Techniques of Circuit analysis
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Transcript of Techniques of Circuit analysis
Menuntut ilmu adalah TAQWA
Menyampaikan ilmu adalah IBADAH
Mengulang-ulang ilmu adalah ZIKIR
Mencari ilmu adalah JIHAD
Chapter 1 : Techniques of DC Circuit
Analysis
Engr. Mohd Riduwan bin Ghazali (Grad. IEM)
1.1 Review Circuit Analysis I:
Most Important thing that should remember:
Ohm Laws, V = IR
(Kirchoff’s) KCL/KVL
Independent/Dependent Source
Nodal/SuperNode
Mesh/SuperMesh Mesh/SuperMesh
Source Transformation
OHM’S LAW
Ohm’s Law state that the voltage, v across a resistor is directly proportional to the current, i flowing through the resistor.
Current and voltage are linearly proportional
V=iR
KIRCHHOFF’S LAW
• States that the sum of currents entering a node (or closed boundary) is zero.
Kirchhoff current law (KCL)
1 2 3 4 5( ) ( ) 0i i i i i+ − + + + − =
1 3 4 2 5i i i i i+ + = +
Total current in = Total current out
KIRCHHOFF’S LAW
• States that the sum of voltages around a closed path (or loop) is zero
Kirchhoff voltage law (KVL)
1 2 3 4 5 0v v v v v− + + − + =
2 3 5 1 4v v v v v+ + = +
CIRCUIT ELEMENTS
Active Elements Passive Elements
IndependentSources
(round shape)
DependantSources
(diamond shape)
• voltage source comes with polarities (+-) in its symbol• current source comes with an arrow
NODAL ANALYSIS
Provides a general procedure for analyzing circuits using node voltagesas the circuit variables.• Steps:
i) select a node as a reference node (ground)ii) assign voltage designations to nonreference nodesiii) redraw the circuit to avoid too much information on the
same circuit
iv) apply KCL for each nonreference node- at node 1: I1 = I2 + i1 + i2 -at node 2: I2 = i3 – i2
v) apply Ohm’s law (current flows from a higher potential to a lower potential)
R
vvi lowerhigher −
=1
11
0
R
vi
−=2
212 R
vvi
−=3
23
0
R
vi
−=
Nodal Analysis With Voltage Sources
We now consider how voltage sources affect nodal analysis:
CASE 1: If a voltage source is connected between the reference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source
E.g.: v1 = 10V
CASE 2: If the voltage source (dependent or independent) is connected CASE 2: If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; apply both KCL and KVL to determine the node voltagesE.g.: nodes 2 and 3 form a supernode
MESH ANALYSIS
Steps to determine mesh currents
• assign mesh currents i1, i2,,…..,in to the nmeshes
• apply KVL to each of the n meshes. Use Ohm’s law to express the
voltages in terms of the mesh currents
• solve the resulting simultaneous equations to get the mesh
currents1,A p p ly in g K V L to m e s h
1 1 1 3 1 2
1 3 1 3 2 1
1,
( ) 0
( )
A p p ly in g K V L to m e s h
V R i R i i
R R i R i V
− + + − =+ − =
2 2 2 3 2 1
3 1 2 3 2 2
2 ,
( ) 0
( )
A p p ly in g K V L to m e s h
R i V R i i
R i R R i V
+ + − =− + + = −
1 1
2 2
1 3 3
3 2 3
i VR R R
R R R i V
+ − = − + −
CASE 1
When a current source exists only in one mesh, we set i2 = -5A and write a mesh equation for the other mesh in the usual way:
-10 + 4i1 + 6(i1 – i2) = 0 i1 = -2A
MESH ANALYSIS WITH CURRENT SOURCES
CASE 2
When a current source exists between two meshes, we create a supermeshby excluding the current source and any elements connected in series with it
MESH ANALYSIS WITH CURRENT SOURCES
Applying KVL in (b),
-20 + 6i1 + 10i2 + 4i2 = 0 6i1 + 14i2 = 20 ………..(i)
Applying KCL to a node where the two meshes intersect,
i2 = i1 + 6 ………… (ii)
Solve (i) and (ii), i1 = -3.2 A, i2 = 2.8A
SOURCE TRANSFORMATION
• series parallel combination and wye-delta transformation help simplifycircuits• source transformation is another tool for simplifying circuits• is the process of replacing a voltage source, Vs in series with a resistor by acurrent source in parallel with a resistor, or vice versa• basic to these tools is the concept of equivalence
Riv ss =
R
vi s
s =
Superposition theorem
For circuit network have more than one independent source
voltage or current produced by a source acting in isolation can be determined by assuming other sources do not work, where the resources should be switched off in the following manner: -be switched off in the following manner: -
Independent voltage source – short circuit (0 V) or internal resistance if have
Independent current source – open circuit (0 A)
Superposition theorem cont.
