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2

2.1 KIRCHHOFFS CURRENT LAW (KCL):-

It states that the algebraic sum of all currents

entering a node is zero. Mathematically:

Currents are positive if entering a nodeCurrents are negative if leaving a node.

Example:


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3

Applying Kirchhoff's current law:

I1 + I2 + I3 + I4 = 0

(the negative sign inI2indicates that I2 has a

magnitude of 3A and is flowing in the direction

opposite to that indicated by the arrow)

Substituting:

5 - 3 + I3 + 2 = 0 Therefore, I3 = - 4A (ie 4A

leaving node)


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4

2.2 KIRCHHOFFS VOLTAGE LAW (KVL):-

It states that the algebraic sum of the voltage drops

around any loop, open or closed, is zero.Mathematically

Example:

Going round the loop in the direction of the current, I,

Kirchhoff's Voltage Law gives:

10- 2I - 3I = 0


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5

-2Iand -3Iare negative, since they are voltage drops

i.e. represent a decrease in potential on proceeding

round the loop in the direction of I. For the same

reason + 10V is positive as it is a voltage rise or increasein potential.

Concluding:

5 I = 10 Therefore, I = 2A


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Loop-current Analysis Loop analysis is systematic method of network

analysis.

It is a general method and can be applied to anyelectrical network, howsoever complicated it may

be. It is based on writing KVL equations for

independent loops.

A loop is a closed path in a network.

A node or a junction is a point in the networkwhere three or more elements have a commonconnection.

Next


7/95, 102013 Ch. 3 Network Analysis- Part II 7

Before the loop analysis can be applied to a

network, we must first check that it has onlyvoltage sources (independent or dependent).

Any current source must be transformed into its

equivalent voltage source. Sometimes, it is a difficult task to identify

independent loops in a network.

The method of loop analysis can be best understood

by considering some examples.

Next



Example1

Find the voltage across the 2- resistance.

Next



Recognize the independent loops (which does not pass

through a current source), and mark the loop currents.

This choice reduces labour, as only one currentI1 is

to be calculated.

Next



Write KVL equations and solve forI1.

A0.435

1

1222

121

79)1)(()3()4(67)2()1)((

I

IIII

III

Next



Counting Independent Loops

It appears to have two loops. But, these two loops are not independent.

Next


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Suppose that we had marked the two loop currentsI1

andI2 in the standard way,

2 1 2AI I

The values of these two currents are constrainedby

the above relation.

Then,

Next


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We identify independent loops by turning off allsources. We are, then, left with one loop containing

two resistances.

Hence, we have only one independent loop

requiring only one KVL equation.

Next


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For determining the current through 5- resistance,

we should choose

Thus, the single KVL equation is

A0.462

1

11 0)2(8510

I

II

Next

Click


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In case, we are to determine the current through 8-

resistance, we should choose

A1.538

1

1108)2(510

I

II

The single KVL equation then becomes

Next

Click


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BenchmarkExample1

Consider the benchmark example, and solve itby using loop-current analysis.

Next


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Solution :

We note that the givencircuit has one independent

loop and two constrained loops.

Our aim is to determine the voltage across 3-

resistance.

So, we should select the unknown loop currentIpassing through 3- resistance (but not through any

current source).

The two known loop currents of 4 A and 5 A aremarked to flow in the two loops as shown.

Next


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Writing KVL equation around the loop ofI, we get

A650)45(1632 IIII

Therefore, the unknown voltage, v = 3I= 2.5 V.

Next

Click


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Example2

Find the currents i1and i2 in the circuit givenbelow.

Next


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Solution :Applying KVL to the two loops,

2

2

3 2

or 2 3

i

i

1 A

Next


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MESH ANALYSIS

In circuit terminology, a loopis any closedpath.

A meshis a special loop, namely, thesmallest loop one can have.

In other words, a mesh is a loop thatcontains no other loops.

Mesh analysis is applicable only to a planar

network. However, most of the networks we shall need

to analyze are planar.

Next


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Once a circuit has been drawn in planar form,

it often looks like a multi-paned window.

Each pane is a mesh.

Meshes provide a set of independent

equations.

Next


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By definition, a mesh-current is that currentwhich flows around the perimeter of a mesh. It is

indicated by a curved arrow that almost closes onitself.

Branch-currents have a physical identity. They

can be measured. Mesh-currents are fictitious.

The mesh analysis not only tells us the minimum

number of unknown currents, but it also ensuresthat the KVL equations obtained are independent.

