7/29/2019 ES100 Lecture 2

1/34

1

Lecture 2Lecture 2

Dr Ahmed AbuDr Ahmed Abu--SiadaSiada

Electrical Systems 100Electrical Systems 100

7/29/2019 ES100 Lecture 2

2/34

2

ContentsSeries-Parallel Circuits

Wye-Delta Conversion

Ladder Networks

Current Sources

Source Conversion

Current Sources in Parallel (Series?)

Mesh Analysis

Nodal Analysis

7/29/2019 ES100 Lecture 2

3/34

3

Series-Parallel CircuitsSeries-parallel circuits are networks where there are

both series and parallel elements

7/29/2019 ES100 Lecture 2

4/34

4

Reduce and Return Approach

Step 1: Take an overall mental look at the problem

Step 2: Examine each section of the network independently beforetying them together

Step 3: Redraw the network as often as you will need to arrive at

reduced branches. Maintain the original unknown quantities to befound for clarity where applicable

Step 4: Take the trip back to the original network to find detailed

solution

(Some time it may help to draw branche/s as blocks and work out

blockwise)

7/29/2019 ES100 Lecture 2

5/34

5

An Example of Reduced and Return Approach

Finding V4?

1R

243 //)(' RRRR T +=

1' RRR TT+=

SI

SI

T

SR

EI = SI

TR'

2V

TS RIV =2

2V43

424

RR

RVV

+=

7/29/2019 ES100 Lecture 2

6/34

6

An Example of Reduced and Return Approach

2k //6k12k

( )//6k12k2kIS

+=

E

54V

S2

S1

I6k12k

12kI

I6k12k

6kI

+

=

+

=

7/29/2019 ES100 Lecture 2

7/34

7

Wye-Delta Conversion

Often we encounter a different kind of network which appears to benot in series or parallel in relation to the rest of the network. Under

these circumstances it is necessary to convert this portion of circuit

from one form to the other to find appropriate branch connectionwhich then appears clearly in series or parallel with rest of the

network.

7/29/2019 ES100 Lecture 2

8/34

8

Y- ConversionThe purpose is then to be able to convert Y to or to Y.

)(

)(

)//(

31

31

CAB

CB

CAB

AB

CAB

CAB

ca

CABca

RRR

RR

RRR

RR

RRR

RRRRRR

RRRRRR

+++++=

++

+=+=

+=+=

7/29/2019 ES100 Lecture 2

9/34

9

-Y Conversion

CBA

CB

RRR

RRR

++=1

)(

)(

)(

)(

)(

)(

32

21

31

CBA

CA

CBA

BA

CBA

CBA

cb

CBA

BC

CBA

AC

BAC

BA

ba

CBA

CB

CBA

AB

CAB

CAB

ca

RRR

RR

RRR

RR

RRR

RRRRRR

RRR

RR

RRR

RR

RRR

RRRcRRR

RRR

RR

RRR

RR

RRR

RRRRRR

++

+

++

=

++

+=+=

+++

++=

++

+=+=

+++

++=

++

+=+=

CBA

CA

RRR

RRR

++=2

CBA

BA

RRR

RRR

++

=3

7/29/2019 ES100 Lecture 2

10/34

10

Y- Conversion

1

313221

RRRRRRRRA

++=

Similarly, for converting Wye quantities to Delta are given as:

2

313221

RRRRRRRRB ++=

3

313221

R

RRRRRRRC

++=

7/29/2019 ES100 Lecture 2

11/34

11

-Y /Y- Conversion

If all resistors in the or Y are the same (RA = RB = RC):

From -Y eq:

12

2

3

3

3

RRR

RR

RRRRRR

A

A

A

CBA

BA

===

=++

=

This shows that for a Y of three equal resistors the value of each

resistor equivalent is 3 times the Y resistor.

Y

Y

3RR:

YFor

3

RR:YFor

=

=

7/29/2019 ES100 Lecture 2

12/34

12

Ladder NetworksA Ladder Network is one where a series-parallel section of a network occur

repeatedly within the network. An example of such network is a Low Pass Filter

circuit. Figure below shows a three section Ladder Network.

