Download Mesh Analysis

KITRC
CIRCUIT & NETWORKS
MADE BY
AGNA PATEL:130260111011
AESHA CHAUHAN:130260111004
28-Apr-17
1
CONTENTS
• Definitions
• Kirchoff’s Voltage Law (KVL)
• Kirchoff’s Current Law (KCL)
• Nodal analysis
• Mesh analysis
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2
BRANCHES AND NODES
Branch:
elements connected end-to-end,
nothing coming off in between (in series)
Node:
place where elements are joined—entire wire
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3
NOTATION: NODE VOLTAGES
The voltage drop from node X to a reference node
(ground) is called the node voltage Vx.
Example:
a
+
Va
_
b
+
+
_
Vb
_
ground
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 Linear network

 Nonlinear network
– A circuit or network to  A circuit whose parameter
parameter are R,L,C are
changes there value with
always
constant
change in time, temp
irrespective of change in
,voltage is known as non
time
.voltage,
temp.,
linear parameter.
known as linear network.
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 Bilateral network –
 Unilateral network
–
A
circuit characteristic
behavior irrespective of
direction of current to
various element is known
as bilateral network
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 A
circuit
whose
operation behavior
is
depended on direction of
current to various element
is known as unilateral
network
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 Active network
 Passive network

 A circuit which contains
A circuit which consist of
at least one source of
energy is called active
network.
 example – voltage source,
ac/dc signal, generator ,
transistor.
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number any kind of source
is called passive network .
 Example – R,L,C.
7
KIRCHOFF’S VOLTAGE LAW (KVL)
The sum of the voltage drops around any closed loop is zero.
We must return to the same potential (conservation of energy).
Path
+
V1 “drop”
-
Path
- “rise” or “step up”
V2 (negative drop)
+
Closed loop: Path beginning and ending on the same node
Our trick: to sum voltage drops on elements, look at the first sign
you encounter on element when tracing path
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KVL EXAMPLE
+ v2 
1
+
va

b

a
+
vb
-
v3
2
+
Examples
of
three closed paths:
c
+
vc

3
Path 1:
 va  v 2  vb  0
Path 2:
 vb  v 3  vc  0
Path 3:
 va  v2  v3  vc  0
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KCL [KIRCHOFF CURRENT LAW]
“Algebraic sum” of currents entering node = 0
where “algebraic sum” means currents leaving are
included with a minus sign
“Algebraic sum” of currents leaving node = 0
currents entering are included with a minus sign
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KIRCHHOFF’S CURRENT LAW EXAMPLE
24 A
-4 A
10 A
i
Currents entering the node: 24 A
Currents leaving the node: 4 A + 10 A + i
Three formulations of KCL:
1:
2:
3:
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24  4  10  i
24  ( 4)  10  i  0
 24  4  10  i  0



i  18 A
i  18 A
i  18 A
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Nodal Analysis
 The selection of mesh current and the application established the
mesh analysis For the solution of networks. this was studied in the
previous section . In this section ,the same solution Is found by
introducing the set of the equation by the application Of kirchoff’s
current law . this method is known as nodal analysis.
Steps
1.
While assuming branch currents, make sure that each unknown
branch Current is considered at least once.
2. Convert voltage source present into their equivalent
current sources for node analysis where possible.
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3.Follow the same sign convention ,currents entering at node
are to be considered positive, while current leaving the node
are to be considered as negative.
4.Select the direction of branch current leaving the respective
nodes.
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Super node
By considering the above figure nodes v2 and v3 are connected
directly
Through a voltage source without any circuit element . The region
surrounding
a voltage source which connects two nodes directly is called
super node .
V2 =v3 + v x
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Super node : circuit
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Nodal Analysis: Concept Illustration:
v1
v2
v3


R4
R2
R1
I
R3
reference node
Figure 6.1: Partial circuit used to illustrate nodal analysis.
V1  V2
R2
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
V1
R1

V1
R3

V1  V3
R4
I
Eq 6.1
16
Nodal Analysis
Clearing the previous equation gives,
 1
 1 
 1 
1
1
1 
 

 V1   V2   V3  I
 R2 
 R4 
 R1 R2 R3 R4 
Eq 6.2
We would need two additional equations, from the
remaining circuit, in order to solve for V1, V2, and V3
4
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Nodal Analysis: Example
Given the following circuit. Set-up the equations
to solve for V1 and V2. Also solve for the voltage V6.
R2
v1
R3

v2
R5

+
R1
I1
R4
v6
_
R6
Figure 6.2: Circuit for Example 6.1.
5
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Nodal Analysis: Example : Set up for solution.
V1
R1  R2
V2  V1
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
V2
R3
V2
Eq 6.3
Eq 6.4
 1
 1 
1 


