
Source transformation
Example…
• Source transformation procedure
From
To
method
Use,
R p  Rs
Vs
Is 
Rs
From
To
method
Use,
Vs  I s R p
Rs  R p
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
THEVENIN EQUIVALENT
CIRCUIT
• Thevenin equivalent circuit
consist of an independent
voltage source, VTh in series
with a resistor RTh.
Thevenin equivalent circuit
RTh
a
VTh


b
• Thevenin voltage, VTh = open circuit
voltage in the original circuit.
• Thevenin resistance, RTh is the ratio of
open-circuit voltage to the short-circuit
current.
VTh
isc 
RTh
VTh
R Th 
isc
Example…
5
25V


20
4
3A
a


V1
Vab

b
• Step 1: node-voltage equation for
open-circuit:
V1  25 V1

3  0
5
20
V1  32V  VTh
• Step 2: short-circuit condition at terminal
a-b
5
25V


4
a

20
3A
V2
I sc

b
• Node-voltage equation for shortcircuit:
V2  25 V2
V2
 3  0
5
20
4
V2  16V
Short-circuit current:
16
I sc 
 4A
4
Thevenin resistance:
VTh 32
RTh 

 8
I sc
4
Thevenin equivalent circuit
8
a
32V


b
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
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
Example
5
4
a
25V


20
3A
b
Step 1: Source transformation
4
5A
5
20
a
3A
b
Step 2: Parallel sources and parallel
resistors combined
4
8A
a
4
b
Step 3: Source transformation, series
resistors combined, producing the
Thevenin equivalent circuit
8
a
32V
THEVENIN
EQUIVALENT
CIRCUIT


b
Step 4: Source transformation, producing
the Norton equivalent circuit
a
4A
NORTON
EQUIVALENT
CIRCUIT
8
b
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
MAXIMUM POWER TRANSFER
• Two basic types of system:
– Emphasizes the efficiency of the power
transfer
– Emphasizes the amount of power
transferred.
• Maximum power transfer is a technique
for calculating the maximum value of p
that can be delivered to a load, RL.
• Maximum power transfer occurs when
RL=RTh.
Example…
RTh
VTh
a

i

b
RL
• Power dissipated by resistor RL
p  i RL
2
 VTh
 
R

R
L
 Th
2

 RL

• Derivative of p with repect to RL






dp
R

R

R

2
R

R
2
Th
L
L
Th
L
 V Th 

4
dRL
RTh  RL 


2
• Derivative is zero and p is
maximum when
RTh  RL 
2
 2RL ( RTh  RL )
RTh  RL
• The maximum power transfer occurs
when the load resistance, RL = RTh
• Maximum pwer transfer delivered to RL:
2
p
VTh RL
2RL 
2
2
VTh

4 RL
CIRCUIT ANALYSIS
METHODS
•
•
•
•
•
•
•
Node-Voltage method
Mesh-current method
Source transformation
Thevenin equivalent circuit
Norton equivalent circuit
Maximum power transfer
Superposition principle
SUPERPOSITION
PRINCIPLE
• In a circuit with multiple independent
sources, superposition allows us to
activate one source at a time and sum the
resulting voltages and currents to
determine the voltages and currents that
exist when all independent sources are
activate.
Step of Superposition principle
1. Deactivated all the sources and only
remain one source at one time. Do
circuit analysis to find voltages or
currents.
2. Repeat step 1 for each independent
sources.
3. Sum the resulting voltages or
currents.
REMEMBER!!
!
1. Independent voltage source will
become short-circuit with 0Ω
resistance.
2. Independent current source will
become open-circuit.
3. Dependent sources are never
deactivated when applying
superposition.
Example…
• Step 1: deactivated all sources except
voltage source
• V0 is calculated using voltage
divider:
 10 
V0  2k    5V
 4k 
• Step 2: Deactivated all sources except
current source
• V0 is calculated by using current
divider:
2k
i0 
(2m)  1mA
4k
V0  (1m)( 2k )  2V
• Step 3: Sum all the resulting voltages:
V0 =2+5=7V.
Question 1 (nodevoltage)
• Calculate the value of Io
Solution
• Node 1:
V1 V1  V2 

 2  4
1
2
 3  V2
V1   
 6
2 2
• Node 2:
V2  V1 V2 V2
 
4
2
2
4
V1
1 1 1
  V2      4
2
2 2 4
V1
5
  V2    4
2
4
 2  6
 1

4
2

V2  3
1

 2
2
 1
5 
4 
 2
63

 1.846
1.625
3
1.846
I 0 
 0.923 A
2
Question 2 (mesh-current)
• Determine the value of currents, I1, I2 and I3.
• Supermesh:
10 I1  5( I 2  I 3 )  0
• Mesh 3:
5I 3  5I 3  I 2   125  0
10 I 3  5I 2  125
• Dependent current source
I1  I 2  2V0
• Vo
V0  5( I 2  I 3 )
• Substitute V0
I1  I 2  10 ( I 2  I 3 )
I1  11 I 2  10 I 3  0
• Use Cramer rule
 0
 125

 0
I1 
10
0

 1
 5

 5 10 
 11 10 
5  5

 5 10 
 11 10 
5
 5
 5
125

 11 10 
  5 10
0 10
0  5 
10 
 5
 5



 11 10
1 10
1  11
 625

625
 1A
• Current I2:
0
 5
10
 0  125 10 


 1
0
10 
I2 
 625
10  5
125

1 10 


 625
13125

 625
 21A
• Current I3:
0 
10 5
 0  5  125


 1  11
0 
I3 
625
10  5
125

1 10 


625
 14375

625
 23 A
Question 3 (thevenin)
• Open-circuit voltage, Voc:
• Node-voltage equation for Voc
Voc  24 Voc

20
2
2
Voc  24  Voc  4  0
2Voc  20
Voc  10V
• Thevenin resistance, RTh:
RTH  2 2  4  5
• Thevenin
equivalent
circuit:
11
V0  (10)  6.88V
16
Question 4 (norton)
• Open-circuit current, Isc:
12
I sc  3 
 6A
4
• Norton resistance, RN:
RN = 4Ω
• Norton equivalent circuit:
V0  64 12  6(3)  18V
Question 5 (superposition)
• Use superposition principle to determine
the voltage Vo.
+
• Deactivated current source
-
+
 2
V0  24   4V
 12 
• Deactivated voltage source
11
0
V
 6 4

2  4V
 12 
• Summing the voltage V0
V0  V0  0V
Question 6 (node-voltage)
• Determine the value of Vo.
node-voltage equation:
V0 V0  5i V0  80
3


0
200
10
20
V

80
Current iΔ: i  0

20
• Thus:
V0 =50V