To solve a ladder network follow the steps:

Calculate the total resistance

Calculate the source current or total current drawn from source

Work back through the ladder until desired current or voltage is obtained

7/29/2019 ES100 Lecture 2

13/34

13

Ladder Networks

Combining parallel and series elements to reduce the circuit we get :

S1

T

S

T

IIbackwards,Working

A30

8

240V

R

EI

835R

=

===

=+=

7/29/2019 ES100 Lecture 2

14/34

14

Ladder NetworksUsing current divider I6 can be found.

A15I

A152IIA,30I

2

S

3S

=

===

Finally,

V20210ARIVand

A10I9

6I

3)(6

6I

666

336

===

==+

=

(Current divider)

7/29/2019 ES100 Lecture 2

15/34

15

Current Sources

A battery supplies fixed voltage and the source current may vary

according to load. Similarly, a current source is one where it

supplies constant current to the branch where it is connected andthe voltage and polarity of voltage across it may vary according to

the network condition.

AII

EVS

437I

KCLApplyingA,3

4

12VI

V12

21

2

===

=

=

==

7/29/2019 ES100 Lecture 2

16/34

16

Source Conversion

A Voltage source can be converted to a current source and vice

versa. In reality, Voltage sources has an internal resistance Rs and

current sources has a shunt resistance Rsh. In ideal cases, Rs equalto 0 and Rsh equal to .

7/29/2019 ES100 Lecture 2

17/34

17

Source Conversion

For us to be able to convert sources, the voltage source must have a series

resistance and current source must have some shunt resistance.

Eg.

7/29/2019 ES100 Lecture 2

18/34

7/29/2019 ES100 Lecture 2

19/34

19

Method of Circuit Analysis

(Mesh Analysis and Nodal Analysis)

Mesh is a closed loop which does not contain any other loops

within it. In most circumstances, a mesh will contain one or more

voltage sources and one or more type of circuit elements. In dccircuit theory, these elements are limited to resistances only in

steady state analysis. Mesh analysis determines the mesh or loop

currents in the circuit.

i1 i2I1 = i1I2 = i2I3 = i1-i2

By solving i1 and i2

7/29/2019 ES100 Lecture 2

20/34

20

Mesh AnalysisSteps to determine Mesh Currents:

Identify the n number meshes in the circuit

Assign mesh currents, in clockwise directions

Apply KVL to each of the n meshes. Use Ohms law to

express the voltages in terms of the mesh currents. Take

appropriate voltage drop polarity (+ve clockwise and veanticlockwise) into consideration in writing these equations.

Solve the n simultaneous equations to get the n mesh

currents.

nn iiiii ,1321 ......,,

7/29/2019 ES100 Lecture 2

21/34

21

Mesh Analysis-An Example

A.1iandA1i

12i36i

12

22

==

=

Find I1, I2 and I3.

i1 i2

(Loop i1)

12i3i

010i15i5

010)i10(i5i15

21

21

211

=

=+

=

12i

1i2i020i10i10

04i6i)i10(i10

21

12

21

2212

=

=

=+

=

i

(Loop i2)

Method 1:Using method of substitution

7/29/2019 ES100 Lecture 2

22/34

22

Mesh Analysis-An ExampleMethod 2: Using Cramers Rule (Also known as Format Approach):

41311

13

42221

21

42621-

2-3

1

1

i

i

21

23

2

1

2

1

=+=

=

=+=

=

===

=

12ii

12i3i

21

21

=+

=From previous 2 loops :

We obtain the determinant as

A1

i

A1

i

22

11

==

==Thus:

7/29/2019 ES100 Lecture 2

23/34

23

Mesh Analysis-Super MeshSometime, there may be a current source in one of the mesh or a

current source in common between two meshes.

If the current source involves only one of the meshes, then the

analysis is easier as we have 1 less equation to solve as the current is

defined by the current source in one of the mesh already.