 V1   V2  I 1
 R1  R2 R3 
 R3 
Eq 6.5
 1 
 1
1
1 



V2  0
  V1   

 R3 
 R3 R4 R5  R6 
Eq 6.6
R4

 I1
0
R3
7

V1  V2
R5  R6
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Mesh Analysis

The branch current is resultant current flowing through
particular branch of circuit mesh current is fictitious
current assumed to be common to all the elements of particular
mesh in ckt.
 Mesh currents are sort of fictitious in that a particular
mesh current does not define the current in each branch
of the mesh to which it is assigned.
I1
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I2
I3
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•We assume the current source to be open circuited and create
a new mesh called the supermesh . A supermesh constitute by
two adjust meshes that have a common current source . We can
apply kvl only to those meshes in modified network . This
technique is useful as it does not increase the number of
eqation to be solved .
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1.certain that n/w is
applicable.
play n/w or mesh analysis is not
2.Make neat simple circuit diagram indicate all element and
source value.
3.Asssuming that the circuit has m mesh assign a clockwise
mesh current
in each mesh.
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4.If circuit contain only voltage sources apply kvl around each
mesh If circuit has only independent voltage source .equate the
clockwise sum of all resistance voltage to counter clockwise
sum of all source voltage and order the terms of I1 to Im.
For each dependent voltage source present ,relate the voltage
source and controlling quantity to variable.
5.If circuit contains current source create a super mesh for each
one that is common to two mesh by applying kvl around large
loop formed by branches not common to mesh.kvl not need be
applied to mesh containing current source that lies on perimeter
of entire circuit.
The assigned mesh current should not be change.
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Basic Circuits
Mesh Analysis:
R1
+
VA
V1
R2
_
V2
_
+
+
_
+
I1
VL1
_
Rx
+
I2
_
VB
Figure 7.2: A circuit for illustrating mesh analysis.
Around mesh 1:
V1  VL1  V A
where V1  I1 R1 ; VL1  I1  I 2 RX
so, ( R1  RX ) I1  RX I 2  V A
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Eq 7.1
24
R1
+
VA
V1
_
+
V2
_
+
+
_
R2
I1
VL1
_
Rx
+
I2
_
VB
Around mesh 2 we have
V L1  V2  V B
with; V L1   ( I 2  I 1 ) R X ; V2  I 2 R2
Eq 7.2
Eq 7.3
Substituting Eq 7.3 in 7.2 gives,
R X I 1  ( R X  R2 ) I 2  V B
or
 R X I 1  ( R X  R2 ) I 2   V B
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Eq 7.4
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We are left with 2 equations: From (7.1) and (7.4)
we have,
( R1  RX ) I1  RX I 2  VA
Eq 7.5
 RX I1  ( RX  R2 ) I 2  VB
Eq 7.6
We can easily solve these equations for I1 and I2.
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The previous equations can be written in matrix form as:
 R X   I1   V A 
( R1  RX )

 R




( RX  R2   I 2   VB 
X

or
Eq (7.7)
1
 RX   V A 
 I1  ( R1  RX )
I     R



( RX  R2   VB 
X
 2 
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Eq (7.8)
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Mesh Analysis
Standard form for mesh equations
Consider the following:
 R11
R
 21
 R31
R12
R22
R32
R13   I1 
  emfs (1) 
R23   I 2    emfs ( 2) 
  emfs ( 3) 
R33   I 3 
R11 =  of resistance around mesh 1, common to mesh 1 current I1.
R22 =  of resistance around mesh 2, common to mesh 2 current I2.
R33 =  of resistance around mesh 3, common to mesh 3 current I3.
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R12 = R21 = - resistance common between mesh 1 and 2
when I1 and I2 are opposite through R1,R2.
R13 = R31 = - resistance common between mesh 1 and 3
when I1 and I3 are opposite through R1,R3.
R23 = R32 = - resistance common between mesh 2 and 3
when I2 and I3 are opposite through R2,R3.
 emfs (1) = sum of emf around mesh 1 in the direction of I1.
 emfs ( 2) = sum of emf around mesh 2 in the direction of I2.
 emfs ( 3) = sum of emf around mesh 3 in the direction of I3.
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Mesh Analysis:
Example
Write the mesh equations and solve for the currents I1, and I2.
4
10V
+
_
2
7
6
I1
2V +_
I2
_
+
20V
Figure 7.2: Circuit for Example 7.1.
Mesh 1
Mesh 2
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4I1 + 6(I1 – I2) = 10 - 2
Eq (7.9)
6(I2 – I1) + 2I2 + 7I2 = 2 + 20
Eq (7.10)
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Simplifying Eq (7.9) and (7.10) gives,
10I1 – 6I2 = 8
Eq (7.11)
-6I1 + 15I2 = 22
Eq (7.12)
I1 = 2.2105
I2 = 2.3509
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