Example for superposition theorem.
Solution.
IB
Solution.1. Current source IB work ( voltage source VB off- short circuit)
Get the value of I’I’
Superposition theorem cont.
2. Voltage source VB work (current source IB off – open circuit)
I”
Get the value of I”
3. So get the value of current flowing at resistance R2 with I = I’ +I”
Record:- Various methods can be used to obtain the value of I 'and I “, such as current divider or mesh analysis or nodal analysis or node.
Superposition theorem cont.
Example 1.10
For the circuit in the figure below, find the value of I, the voltage across
the resistor 2Ω and the power absorbed by the resistor
0.5 Ω 0.5 Ω
I
0.5 Ω2 Ω
5 V
0.5 Ω
10 VI
V2Ω
Superposition theorem cont.
1. Voltage source 5V work ( voltage source 10 V off – short circuit)
Solution
Find the value of I’ ( )1 2 //1 1.67
53
1.67
1' 1
2 1
T
T
T
R
I A
I xI A
= + = Ω
= =
= =+
Superposition theorem cont.2. Voltage source 10 V work ( voltage source 5 V off – short circuit)
Find the value of I” ( )1 2 //1 1.67R = + = ΩFind the value of I” ( )1 2 //1 1.67
106
1.67
1" 2
2 1
T
T
T
R
I A
I xI A
= + = Ω
= =
= =+
3. So the value of current I (make sure the direction of I’ and I”)
2
2 22
' ( ") 1
( 1)(2) 2
( ) ( 1) (2) 2
I I I A
V IR V
P I R W
Ω
Ω
= + − = −= = − = −
= = − =
Superposition theorem cont.
Example 1.11Refer to the figure below, calculate current flowing and voltage across resistance 4Ω.
8 Ω1 Ω
I
5 A2 Ω4 Ω
1 Ω
Superposition theorem cont.
Example 1.12
Refer to the figure below, determine voltage across resistance 4 Ω.
5 Ω 2 Ω 4 Ω
1 Ω3 Ω 3 A 4 Vx
-+Vx
Thevenin theorem
This theorem is in use to facilitate a complex circuit network to a simple circuit called the Thevenin equivalent circuit.
The equivalent circuit contains a voltage source Vth in series with a resistor Rth
Complex Circuit
a
b
Thevenin theorem cont.
The steps to get the Thevenin equivalent circuit:-a. Remove section of the network where to find the thevenin
equivalent circuit and mark clearly the two terminals as a-b
Complex Circuit
I=0 A
VTH
a
b
OFF Circuit
a
bRTH
equivalent circuit and mark clearly the two terminals as a-b
b. Determine the Thevenin equivalent resistance seen from the terminal a-b with independent sources is turn off
c. Get the values of Thevenin voltage on the voltage across the terminal a-b when the terminal at open circuit. (various method can be used to obtain Vth, whether to used loop analysis/nod analysis)
d. Draw the Thevenin equivalent circuit and connect the back portion removed from the (a) above
Thevenin theorem cont.
Example 1.13For circuit below, sketch the Thevenin equivalent circuit at terminals a-b.
6 Ω 2 Ωa
20 V
4 Ω
5 A RL
b
Thevenin theorem cont.
Solutiona. Remove RL from circuit
b. Determine RTH seen from terminal a-b with all independent sources are turn off.
Thevenin theorem cont.
c. Get VTH at terminal a-b
5
205
1070
70
X
X
TH A X
V
V V
V V V
− =
== = =
VTH
I=0 A
d. Draw the Thevenin equivalent circuit and connect the back portion removed from the (a) above
VTH=70 V RL
RTH=12Ω
a
b
Thevenin theorem cont.
Example 1.14
Refer to the circuit below, sketch Thevenin equivalent circuit at terminal
a-b, next calculate the current flowing, I3Ω and voltage across, V3Ω the resistor 3Ω,
5 Ω 1 Ωa
4 Ω 3 ΩV3Ω
a
b
28 V
I3Ω
Thevenin theorem cont.
Example 1.16Refer to the circuit below, get the value of V1/3Ω
½ Ω
½ Ω3 V
¼ Ω 1/3 Ω V1/3Ω2 A
Norton theorem
This theorem is in use to facilitate a complex circuit network to a simple circuit called the Norton equivalent circuit.
This equivalent circuit consists of a current source IN connected in parallel with a resistor RN.
Complex Circuit
a
b
Norton theorem cont.
