Chapter 4 Analysis DC Circuit Norton’s...
Transcript of Chapter 4 Analysis DC Circuit Norton’s...
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BENE 1133
ELECTRICAL PRINCIPLES
Chapter 4
Analysis DC Circuit
copyright@sharatulizah2
TOPICS COVERED
Introduction to branch, node and mesh
Nodal analysis
Mesh analysis
Superposition Theorem
Source Transformation
Thevenin’s Theorem
Norton’s Theorem
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Chapter Objectives
At the end of this Chapter 4, students should
be able to:
1. Apply nodal analysis to find unknown
quantities in electric circuit.
2. Apply mesh analysis to find unknown
quantities in electric circuit.
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Chapter Objectives
3. Apply source transformation technique to
find unknown quantities in a circuit.
4. Apply Thevenin’s theorem to find Thevenin
equivalent circuit.
5. Apply Norton’s theorem to find Norton
equivalent circuit
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4.1 Introduction to Branches, Nodes
and Loops
Branches
a branch represents a single element such
as voltage source or a resistor.
a branch represents any 2-terminal element.
5 branches: 10-V voltage
source, the 2-A current source,
and the 3 resistors
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4.1 Introduction to Branches, Nodes and
Loops
Nodes
a node is the point of
connection between two or
more branches.
usually indicated by a dot in a
circuit.
if a connecting wire connects 2
nodes, the 2 nodes constitute
a single node.
3-nodes
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Exercise 1:
How many branches and nodes does the
circuit?
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4.1 Introduction to Branches, Nodes and
Loops
Loops
a loop is any closed path in a circuit
formed by a starting at a node, passing
through a set of nodes, and returning back
to the starting node without passing through
any node more than once.
A network with b branches, n nodes and l loops
will satisfy the fundamental theorem or network
topology, b= l + n –1
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4.2 Introduction to Nodal and Mesh
Analysis
There are two techniques for circuit analysis:
Nodal Analysis –
based on a systematic application
of Kirchhoff’s current law (KCL).
Mesh Analysis –
based on a systematic application
of Kirchhoff’s voltage law (KVL).
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4.2 Introduction to Nodal and Mesh
Analysis
These two techniques:
Used to analyze linear circuits by obtaining aset of simultaneous equations that are thensolved to obtain the required values ofcurrent or voltage.
Simultaneous equations can be solved using:
Substitution eq
Cramer’s rule
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4.2.1 Nodal Analysis
Using node voltages as the circuit variables.
How to find the node voltages?
1.Determine the node, n
2.Select a node as the reference node.
3.For remaining n-1 nodes, assign voltages
v1, v2, ….., vn-1.
–The voltages are referenced with
respect to the reference node.12
4.2.1 Nodal Analysis
4. Apply KCL to each of the n-1 nodes.
–Use Ohm’s law to express the branch
currents in terms of node voltages.
5. Solve the resulting simultaneous
equations to obtain the unknown node
voltages.
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4.2.1 Nodal Analysis
Original circuit Analysis circuit
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4.2.1 Nodal Analysis
4. Applying KCL gives
I1 = I2 + i1 + i2 …..at node 1
I1+ i2 = i3….. at node 2
R
vvi
lowerhigher
we obtain,
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2
212
1
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0
0
R
vi
R
vvi
R
vi
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4.2.1 Nodal Analysis
2
21
1
121
R
vv
R
vII
3
2
2
212
R
v
R
vvI
Since at node 1, I1 = I2 + i1 + i2 then
Since at node 2, I2+ i2 = i3 then
(1)
(2)
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Example 1
Calculate the node voltages in the circuit.
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Solution Example 1
At node 1
At node 2
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Solution Example 1
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06053
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21
vv
vv
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We have two simultaneous equation
Method 1: Substitution eq
Method 2: Cramer Rule
VvVv
vv
vvvv
33.1320
60520
)2(6053)1(203
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2121
60
20
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2
1
v
v
Vv
Vv
2012
60180603
203
33.1312
60100560
120
22
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121231553
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(2)(1)
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Example 2
Determine the voltages at the nodes given.
Original circuit Analysis circuit 22
4.2.1 Nodal Analysis with Voltage Sources
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4.2.1 Nodal Analysis with Voltage Sources
Consider 2 possibilities:
Case 1
If a voltage source is connected between
the reference node and a non-reference
node, the voltage at the non-reference
node = voltage of the voltage source.
For example,
v1 = 10V
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4.2.1 Nodal Analysis with Voltage Sources
Case 2
If the voltage source is connected between two non-reference nodes, the two non-reference nodes form a generalized node or supernode.
Then, apply both KCL and KVL to determine the node voltages.
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4.2.1 Nodal Analysis with Voltage Sources
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0
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0
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vvvvvv
(i) KCL at node2 and node 3;
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4.2.1 Nodal Analysis with Voltage Sources
532 vv05 32 vv
(ii) KVL at supernode;
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Example 3
Find the node voltages using nodal analysis.
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Example 4
Find the node voltages using nodal analysis.
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4.2.2 Mesh Analysis
Using mesh currents instead of element
currents as circuit variables.
A mesh is a loop that does not contain
any other loop within it.
Apply KVL to find circuit variables (mesh
current).
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4.2.2 Mesh Analysis
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4.2.2 Mesh Analysis
• Paths abefa and bcdeb are meshes, but path abcdefa is not a mesh.
• The current through a mesh is known as mesh current32
4.2.2 Mesh Analysis
Three (4) steps to determine Mesh Currents:1.Assign mesh currents i1, i2……in to the
n meshes.
