Chapter 4 Techniques of Circuit Analysis

So far we have analyzed relatively simple resistive circuits by applying KVL and KCL in combination with Ohm’s law.

However as circuit become more complicated and involve more elements, this direct method become cumbersome ( , سهل مرهقةغير )

+-

2R

sv

1R

3R

si

1 2 30

sv iR iR iR+ +-

Applying KVL and Ohm’s law, we have,

Solving for i , we have,

1 2 3

sv

iR R R+ +

+-

1R

2R

3R

6R

7R

8R

sv 7

R

In this chapter we introduce two powerful techniques of circuit analysis that aid in the analysis of complex circuit which they are known as

Node-Voltage Method or Nodal Analysis Mesh-Current Method or Mesh Analysis

In addition to these method we will develop other method such as source transformations,Thevenin and Norton equivalent circuit.

However as circuit become more complicated and involve more elements, this direct method become cumbersome ( , سهل مرهقةغير )

4.1 Terminology

+-

1R

2R

3R

6R

7R

8R

sv 7

R

Planar Circuit : a circuit that can be drawn on a plane with no crossing branches as shown

The circuit shown can be redrawn which is equivalent to the above circuit

Planar Circuit

5R 4

R

3R

7R

8R

+-

6R

sv

1R

2R

5R 4

R

3R

7R

8R

+-

6R

sv

1R

2R

9R

10R

11R

12R

The circuit shown is non planar circuit because it can not be redrawn to make it planar the circuit

Describing a Circuit – The vocabulary

Consider the following circuit

Node A point where two or more circuit elements join a

Essential node A node where three or more circuit elements join b

path A trace of adjoining basic elements with no elements included

branch A path that connects two nodes

1 1 5 6v R R R- - -

1R

Essential branch1 1

v R-A path that connects two essential nodes without passing through an essential node

loop A path whose last node is the same as the starting node

1 1 5 6 4 2v R R R R v- - - - -

mesh A loop that does not enclose any other loops1 1 5 3 2

v R R R R- - - -

Planar Circuit A circuit that can be drawn on a plane with no crossing branches

Example 4.1 For the circuit identify the following

The nodes are a, b, c, d, e, f and g

)a (All nodes ?

The essential nodes are b, c, e and g

)b (The essential nodes ?

A node where three or more circuit elements join

The branches are

1 2 1 2 3 4 5 6 7, , , , , , , , and Iv v R R R R R R R

)c (All branches ?

A path that connects two nodes

The essential branches are

1 1v R-

)d (All essential branches ?

A path that connects two essential nodes without passing through an essential node

2 3 R R- 2 4v R- 5R 6R 7R I

The meshes are

1 1 5 3 2v R R R R- - - -

)e (All meshes ?

2 2 3 6 4 v R R R R- - - - 5 7 6R R R- - 7 IR -

1 5 6R R R- -

)f (Two paths that are not loops or essential branches ?

is a paths , but it is not a loop because it does not have the same starting and ending nodes

Nor is it an essential branch because it does not connect two essential nodes

22v R- Is a paths that are not loop or essential branches ?

1 5 6 41 2v vR R R R- - - --

)g (Two loops that are not meshes ?

is a loop , but it is not a mesh because there are two loops within it

5 6I R R- - is a loop , but it is not a mesh because there are two loops within it

Simultaneous Equations- How many

Let the circuit as shown

There are nine branches in which the current is unknown

Note that the current I is known

Since there are nine unknown current Therefore we need nine independent equations

Since there are 7 nodes There will be 6 independent KCL equations since the 7th one can be written in terms of the 6 equations

We still need 3 more equations

1i

3i

5i

6i

9i

8i 7i

2i

4i

They will come from KVL around three meshesor loops

Note the meshe or loop we apply KVL around should not contain the current source I since we do not know the voltage across it

1i

3i

5i

6i

9i

8i 7i

2i

4i

Since KCL at the non essential (connecting two circuit elements only) like a , d , f will have the following results

9 1i i 3 7i i

Therefore we have 6 currents and 4 essential nodes (connecting three or more circuit elements) like b, c, e, g

5 8i i

1i

3i

5i

6i

2i

4i

Applying KCL to 3 of the 4 essential nodes since KCL on the 4th node will have an equation that is not independent but rather can be derived from the 3 KCL equations