If however, the current source is common between two meshes,

then we need to form a super mesh. This is necessary as we need toapply KVL in solving mesh equations and we do not know the

voltage across the current source in advance.

7/29/2019 ES100 Lecture 2

24/34

24

Mesh Analysis-Super MeshCase 1:

In this example, A.52 =i

i1 i2

A

ii

2i

0)(64i10-

1

211

=

=++

Applying KVL in Mesh 1

7/29/2019 ES100 Lecture 2

25/34

25

i1 i2

Mesh Analysis-Super MeshCase 2:

we create a super-mesh by excluding the current source and any

element connected in series with it as shown.

A8.2i2.3i

20146i

0410620

21

21

221

==

=+

=+++

A

i

iii

A6ii 12 =

i1 i2

In super-mesh

super-mesh

7/29/2019 ES100 Lecture 2

26/34

26

Nodal AnalysisA Node is defined as the junction of two or more branches. In a n

node circuit, 1 node is taken as the reference (usually the ground is

taken as reference) and we need to solve node voltages using KCL.

By solving node v1 and v2

We can solve :

2

212

R

VVI

=

1

11

RVI =

3

23

R

VI =

7/29/2019 ES100 Lecture 2

27/34

27

Nodal AnalysisSteps to determine Nodal voltages:

Select a node as reference node

Assign voltages to the remaining nodes

Apply KCL to each of the n-1 non-reference nodes. Use Ohms

law to express currents in terms of node voltages.

Solve the n-1 simultaneous equations to get the unknown nodevoltages.

121 ,...., nvvv

7/29/2019 ES100 Lecture 2

28/34

28

Nodal AnalysisApplying KCL to each nodes

Example in node v1

00

12

2

21

1

1=+

+

II

R

VV

R

V

node v2

00

2

2

12

3

2=

+

I

R

VV

R

V

7/29/2019 ES100 Lecture 2

29/34

29

Nodal Analysis- An ExampleAt node 1, applying KCL,

6053v-

01054

v

6

0

21

122

=+

=+

+

v

vv

v1v2

20v3v

vv2v20

054

vv

2

0v

21

211

211

=

+=

=

+

At node 2, applying KCL,v2v1

7/29/2019 ES100 Lecture 2

30/34

30

Nodal Analysis- An ExampleUsing Cramers Rule or Format Approach:

Vv

Vv

v

v

2012

60180603

203

33.1312

60100560

120

1231553

13

60

20

53

13

22

11

2

1

=+

=

=

=

=+

=

=

=

==

=

=

7/29/2019 ES100 Lecture 2

31/34

31

Nodal Analysis-SupernodeSometime, there may be a voltage source connected between a

reference and nonreference node. If this is the case, the voltage of

the nonreference node is simply set equal to the voltage source and

we have 1 less equation to solve.If however, the voltage source is common between two or more

unknown nodes, then we need to form a super node. This is

necessary as we need to apply KCL in solving node equations and

we do not know the current through the voltage source in advance.

A Supernode is formed by enclosing the voltage source between

two nonreference nodes and any branches in parallel with it.

7/29/2019 ES100 Lecture 2

32/34

32

Nodal Analysis-SupernodeConsider the following circuit. Nodes 2 and 3 form a supernode. At

Supernode we get,

10Vv

04

vv6

0v8

0v2

vv

1

133212

=

=+++

v1 v3v2 v1

v2 v3

At Supernode, applying KCL,

7/29/2019 ES100 Lecture 2

33/34

33

Nodal Analysis-SupernodeApplying KVL to supernode we get,

5v,

05

32

32

=

=++

vor

vv

v

2

v

3

7/29/2019 ES100 Lecture 2

34/34

34

Nodal Analysis-SupernodeWe note the following properties of a supernode,

The voltage source inside the supernode provides a constraint

equation to solve for the node voltages

A supernode do not have a voltage of its own

A supernode requires the application of both the KCL and KVL