Next


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Loop (Mesh) Analysis

Next
http://localhost/var/www/apps/conversion/tmp/scratch_10/Loop%20Analysis%20for%20Ch.%203-Part-II.swfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Loop%20Analysis%20for%20Ch.%203-Part-II.swfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Loop%20Analysis%20for%20Ch.%203-Part-II.swf


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Example 2

Let us consider a simple network having only two

meshes.

Although the directions of the mesh currents are

arbitrary, we shall always choose clockwise mesh

currents.

This results in a certain error-minimizing

symmetry.

Note that by taking mesh currents, the KCL isautomatically satisfied.

Next


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, 102013 Ch. 3 Network Analysis- Part II 26Next


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Resistance Matrix

Mesh current matrix

Source matrix

Next


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ApplyingCrammers rule :

The current in 3-ohm resistor is I1I2 = 64 = 2A

Next


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Three-mesh Network Write the three equations for the three meshes and

put them in a matrix form.

Next


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Self-resistance of mesh 1

Mutual resistance

between mesh 1 and 2.

Next


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The Resistance Matrix

It is symmetrical about the major diagonal, asR12 = R21, R13 = R31, etc.

All the elements on the major diagonal have

positive values. The off-diagonal elementshavenegative values.

The mutual resistance between two meshes will

be zero, if there is no resistance common to them.

Next


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Mesh Analysis Limitations

It is applicable only to those planar networks

which contain only independent voltage sources.

If there is a practical current source, it can be

converted to an equivalent practical voltage

source.

Next


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Planar Network If a network can be drawn on sheet of paper

without crossing lines, it is said to beplanar.

Is it a planar network ?

Yes, it is. Because it can be drawn in a plane,

as shown in the next figure.

Next

Click


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This is definitely non-planar.

Next


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Procedure for Mesh Analysis

1. Make sure that the network is planar.

2. Make sure that it contains only independent

voltage sources.

3. Assign clockwise mesh currents.

4. Write mesh equations in matrix form by

inspection. An element on the principal diagonal

is the self-resistance of the mesh. These

elements are all positive. An element off themajor diagonal is negative (or zero), and

represents the mutual resistance.

Next


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5.Check the symmetry of resistance matrix aboutthe major diagonal.

6. An element of the voltage source column matrix

on the right side represents the algebraic sum of

the voltage sources that produce current in the

same direction as the assumed mesh current.

7. Solve the equations to determine the unknown

mesh currents, using Calculator.8. Determine the branch currents and voltages.

Next


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Example 3

Determine the currents in various resistances of

the network shown.

Next


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Writing the mesh equationsby inspection,

Solving,

Next

Solution :

Click

we get I1 = 2.55 A, I2= 3.167 A

Click


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Example 4 Find the current drawn from the source in the

network, using mesh analysis.

Next


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Using Calculator, we get Click

1I 6 A


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How to Handle Current Sources

If a circuit has current sources, a modestextension of the standard procedure is

needed.

There are three possible methods.

Next


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First Method

If possible, transform the current sources into

voltage sources.

This reduces the number of meshes by 1 for

each current source.

Apply the standard procedure of meshanalysis to determine the assumed mesh

currents.

Go back to the original circuit, and getadditional equations, one for each current

source,

Next


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Example 5

Solve the following circuit for the three meshcurrents.

Next


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Solution :

We convert the 13-A current source in parallel with 5-

resistor into an equivalent 65-V voltage source inseries with 5- resistor.

This reduces the number of meshes to two.

Next


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We can write the mesh equations in the matrix form just

by inspection,

1

2

9 5 10

5 11 52

I

I

1 2andI I 5 A 7 A

We now go back to the original circuit. Obviously, the

current through the current source is

2 3 3 213A 13 7 13I I I I 6 A

Next

Click


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Second Method

We can assign unknown voltages to each

current source.

Apply KVL around each mesh, and

Relate the source currents to the assumed

mesh currents.

This is generally a difficult approach.

Next


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Third Method

(Supermesh Method) Create a supermeshfrom two meshes that

have a current source as a common element.

The current source is in the interior of thesupermesh.

Thus, the number of meshes is reduced by 1for each current source present.

If the current source lies on the perimeter of

the circuit, then ignore the single mesh inwhich it is found.

Apply KVL to the meshes and supermeshes.

Next


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Example 6

Solve the circuit of Example 5, using

supermesh method.

Solution :

Next


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Going along the dotted arrow, the KVL equation for

this supermesh is

3 1 2

1 2 3

5( ) 6 13 0

or 5 6 5 13

I I I

I I I

The KVL equation for mesh 1 is

1 2 39 0 5 75I I I

We have only two equations for three unknowns.