Therefore

Applying KCL to the essential nodes b, c, and e ( we could have selected any three essential nodes) we have

KCL at node b we have 1 2 6 0Ii i i+ + - -KCL at node c we have 1 3 5 0i i i- -

KCL at node e we have 3 4 2 0i i i+ -

Note applying KCL on node g 5 4 6 0Ii i i- - + Which is a linear combination of the other KCL’s

1i

3i

5i

6i2i

4i

Therefore the remaining 3 equations will be derived from KVL around 3 meshes

KVL around mesh 1 1 5 21 32 13( 0)R R R Ri i i v+ + - +

Since there are 4 meshes only, we will use the three meshes with out the current source

2 3 63 4 4 25( 0)R R R Ri i vi- + + - +KVL around mesh 2

5 7 62 6 40R R Ri i i- + - KVL around mesh 2

1i

3i

5i

6i2i

4i

1 3 5 0i i i- -

3 4 2 0i i i+ -

1 5 21 32 13( 0)R R R Ri i i v+ + - +

2 3 63 4 4 25( 0)R R R Ri i vi- + + - +

5 7 62 6 40R R Ri i i- + -

6 equations and 6 unknown which can be solved for the currents

1 2 3 4 6, , , , i i i i i

KCL Equations

1 2 6 0Ii i i+ + - -

KVL Equations

4.2 Node-Voltage Method

Nodal analysis provide a general procedure for analyzing circuits using node voltages as the circuit variables

Node-Voltage Method is applicable to both planar and nonplanar circuits

Using the circuit shown, we can summarize the node-voltage methods as shown next:

+-10 V

1 W

5 W

2 W

10 W 2 A

Step 1 identify all essentials nodes

+-10 V

1 W

5 W

2 W

10 W 2 A

1 2

3 3

Do not select the non essentials nodes

Step 2 select one of the essentials nodes ( 1, 2, or 3) as a reference node

Although the choice is arbitrary the choice for the reference node is were most of branches, example node 3

Selecting the reference node will become apparent as you gain experience using this method (i,.e, solving problems)

Step 3 label all nonrefrence essentials nodes with alphabetical label as v1, v2…

+-10 V

1 W

5 W

2 W

10 W 2 A

v1

A node voltage is defined as the voltage rise from the reference node to a nonreference node

v2

1v

+

-2v

+

-

Step 4 write KCL equation on all labels nonrefrence nodes as shown next

KCL at node 1 1i + 2i + 3i = 0

+-10 V

1 W

5 W

2 W

10 W 2 A

v1 v2

1v

+

-2v

+

-

i1 i2

i3

By applying KVL 1 1(1) 10 0i v+ - 11

10

(1)

vi

-

Let us find 1i 2i 3i, ,

similarly 1 30 (5)v i- 13 (5)

vi

KVL on the middle mesh, we have 1 2 2(2) 0v i v- + + 1 22 2

v vi

-

Therefore

+-10 V

1 W

5 W

2 W

10 W 2 A

v1 v2

1v

+

-2v

+

-

i1 i2

i3

1i + 2i + 3i = 0 We have

1

5

v1 2

2

v v-1 10

1

v -+ + = 0

If we look at this KCL equation, we see that the current we notice that the potential at the left side of the 1 W resistor which is tied to the + of the 10 V source is 10 V because

the – is tied to the reference

10

1 W

5 W

2 Wv1 v2

i1 i2

i3

10

13 5

vi

1 22 2

v vi

-

11

10

1

vi

-

1 2

2

v v-1 10

1

v -+ + 01

5

v=

Therefore we can write KCL at node 1 without doing KVL’s as we did previously

2 W

10 W 2 A

v1 v2

i1 i3

i2

2

10

v2 1

2

v v-02- =KCL at node 2

+

Similarly

+-10 V

1 W

5 W

2 W

10 W 2 A

v1 v210

1 2

2

v v-1 10

1

v -+ + 01

5

v=

2

10

v2 1

2

v v-02- =+

Two equations and two unknowns namely v1 , v2 we can solve and have

1

1009.09

11v V 2

12010.91

11v V

4.3 The Node-Voltage Method and Dependent Sources

If the circuit contains dependent source, the node-voltage equations must be supplemented with the constraint equations imposed by the dependent source as will be shown in the example next

