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CIRCUIT ANALYSIS
METHOD
TOPIC
• Node-Voltage Method
• Mesh-current Method
• Source of embodiment
principle
• Thevenin’s Circuit
• Norton’s Circuit
• Maximum Power Transfer
• Superposition Principle
INTRODUCTION TO
NODE-VOLTAGE
METHOD
• Base on Kirchhoff’s Current
Law
• Important step: select one
node as a reference.
Example:Node-Voltage method
• Previous circuit, set node 3 as
a reference. By using
Kirchhoff’s Current law at
node 1,
V1  10 V1 V1  V2
0
 
1
5
2
Node-voltage equation at
node2
V2  V1 V2
0
 2
2
10
• Solve previous equation
100
V1 
 9.09V
11
120
V2 
 10.91V
11
NODE-VOLTAN METHOD
THAT CONTAIN DEPENDENT
SOURCE
• If the circuit contains dependent
source, the node-voltage equation
imposed by the presence of the
dependent source.
Find power that absorb by 5Ω
Resistor using node-voltage
method.
• Circuit have 3 node.
• Need 2 node-voltage equations.
• Summing the currents away
from node 1 generates the
equation,
V1  20 V1 V1  V2


0
2
20
5
• Summing the currents away
from node 2 yields
V2  V1 V2 V2  8 i
 
0
5
10
2
• These two node-voltage
equations contain three
unknowns, namely, V1, V2 and iø .
To eliminate iø, we must
express this controlling current
in terms of the node-voltage,
V1  V2
i 
5
• Substituting this relationship
into the node 2 equation
simplifies the two nodevoltage equations
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
• Power absorb by 5Ω resistor
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 are three essential
nodes in this circuit, which
means that two simultaneous
equations are needed.
• There is only one unknown
node-voltage V2, but V1 =100V.
• Solution of this circuit thus
involves only a single nodevoltage equation at node 2.
V2  V1 V2

5  0
10
50
Have V1 =100V, and solved
V2 =125V.
SUPERNODE
• A supernode is formed by
enclosing a (dependent or
independent) voltage source
connected between two
nonreference nodes and any
elements connected in parallel
with it.
Example:supernode
• Select node:
• Node voltage equation at
node 2 and node 3
V2  V1 V2

i  0
5
50
V3
i 4  0
100
• Add previous equation
V2  V1 V2 V 3


4  0
5
50 100
• Previous equation could get
when use supernode concept
at node 2 and node 3.
Supernode
• From 5Ω resistor
V2  V1 V2 V3


4  0
5
50 100
• Pink equation was equal to
green equation.
• Using supernode at node 2
and 3 make it simple to
analyse the circuit.
• Have V1 =50V and V3 can be
describe with V2,
V3  V2  10 i
V2  50
i 
5
• Replace V1 =50, V3 and iø, pink
equation become
1
10 
 1 1
V2   