The third equation is obtained by applying KCL to

either node of the current source

Next

Click

Click


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Thus, we have

1 2 30 13I I I

These three equations can be put in the matrix form,

1

2

3

5 6 5 13

9 0 5 75

0 1 1 13

I

I

I

Using Casio fx-991ES, we directly get

1 2 3, and .I I I 5 A 7 A -6 A

Which is same result as obtained in Example 5.

Next

Click


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Example 7

Apply mesh analysis to determine current

through 7- resistance in the given network.

Next


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Solution :

The given network is a planar networkhaving

independent voltage sources.

It has three meshes for which the mesh currentsI1,I2,

andI3 are marked all with clockwise directions.

By inspection, the matrix equation is written as

1

2

3

3 4 4 0 42 25

4 4 5 6 6 25 57 70

0 6 6 7 70 4

I

I

I

Next

Click


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1

2

3

7 4 0 67

or 4 15 6 1520 6 13 74

I

I

I

7 3

I I 2 A

Solving the above equation forI3,

Next

Click


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Node-Voltage Analysis

Next

N d V l A l i
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Node Voltages Analysis

It is dual of the Mesh Analysis.

It involves the application of KCL

equations, instead of KVL.

One of the nodes is taken as reference ordatum or groundnode.It is better to select

the one that has maximum number of

branches connected. The reference node is assumed to be at

ground or zero potential.

Next


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The potentials of all other nodes are defined

w.r.t. the reference node.

KCL equations are written, one for each

node, except the reference node.

The equations are solved to give node

voltages.

Current through any branch and voltage atany point of the network can be calculated.

Next

E l 8


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Example 8

Solve the circuit given, using the node voltagemethod.

Next


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Solution : It has only two nodes. Node 2 has been

taken as reference node. The currents in various

branches have been assumed. Writing the KCLequations,

A

V

67

60,Now

18

047

6012

0

12

1

111

321

VI

V

VVV

III

Next

H H dl V l S


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How to Handle Voltage Sources

If one terminal of a voltage source with a

series resistance is grounded (as in theExample 8), the KCL equation can bewritten in terms of this voltage.

Difficulty arises, if a circuit containsfloating voltage sources.

A voltage source is floating if its neitherterminal is connected to ground.

If possible, first transform the voltagesources into current sources.

Next

C i d N d SUPERNODE


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Constrained Node or SUPERNODE

There is another way which uses the concept of

constrained nodeorsupernode. This method is especially suitable for the circuits having

a floating voltage source with no series resistance.

The two ends of a voltage source cannot make two

independent nodes. Hence, we treat these end nodes together as a

supernode.

The supernode is usually indicated by the region

enclosed by a dotted line.

The KCL is then applied to both nodes at the same

time.

Next

C i I d d N d


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Counting Independent Nodes

Itis a node whose voltage cannot be derived

from the voltage of another node.

First turn off all sources, and then counting all

the nodes separated by resistors.

The number of independent nodes is oneless than this number.

Next

E l 9


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Example 9 Determine the current through 4- resistor in the

circuit given below.

Next

S l i


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Solution :

Here, the voltages at nodes a and b are not independent.

The two node voltages are related as

a b a b c6 or 0 6v v v v v

We can treat the two constrained nodes a and b, as asupernode.

Now, writing KCL for this supernode, we get

a b c

a b c

23 4

or 0.33 0.25 0.25 2

v v v

v v v

Next

Click

A l i KCL d


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Applying KCL to node c

c c b

a b c

7

5 4or 0 0.25 0.45 7

v v v

v v v

Above equations can be written in the matrix form,

a

b

c

1 1 0 6

0.33 0.25 0.25 2

0 0.25 0.45 7

v

v

v

Solve the above equation

Next

Click


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We solve the above equations using calculator to get

b c8.77V and 20.43Vv v

Finally, the current through 4- resistor is

b c 8.77 ( 20.42)4 4

v v 2.9125 A

Next

Click

B h k E l 10


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BenchmarkExample 10

Consider the benchmark exampleand solve it

by using node-voltage analysis.

Next


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Solution :

Nodes c and dare constrained to one another.

To find the number of independent nodes, we turn off

the sources to get the circuit,

Next


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There are three nodes, two of which are independent.

However, if we add the two series resistors to make a 5-

resistor we will have only one independent node (node

a).

Hence we will have to solve only one equation.

The unknown voltage across 3- resistor can then be

determined by applying voltage divider rule.