Example 4.3 Use the node voltage method to find the power dissipated in the 5 W resistor

i

i

1 2

33

The circuit has three essentials nodes 1, 2, and 3 and one of them will be the reference

We need two node-voltage equations to describe the circuit

We select node 3 as the reference node since it has the most branches (i.e, 4 branches)

i

v1 v220 8i

The circuit has two non essentials nodes which are connected to voltage sources and will impose the constrain imposed by the value of the voltage sources on the non essential nodes voltage

Applying KCL at node 11 1 1 220

02 20 5

v v v v- -+ +

Applying KCL at node 222 1 2

8 0

5 10 2

vv v v i-- + +

i

v1 v220 8i

1 1 1 220 0

2 20 5

v v v v- -+ + 22 1 28

05 10 2

vv v v i-- + +

The second node equation contain the current i which is related to the nodes voltages as

1 2

5

v vi

- Substituting in the second node equation, we have the following nodes equations

1 20.75 0.2 10 v v-

1 21.6 0 v v- + Solving we have

1 216 V 10 Vv v

2

516 10

= ( ) (5)5

A W i p W- 1.2 1 7.2.2

4.4 The Node-Voltage Method : Some Special Cases

Let us consider the following circuit

The circuit has three essentials nodes 1, 2, and 3 which means that two simultaneous equations are needed.

From the three essentials nodes, a reference node has been chosen and the other two nodes have been labeled

1 2

33

+-100 V 25 W

10 W

50 W 5 A

However since the 100 V source constrains the voltage between node 1 and the reference node to 100 V.

1 Vv 100

This mean that there is only one unknown node voltage namely (v2)

v1 v2

+-100 V

25 W

10 W

50 W5 A

100

Applying KCL at node 22 1 2 5 010 50

v v v- + + -

1 Vv 100Since 2 Vv 125

Knowing v1 and v2 we can calculate the current in every branch

This voltage source is connected between non reference and

reference

Nodal Analysis with voltage sources Some Special Cases

This voltage source is connected between two non reference nodes reference

Enclose the voltage source in a region called supper node

Apply KCL to that region

KCL on the Supper node

2 23 3 36v v+

KCL on the Supper node2 2

3 3 36v v+ The second equation come from the constraint on the voltage source

2 35v v-

Two equations and two unknowns , we can solve

Two voltage sources are connected between two non reference nodes reference

Two supper nodes touching or intersection

Dependent Voltage source supper node apply as will to dependent sources

Combine the two supper nodes to supper supper node

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

4.5 Introduction to the Mesh-Current Method

Consider the following planar circuit

There are 4 essentials nodes

+-

1R

2R

3R

6R

8R

7R

5R

4R

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

There are 7 essentials branches were the current is unknown

Therefore to solve via the mesh-current method we must write 4 mesh -current

( 4 17 )- - Number Number Exxentia Exxential Nodesl Branches

4

1i

2i

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

3i 4

i

A mesh current is the current that exists only in the perimeter of a mesh as shown in the circuit below

The mesh current satisfy Kirchhoff’s Current Law (KCL)

At any node in the circuit, a given mesh current both enters and leaves the node

Mesh current is not always equal branch current

Mesh current is not always equal branch current

1i

2i

+-

1R

2R

3R

6R

8R

sv 7

R

5R

4R

3i 4

i

aI bI

The branch current aI Equal the mesh current 1i

The branch current Doesn't equal any of the mesh currents

bI

+-1

v

1R

+- 2

v

2R

3R

1i 2i

3i

Consider the following circuit were we want to solve for the currents , , 1 2 3i i i

The circuit has 3 essential branches and 2 essentials nodes

( 2 13 )- - Number Number Exxentia Exxential Nodesl Branches

2

Therefore to solve via the mesh-current method we must write 2 mesh -current

+-1

v

1R

+- 2

v

2R

3R

1i 2i

3i

Applying KCL to the upper node (one KCL for two essentials nodes)