  10  4  1
 50 5 100 500 
V2 (0.25)  15
V2  60V
• Insert V2
60  50
i 
 2A
5
V3  60  20  80V
INTRODUCTION TO
MESH-CURRENT METHOD
• One mesh mean a loop that no
others loop inside.
• This mesh-current method
used Kirchhoff’s voltage law
to find current each mesh.
Example:Mesh-current
• From Kirchhoff’s law
i1  i2  i3
(1)
V1  i1R1  i3 R3
 V2  i2 R2  i3 R3
(2)
• Use i3 from equation (1) and
insert to equation (2)
V1  i1 ( R1  R3 )  i2 R3
 V2  i1R3  i2 ( R2  R3 )
• Mesh-current circuit with mesh
current ia and ib.
• Using KVL at those two mesh
V1  ia R1  ia  ib R3
 V2  ib  ia R3  ib R2
• After ia and ib known, then
we can calculate voltage at
power at each resistor.
MESH-CURRENT
METHOD THAT HAVE
DEPENDENT SOURCE
• When circuit have dependent
source, mesh-current
equation will have constant
value related to dependent
source.
Example:mesh method with
dependent source
• Find power that obserb by 4Ω
resistor using mesh-current
method.
• From Kirchhoff’s voltage law
50  5i1  i2   20i1  i3 
0  5i2  i1   1i2  4i2  i3 
0  20i3  i1   4i3  i2   15i
• Have i  i1  i3
• Insert equation iø to related
equation,
50  25i1  5i2  20i3
0  5i1  10i2  4i3
0  5i1  4i2  9i3
• By using Cramer law, i2 and i3
can be calculated as below,
 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 that absorb by 4Ω
resistor
pi R
2
 (28  26) 4
 16W
2
SPECIAL CASE
(SUPERMESH)
• When a branch of current
source can be remove and
use supermesh concept
(current source assume as
open circuit)
• Assume that current source as
open circuit
• Supermesh equation
 100  3ia  ib   2ic  ib 
 50  4ic  6ia  0
50  9ia  5ib  6ic
• Mesh-current equation for
mesh 2
0  3ib  ia   10ib
 2ib  ic 
• Known
ic –ia= 5A
By using Cramer law at those
three equation, value for
those three mesh current
could be calculated.
SOURCE
TRANSFORMATIONS
• Source transformation is the
process of replacing a voltage
source vs in series with a
resistor R by a current
source is in parallel with a
resistor R, or vice versa.
Source Transformation
Example:Source transformation
• When resistor R=0, terminal
a-b become close circuit.
Beginning, close circuit
current should be same.
Therefore,
Vs
Is 
Rs
• Close circuit current for
second circuit was Is.
Therefore,
Vs
Is 
Rp
• When resistor R = ∞, these
circuit become open circuit.
• From first circuit, we have
Vab =Vs . Therefore, it was
voltage for open circuit.
Vab  I s R p
• Vab for those two circuit should
be same. Therefore, Vs = Is Rp .
• Replace Is
 Vs 
Vs    R p
R
 s
 Rs  R p
• Summarize for Source
transformation
Before
After
Method
Tetapkan
R p  Rs
Vs
Is 
Rs
Before
After
Method
Tetapkan
Vs  I s R p
Rs  R p
THEVENIN EQUIVALENT
CIRCUIT
• Introduced in 1883 by M.
Leon Thevenin (1857-1926), a
French telegraph engineer.
• Thevenin’s theorem states that a
linear two-terminal circuit can be
replaced by an equivalent circuit
consisting of a voltage source VTh in
series with a resistor RTh where VTh
is the open-circuit voltage at the
terminals and RTh is the input or
equivalent resistance at the
terminals when the independent
sources are turned off.
• This theorem usually used to
replace large sequence part
(complex) with one simple
equivalent circuit. This simple
circuit makes voltage, current
and circuit power could be
calculated easily.
Thevenin equivalent circuit
• Thevenin voltage, VTh = open
circuit voltage for origin
circuit.
• When load decrease until
zero, circuit become close
circuit and current become:
VTh
isc 
RTh
VTh
R Th 
isc
Example
• Step 1: node-voltage equation
for open circuit:
V1  25 V1

3  0
5
20
V1  32V  VTh
• Step 2: replace close circuit at
a-b terminal
• Node voltage equation for
close circuit:
V2  25 V2
V2
 3  0
5
20
4
V2  16V
Close circuit current:
16
isc 
 4A
4
Thevenin resistance
VTh 32
RTh 