2.Apply KVL to each of the n meshes
3.Use Ohm’s law to express the voltages n terms of mesh currents.
4.Solve the resulting n simultaneous equations to get the mesh currents.
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4.2.2 Mesh Analysis
For mesh 1 0)( 213111 iiRiRV
For mesh 2 0)( 123222 iiRViR
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4.2.2 Mesh Analysis
If a circuit has n nodes, b branches and l independent loops or meshes,
then l = b – n + 1.
Hence, l independent simultaneous equations are required to solve the circuit using mesh analysis.
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Example 5
For the circuit below, find the branch currents
I1, I2, and I3 using mesh analysis.
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Example 6
Use mesh analysis to find the current I0.
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4.2.2 Mesh Analysis with Current Sources
Case 1
When the current source exits only in one
mesh.
0)(6410 211 iii
Ai 21 Ai 52
At mesh 1
At mesh 2Then,
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Case 2
When a current source exits between two meshes.
Create a supermesh
exclude the current source and any elements
connected in series with it
use both KVL and KCL
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21
221
ii
iii612 ii
Using KVL,Using KCL at reference node,
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Example 7
Use mesh analysis to determine i1, i2 and i3.
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Example 8
For the figure below, find i1 to i4 using mesh analysis.
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Solution E
xam
ple 8
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Disadvantage of Kirchhoff’s Laws – tediouscomputation is involved when dealing with alarge and complex circuit.
To handle the complexity, engineers over theyears have developed some theorems tosimplify circuit analysis:Superposition Theorem
Source Transformation
Thevenin’s Theorem
Norton’s Theorem
Applicable to linear circuit.44
4.3 Superposition
This is another way to determine the contribution of each independent source to the variable.
Superposition principle: Voltage across an element in a linear circuit is the
algebraic sum of the voltage across that element due to each independent source acting alone.
Current through an element in a linear circuit is the algebraic sum of the current through that element due to each independent source acting alone.
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4.3 SuperpositionSteps to apply the principle: Turn off all independent sources except one source.
Replace the voltage source by 0V (short circuit) & current source by 0A (open circuit).
Using Basic Laws n techniques in chapter 2 & 3, find the output (voltage & current) due to that active source.
Repeat steps 1, 2, 3 for other independent sources.
Find the total contribution by adding algebraically all the contributions due to the independent sources.
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Example 9
Use the superposition theorem to find vin the circuit in figure below.
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Solution Example 9
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Solution Example 9
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Example 10
Using the superposition theorem, find vo
in the circuit.
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4.4 Source Transformation
This is another tool to simplify the circuits.
The basic tools is the concept of equivalence.
HOW?
Replacing a voltage source, vs, in series with a
resistor by a current source, is, in parallel with a
resistor, or vice versa.
Riv ss R
vi s
s
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4.4 Source Transformation
This tool also applicable to dependent sources
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Example 11
Use source transformation to find vo in the circuit figure below.
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Solution Example 11
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Example 12
Use source transformation to find i0 in the circuit in figure below.
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Example 13
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4.5 Thevenin’s Theorem
A particular element is variable (usually refers to as a load) while other elements are fixed.
Each time the load is changed the entire circuit has to be analyzed all over again.
Thevenin’s theorem provides a technique by which the fixed part of the circuit is replacedby an equivalent circuit.
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4.5 Thevenin’s Theorem
The theorem states that:
“A linear two-terminal circuit can be replaced by an equivalent circuit consisting 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.”
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4.5 Thevenin’s Theorem
RTH: the equivalent resistance
at the terminals when the
independent sources turned off
VTH: the open-circuit
voltage at the terminals
fixed part variable part
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4.5 Thevenin’s Theorem
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4.5 Thevenin’s Theorem
Two cases need to be considered in finding Thevenin’s resistance,RTH
Case 1:
Network has no dependent sources
All independent sources are TURN OFF
RTH is the input resistance of network looking between terminal a and b.
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4.5 Thevenin’s Theorem
Case 2:Network has dependent sources
Independent sources → TURN OFF
Dependent sources → not to be TURN OFF
Apply vo at terminal a & b and determine the resulting current io. Resulting in :
o
o
THi
vR
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4.5 Thevenin’s Theorem
The equivalent Thevenin network behaves externally the same way as the original circuit.
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Example 14
Find the Thevenin equivalent circuit of the circuit in figure below, to the left of the terminals a-b.
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Example 15
Using Thevenin’s theorem, find the equivalent
circuit to the left of the terminals in the circuit.
Hence find i.
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Example 16
Find the Thevenin equivalent circuit in figure below.
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4.6 Norton’s Theorem
Similar to Thevenin’s theorem.
“A linear two-terminal circuit can
be replaced by an equivalent
circuit consisting of a current
source IN in parallel with a
resistor RN, where IN is the
short-circuit current through the
terminals and RN is the input
resistance at the terminals when
the independent sources are
turned off”.68
4.6 Norton’s Theorem
IN : the short circuit current
through the terminals
RN: the equivalent resistance
at the terminals when the
Independent sources are
turned off
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4.6 Norton’s Theorem
To convert to a Norton circuit, first remove the load from the circuit.
Set all independent sources to zero.
Determine the open circuit resistance. This is the Norton resistance, RN
Replace the sources and determine the current which would occur in a short between the two terminals, IN
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4.6 Norton’s Theorem The Norton equivalent circuit may be
determined directly from a Thevenin’s circuit.
The Thevenin’s and Norton’s equivalent circuits are related by a source transformation.
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