1 1 1 3 3v R i R i +

+1 2 3i i i

Applying KVL around the two meshes 2 2 2 3 3v R i R i- -

Solving the KCL for i3 and substituting in the KVL equations we have

1 1 3 1 3 2( )v R R i R i + -

2 3 1 2 3 2( )v R i R R i- - + +

+-1

v

1R

+- 2

v

2R

3R

1i 2i

3i

1 1 3 1 3 2( )v R R i R i + -

2 3 1 2 3 2( )v R i R R i- - + +

Now consider the mesh currents ia, ib , and applying KVL around the two meshes we have

+-1

v

1R

+- 2

v

2R

3R

ai

bi

1 1 3( )a a bv R i R i i + -

2 3 2( )b a bv R i i R i- - +

1 1 3 3 ( ) a bv R R i R i + -

2 3 2 3 ( )a bv R i R R i- - + +

The two equations are identical if we replace the currents i1, i2 with the mesh currents ia, ib

+-1

v

1R

+- 2

v

2R

3R

1i 2i

3i

+-1

v

1R

+- 2

v

2R

3R

ai

bi

1 1 3 3( ) a bv R R i R i + -

2 3 2 3( )a bv R i R R i- - + +

the branches currents i1, i2, i3 can be written in terms of the mesh currents ia, ib

1 ai i 2 bi i 3 a bi i i -

Once we solve for the mesh current we can solve for any branch currentOnce we know the branch current we can solve for any voltage or power of interest current

+-

+-

2 W 6 W 4 W

8 W 6 W40 V 20 Vo

v

+

-

+-

+-

2 W 6 W 4 W

8 W 6 W40 V 20 V

ai

bi

ci

Example 4.4

i

1i

2i

3i

i

4.6 The mesh-Current Method and dependent sources

Example 4.5Use the mesh-current method to determine the power dissipated in the 4 W resistor

1i

2i

3i

i

1 2 1 350 5( ) 20( )i i i i +- -

KVL on mesh 1

2 1 2 2 30 5( ) 1 4( )i i i i i + +- -

KVL on mesh 2

3 1 3 20 20( ) 4( ) 15i i i i i + +- -

KVL on mesh 3

1 3Since i i i -

1 2 350 25 5 20i i i - -

1 2 30 5 10 4i i i- + -

1 2 30 5 4 9i i i- - +

1

2

3

5025 5 20

5 10 0

0

4

5 4 9

i

i

i

- -- -- -

1i

2i

3i

i

1

2

3

25 5 20

5 10

50

0

4

5 4 9

0

i

i

i

- -- -- -

To solve, we use Cramer method as follows ,

2

25 20

5 4

5 9 3250 = = 26 A

25 5 20

50

0

0125

5 10 4

5 4 9

i

-- --

- -- -- -

we need only i2 and i3

3

25 5

5 10

5 4 3500 = = 28 A

25 5 20

50

0

0125

5 10 4

5 4 9

i

--- -

- -- -- -

2

3 2

2 4( =4(28 26) =) P i i4W

- - 16 W

4.7 The Mesh-Current Method : Some Special Cases

bi

ai

ci

v

+

-

bi

ai

ci

Supermesh

Bi

ai

ci

bi

The Mesh-Current Analysis of the Amplifier Circuit

Bi

ai

bi

ci

Supermesh

Δi

ai

ci

di

ei

bi

4.8 The Node-Voltage Method Versus the Mesh-Current Method

1v

2v

3v

Supernode

Δi

Δi

av b

vc

v

Supernode

ai b

ic

i

4 W 2.5 W

193 V+-

ov

+

-

0.8 v

2 W

Δ

v+ -

0.4Δ

v 0.5 A+-

6 W 7.5 W

v+ -

8 W

ai

bi

ci

Supermesh

4 W 2.5 W

193 V+-

ov

+

-

0.8 v

2 W

Δ

v+ -

0.4Δ

v 0.5 A+-

6 W 7.5 W

v+ -

8 W

av

bv

ov

193 V

4 W 2.5 W

+-

ov

+

-

2 W

Δ

v+ -

0.4Δ

v 0.5 A+-

v+ -

193

By inspection

4.9 Source Transformations

The Node-Voltage Method and the Mesh-Current Method are powerful techniques for solving circuits.