 8
isc
4
Thevenin equivalent
circuit
Norton equivalent circuit
• In 1926, about 43 years after
Thevenin published his theorem,
E. L. Norton, an American
engineer at Bell Telephone
Laboratories, proposed Norton’s
theorem.
• This equivalent circuit have one
independent source that parallel
with one resistor.
• Norton equivalent circuit could
have from Thevenin equivalent
circuit by source
transformation.
Example
Step 1: Source transformation
Step 2: Combine source and
parallel resistors
Step 3: Source transformation,
Series resistors combined,
producing the Thevenin equivalent
circuit
Step 4: Source transformation
and Producing the Norton
equivalent circuit
Norton equivalent circuit
TOPIC
• Node-Voltage method
• Mesh-current method
• Source transformation
principle
• Thevenin equivalent circuit
• Norton equivalent circuit
• Maximum power transfer
principle
• Superposition principle
MAXIMUM POWER
TRANSFER
• Power system designed to provide
power to load at high-efficiency
and decrease power loss when
delivered to load. Therefore, we
need to decrease source resistance
and delivering resistance.
• Definition for Maximum
power transfer tell that
power that transfer from one
source was represent by
Thevenin equivalent circuit
become max when load
resistor RL and Thevenin
resistor RTh was
Example
• Power absorb by resistor RL
p  i RL
2
 VTh
 
R

R
L
 Th
2

 RL

• Differentiate p with RL
 RTh  RL   RL  2RTh  RL  
dp
2
 V Th 

4
dRL
RTh  RL 


2
• Differential was zero and p
become maximum
RTh  RL 
2
 2RL ( RTh  RL )
• Solve
RTh  RL
• Therefore, for maximum
power transfer, RL must equal
with RTH .
• Maximum power transfer
equation:
2
p
VTh RL
2RL 
2
2
VTh

4 RL
SUPERPOSITION PRINCIPLE
• The superposition principle states
that the voltage across (or current
through) an element in a linear
circuit is the algebraic sum of the
voltages across (or current
through) that element due to each
independent source acting alone.
Step to Apply Superposition
Principle
1. Turn off all independent
sources except one source. Find
the output (voltage or current)
due to that active.
2. Repeat step 1 for each of
the other independent
sources.
3. Find the total contribution
by adding algebraically all
the contributions due to the
independent sources.
REMEMBER
!
1. Independent voltage
source become close circuit
with zero volt.
2. Independent current
source become open
circuit.
3. If dependent source exist,
it should be active while
superposition process.
Example
• Step 1: turn off current
source
• Use voltage divider law to
calculate V0 :
 10 
V0  2k    5V
 4k 
• Step 2: turn off voltage
source
• Use current divider law to
calculate V0 ,
2k
i0 
(2m)  1mA
4k
V0  (1m)( 2k )  2V
Total V0 :
V0 =2+5=7V.
Question 1
Answer
• 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
• 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
• Current source
I1  I 2  2V0
• known
V0  5( I 2  I 3 )
• Replace V0
I1  I 2  10 ( I 2  I 3 )
I1  11 I 2  10 I 3  0
• By using Cramer law
 0
 125

 0
I1 
10
0

 1
5  5

 5 10 
 11 10 
5  5

 5 10 
 11 10 
 5
 5
125

 11 10 
  5 10
0 10
0  5 
10
 5
 5



 11 10
1 10
1  11
 625

625
 1A
0
 5
10
• Current I2: 

0

125
10


 1
0
10 
I2 
 625
10  5
125

1 10 


 625
13125

 625
 21A
0 
10 5
• Current I3: 

0

5

125


 1  11
0 
I3 
625
10  5
125

1 10 


625
 14375

625
 23 A
Question 3
• Open circuit voltage:
• Node-voltage equation for
Voc
Voc  24 Voc

20
2
2
Voc  24  Voc  4  0
2Voc  20
Voc  10V
• Thevenin resistance:
RTH  2 2  4  5
• Get:
11
V0  (10)  6.88V
16
Question 4
• Close circuit current
12
I sc  3 
 6A
4
• Norton resistance
RTH = 4Ω
• Norton equivalent circuit:
V0  64 12  6(3)  18V
Question 5
• Turn off current
source
2
V  24   4V
 12 
1
0
• Turn off voltage source
11
0
V
 6 4 

2  4V
 12 
• Total V0
V V
1
0
11
0
 0V
Question 6
Node-voltage equation:
V0 V0  5i V0  80
3


0
200
10
20
V0  80
Known:
i 
20
• Get:
V0 =50V
• Finally we get:
V0 =50V