Next


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Writing KCL equation for node a,

a a

a

(6) (0)4 5

1 5

6 14.17V

1.2

v v

v

Using the voltage divider, the voltage across 3-

resistor is

34.17

2 3v

2.5 V

Next

Click

Click

Example 11


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Example 11

Apply KCL to determine currentISin the circuit

shown. Take Vo = 16 V.

Next

Solution : Applying KCL at nodes 1 and 2


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Solution : Applying KCL at nodes 1 and 2,

Next

Click


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Therefore, the current,

Next

Example 12


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Example 12

Using nodal analysis, determine the current

through the 2- resistor in the network given.

Next

Solution : It is much simpler to write the KCL


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Solution : It is much simpler to write the KCL

equations, if the conductance (and not the

resistances) of the branches are given.

Next

It has 3 nodes. So, we have to write KCL


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It has 3 nodes. So, we have to write KCL

equations for only 2 nodes.

We just equate the total current leaving the nodethrough several conductances to the total

source-current entering the node.

At node 1,

At node 2,

Next

Writing the above equations in matrix form


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Writing the above equations in matrix form,

Next

V51 V

2

3

2.12.0

2.07.0

2

1

V

V

Solving for V1, using Calculator, we getClick

Finally, the current in the 2- resistor,

2.52

5

2

1VI

Nodal Analysis


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Nodal Analysis

The above examples suggests that it ispossible to write the nodal analysis

equationsjust byinspection of the network.

Such technique is possible if the networkhas only independent current sources.

All passive elements are shown as

conductances, in siemens (S).

Next


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In case a network contains a practicalvoltage source, first convert it into an

equivalent practical current source.

Write the Conductance Matrix, Node-Voltage Matrix and the Node-Current

Source Matrix, in the same way as in the

Mesh Analysis.

Next

Example 13


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Example 13

Let us again tackle Example 12, by writing the

matrix equations just by inspection.

Next


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Conductance matrix.

G11 =Self-conductance

of node 1.G12=Mutual conductance

between node 1 and 2.


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Node-voltage Matrix.

Node current-source Matrix.

Note that all the elements on the major diagonalof matrix G are positive.

All off-diagonal elements are negative or zero.

Next

Example 14


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Example 14

Solve the following network using the nodal

analysis, and determine the current through the 2-Sresistor.

Next

Solution :


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Solution :

Next

We can write the nodal voltage equation in matrix


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We can write the nodal voltage equation in matrix

form, directly by inspection :

25

3

11

1124

263

437

or

)25(

)3(

)8()3(

)524(24

2)123(3

43)34(

3

2

1

3

2

1

V

V

V

V

V

V

Next


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Finally, the current through 2-S resistor is

Next

Using Calculator, we get

V3andV2 32 VV

Example 15


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Example 15

Find the node voltages in the circuit shown.

Next

Solution :


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Solution :First Method

Transform the 13-V source and series 5-S resistor to

an equivalent current source of 65 A and a parallel

resistor of 5 S

Next


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Now, we can write the nodal equations in

matrix form for the two nodes just by

inspection,

1

2

9 5 10

5 11 52

V

V

1 2

andV V 5 V 7 V

Now, from the original circuit shown, we get

3 2 13 7 13V V 6 V

Next

Second Method


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We use the concept of supernode. The voltage source is

enclosed in a region by a dotted line, as shown in figure.

The KCL is then applied to this closed surface:

2 3 16 5( ) 13V V V

The KCL equation for node 1 is

1 39 5 75V V

For three unknowns, we need another independent

equation. This is obtained from the voltage drop across

the voltage source,

2 3 13V V

Next

Writing the above equations in matrix form,

Click

13565 V


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Solving, we get

1 2 3, , andV V V 5 V 7 V 6 V

Which are the same as obtained by first method.

In general, for the supernode approach, the KCL

equations must be augmented with KVL equations the

number of which is equal to the number of the floating

voltage sources.

Next

13

75

13

110

509

565

3

2

1

V

V

V

Click

Choice Between the TWO


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Choice Between the TWO

We select a method in which the number of

equations to be solved is less.

The number of equations to be solved in

mesh analysis isb(n1)

The number of equations to be solved in

nodal analysis is

(n1)

Next

Review


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Review

Loop-current Analysis.


Loops.

Mesh Analysis.

Supermesh Method. Limitations of Mesh

Analysis.

Planar Network.

Procedure for MeshAnalysis.

Node Voltages Analysis. Supernode.


Nodes.

Nodal Analysis.

Choice Between theTWO.

Unit-2 Network Analysis Part II

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Transcript of Unit-2 Network Analysis Part II