We are still interested in in methods that can be used to simplify circuits like to what we did in parallel and series resistors and to Y transformations

A method called Source Transformations will allow the transformations of a voltage source in series with a resistor to a current source in parallel with resistor.

+-sv

a

b

R

The double arrow indicate that the transformation is bilateral , that we can start with either configuration and drive the other

si

a

b

R

+-sv

a

b

R

LR si

a

b

R LR

Li Li

s

L

v

R R

+LiL

RR R

+L si i

Equating we have ,

s

L L

v RR R R R

+ + si s

s

v i

R OR s s v Ri

Example 4.8 (a) find the power associated with the 6 V source) b (State whether the 6 V source is absorbing or

delivering power

We are going to use source transformation to reduce the circuit, however note that we will not alter or transfer the 6 V source because it is the objective.

(20 || 5 )W W 4Ω

408

5A

(8 )( ) 32 VA 4Ω

(6 4)+ 10Ω

( 10)+ 10 20Ω

321.6

20A

(30 || 20 )W W 12Ω

(1.6 )( ) 19.2 VA 12Ω

(4 )+ 12 16Ω

i = 0.825 A1.2- 64+12

i

6 (0.825)(6) 4.95 WVP

It should be clear if we transfer the 6V during these steps you will not be able to find the power associated with it

ov

+

-

Example 4.9 (a) use source transformations to find the voltage vo?

Since the 125 W resistor is connected across or in parallel to the 250 V source then we can remove it without altering any voltage or current on the circuit except the 250 V current which is not an objective any how

Therefore we remove the 125 WOur objective is vo

Similarly the 10 W resistor is connected in series with the 8 A source then we can remove it without altering any voltage or current on the circuit

Now the circuit become

ov

+

-

We now use the source transformation to replace the 250 V with the 25 W resistor with a current source and parallel resistor

25010

25A

ov

+

-

(25 ||100 || 20 )W W W 10ΩWe now combine the parallel resistors

ov

+

-

(25 ||100 || 20 )W W W 10ΩWe now combine the parallel resistors

ov

+

-

The circuit now become

1(2 )( ) 20 V0ov A W

Let us find the Thevenin Equivalent of the following circuit

1 125 3 0

5 20

v v- + -

1v

25

By inspection

1v

1 32v

Therefore the Thevenin voltage for the circuit is 32 V

(4 (5 || 20)) +4

ab ThR R (4 4) 8 + W

To find the RTh we deactivate the sources by shortening independent voltage sources and open independent current source

ab Th

R R20 W

5 W 4 Wa

b

(5 || 20 )W W 4Ω

8 A 4 W

4 W

32 V

4 W

+-

4 W

32 V+-

8 W

255

5A

Let us find the Thevenin Equivalent using source transformations

b

+-Th

V

ThR

a

The Thevenin Equivalent is represented as an independent voltage source VTh and a series resistor RTh

Now let us put a short across the terminal and calculate the resulting current

b

+-Th

V

ThR a

SCi

Th

Th

VR

SC

i

And this another method of finding RTh as Th

SC

Vi

Th

R

Let us find the iSC for the circuit shown before

iSC can be found easily once

we know the voltage v2

2 2 225 3 0

5 20 4

v v v- + - +

2

4sc

vi

To find v2 select the reference as the lower node and write KCL

2v

25 By inspection

2 16 v V

164sci 4 A

324

Th

SC

Vi

8 ΩTh

R

Norton Equivalent

i RTh

4.64 Find Thevenini equivalent with respect to the terminals

a

b

a band

A Using short circuit current

B Deactivating the independent sources

A Using short circuit current

Since Th OC

SC SC

V Vi i

Th

R

17.4

B Deactivating the independent sources

4.12 Maximum Power Transfer

1i

2i

3i

4i

'1

i '3

i

'2

i '4

i

1v

''1

i ''3

i

''2

i ''4

i

3v

4v

P488 Use the principle of superposition to find the voltage v in the circuit

70 V

20 W

+-

bi

+-

4 W

10 W

50 V

+ -

2 W

2b

i

+

-

v

70-V Source acting alone

70

Therefore from 2 and 3 we have

1

2

3

Since

4

From 4 OR

Therefore from 1 and 5 we have

5

OR