The steps to get the Norton equivalent circuit:-a. Remove section of the network where to find the Norton
equivalent circuit and mark clearly the two terminals as a-b
Complex Circuit
IN
a
b
OFF Circuit
a
bRN
equivalent circuit and mark clearly the two terminals as a-b
b. Determine the Norton equivalent resistance seen from the terminal a-b with independent sources is turn off
c. Get the Norton current value of current flowing through the terminals a-b when a short circuit in the terminal. (various method can be used to obtain IN, whether to used loop analysis/nod analysis)
d. Draw the Norton equivalent circuit and connect the back portion removed from the (a) above
Norton theorem cont.
Examples
Determine Norton equivalent circuit at terminals a-b for circuit below.
Next calculate current flowing and voltage across resistance 3Ω
5 Ω 1 Ωa
I
4 Ω 3 ΩV3Ω
b
28 V
I3Ω
Norton theorem cont.
Solution.a. Remove RL from circuit
b. Determine RN seen from terminal a-b with all independent sources are turn off.
RN=(1+(5//4))=3.22 Ω
Norton theorem cont.
c. Get IN at terminal a-b (short a-b)
IN
IT
4 28 43.86
5 (5 (4 //1)) 5N TI I A = = = +
d. Draw the Norton equivalent circuit and connect the back portion removed from the (a) above
IN=3.86 A RN=3.22Ω 3Ω
Current flowing 3Ω,
( )3
3.223.86 2
(3 (3.22))I AΩ = =
+
Voltage across 3Ω
( )3 3 3 6V I VΩ Ω= =
Norton theorem cont.
Example 1.19
½ Ω
½ Ω3 V
Get the value of V1/3Ω
¼ Ω 1/3 Ω V1/3Ω2 A
Thevenin and Norton theorem with dependent sources
To analyze circuits with independent sources, (IN) and (VTH) may be obtained by using the analysis as before.
However, the Thevenin and Norton resistance can not be obtained directly from the network because of dependent sources can not be turned off as an independent source.independent source.
therefore, to solve the circuit dependent sources, two ways:
1. Determine the value of VTH and IN, so
THN TH
N
VR R
I= =
Thevenin and Norton theorem with dependent
sources CONT.
2. Introduce an independent voltage source,VT or an independent current source, IT at the root a-b. VT
and IT value is any value. However, free resources available on the network must be turned off prior circuit.
I
Off Circuit
a
b
I
VTOff Circuit
a
bIT
TTH N
VR R
I= = Ω ab
TH NT
VR R
I= = Ω
Vab
Thevenin and Norton theorem with dependent
sources CONT.
Example 1.20Sketch the Thevenin equivalent circuit at terminal a-b, next calculate value Iab
Thevenin and Norton theorem with dependent
sources CONT.
Solution.
a) Remove resistance 3 Ω from circuit
b) Get the value VTH- +
Write equation every loop
Thevenin and Norton theorem with
dependent sources CONT.
Solve the equation above to get value I2, next find value VTH
c) Get resistance value of Thevenin equivalent, RTH
RTH can be solve in two way.
i) Get value Ii) Get value IN- +
Thevenin and Norton theorem with
dependent sources CONT.
ii) Introduce an independent source.
1. introduce independent voltage source
- +
Thevenin and Norton theorem with
dependent sources CONT.
II. introduce independent current source
Thevenin and Norton theorem with
dependent sources CONT.
So, sketch the Thevenin equivalent circuit
VTH=48.2 V
Relationship between the Thevenin and Norton theorem
Thevenin equivalent circuit can be converted into the Norton equivalent circuit or vice versa by changing the concept (super transformation)of supply where: -
a) Thevenin resistance (RTH) value is equal to the Norton resistance (RN)
Maximum power transfer
a circuit will supply maximum power to the load if the load resistance RL is equal to the equivalent resistance seen by the load
Maximum power transfer can be obtained by replace a complex circuit with the Thevenin equivalent circuit or Norton equivalent circuitcircuit or Norton equivalent circuit
Maximum power transfer cont.
Power to the load RL,
Condition maximum power transfer RL = RTH
Therefore, maximum power supplied to the load is:
2
2 THRL L L
TH L
VP I R R
R R
= = +
2 2V V2 2
4 4TH TH
RLmakL TH
V VP
R R= =
Maximum power transfer cont.
Th
ThThL R
VpRR
4
2
max ==
Maximum power transfer cont.
ExamplesFrom circuit below, calculate:-
a) Value RL when maximum output power
b) Maximum power absorb by load RL
Maximum power transfer cont.
Solution.a) Remove RL from circuit and off all source from circuit to find RTH
b) Get the thevenin voltage at terminal a-b when RL removed from circuit
Maximum power transfer cont.
So, draw Thevenin equivalent circuit when maximum power transfer happen.
Maximum power transfer cont.
Examples 1.22
From circuit below, calculate:-
a) Value RL when maximum output power
b) Maximum power absorb by load RL