circuit analysis methods chapter 3 mdm shahadah ahmad

CIRCUIT ANALYSIS METHODS

Chapter 3

Mdm shahadah ahmad

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

INTRODUCTION OF NODE-INTRODUCTION OF NODE-VOLTAGE METHODVOLTAGE METHODINTRODUCTION OF NODE-INTRODUCTION OF NODE-VOLTAGE METHODVOLTAGE METHOD

• Use KCL.

• Important step: select one of the node as reference node

• Then define the node voltage in the circuit diagram.

Node-voltage example

• In the diagram, node 3 is define as reference node and node 1 and 2 as node voltage V1 and V2.

• The node-voltage equation for node 1 is,

• In the diagram, node 3 is define as reference node and node 1 and 2 as node voltage V1 and V2.

• The node-voltage equation for node 1 is,

251

100 2111 VVVV

• Node-voltage equation of node 2,

2102

0 212

VVV

• Solving for V1 and V2 yeilds

VV

VV

91.1011

120

09.911

100

2

1

THE NODE-VOLTAGE METHOD AND DEPENDENT SOURCES

• If the circuit contains dependent sources, the node-voltage equations must be supplemented with the constraint equation imposed by the presence of the dependent sources.

example…

Use the node-voltage method to find the power dissipated in the 5Ω resistor.

• The circuit has 3 node. • Thus there must be 2 node-voltage

equation.• Summing the currents away from node 1

generates the equation,

05202

20 2111

VVVV

• Summing the current away from node 2 yields,

02

8

1052212

iVVVV

• As written, these two equations contain three unknowns namely V1, V2 and iØ.

• To eliminate iØ, express the current in terms of node-voltage,

521 VV

i

• Substituting this relationship into the node 2 equation,

06.1

102.075.0

21

21

VV

VV

• Solving for V1 and V2 gives,

VV 161 VV 102

• Then,

Ai 2.15

1016

W

Rip

2.7

544.12

SPECIAL CASE

• When a voltage source is the only element between two essential nodes, the node-voltage method is simplified.

Example…

• There is three essential nodes, so two simultaneous equation are needed.

• Only one unknown node voltage, V2 where as V1=100V.

• Therefore, only a single node-voltage equation is needed which is at node 2.

055010

212 VVV

Using V1 =100V, thus V2=125V.

SUPERNODESUPERNODE

• When a voltage source is between two essential nodes, those nodes can be combine to form a supernode (voltage sourse is assume as open circuit).

Supernode example…

• Nodes chosen,

• Node-voltage equation for node 2 and 3,

0505

212

iVVV

04100

3 iV

• Summing both equation,

04100505

3212 VVVV

Above equation can be generates directly using supernode approach

Supernod

• Starting with resistor 5Ω branch and moving counterclockwise around the supernode,

04100505

3212 VVVV

• Using V1 =50V and V3 as a function of V2,

iVV 1023

5

502 V

i

• Substituded into the node-voltage equation,

1410500

10

100

1

5

1

50

12

V

VV

V

60

15)25.0(

2

2

• Using V2 value, gives

Ai 25

5060

VV 8020603

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

INTRODUCTION OF MESH-INTRODUCTION OF MESH-CURRENT METHODCURRENT METHOD

• A mesh is a loop with no loop inside it.

• A mesh current is the current that exist only in the perimeter of a mesh.

• Mesh-current method use KVL to generates equation for each mesh.

Mesh-current example…

• Mesh-current circuit with mesh current ia and ib.

• Use KVL on both mesh,

311 RiiRiV baa

232 RiRiiV bab

• Solving for ia and ib, and you can compute any voltages or powers of interest.

THE MESH-CURRENT METHOD AND DEPENDENT SOURCES

• If the circuit contains dependent sources, the mesh-current equations must be supplemented by the appropriate constraint equations.

Example…

• Use the mesh-current method to determine the power dissipated in the 4Ω resistor.

• Using KVL,

iiiii

iiiii

iiii

154200

4150

20550

2313

32212

3121

• But

• Substituting into the mesh-current equation,

31 iii

321

321

321

9450

41050

2052550

iii

iii

iii

• Using Cramer rule, the values of i2 and i3 can be determine,

945

4105

20525

905

405

205025

2i

45

5254

95

202510

94

2055

95

4550

2i

Ai

i

26125

32505001250625

3250

)125(4)125(10)125(5

)65(50

2

2

A

i

28125

3500125

45

10550

125

045

0105

50525

3

• Power dissipated by 4Ω resistor is

W

Rip

16

4)2628( 2

2

SPECIAL CASE (SUPERMESH)

• When a branch includes a current source, the mesh-current method can be simplified.

• To create a supermesh, remove the current source from the circuit by simply avoiding the branch when writing the mesh-current equations.

• Supermesh equation,

06450

23100

ac

bcba

ii

iiii

cba iii 65950

• Mesh 2 equation,

cb

bab

ii

iii

2

1030

• From the circuit,

ic –ia= 5A• Using Cramer rule, the three

mesh current can be obtain.

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

SOURCE SOURCE TRANSFORMATION TRANSFORMATION

• Source transformation allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa.

Sorce transformation

Example…

• Source transformation procedure

s

ss R

VI

sp RR

From To methodUse,

pss RIV

ps RR

From To method

Use,

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

THEVENIN EQUIVALENT THEVENIN EQUIVALENT CIRCUITCIRCUIT

• Thevenin equivalent circuit consist of an independent voltage source, VTh in series with a resistor RTh.

Thevenin equivalent circuit

ThV

ThRa

b

• Thevenin voltage, VTh = open circuit voltage in the original circuit.

• Thevenin resistance, RTh is the ratio of open-circuit voltage to the short-circuit current.

sc

ThTh

Th

Thsc i

VR

R

Vi

Example…

V25

5a

b

20

4

A3

1V

abV

• Step 1: node-voltage equation for open-circuit:

ThVVV

VV

32

03205

25

1

11

• Step 2: short-circuit condition at terminal a-b

V25

5 a

b

20

4

A3

2V scI

• Node-voltage equation for short-circuit:

VV

VVV

16

04

3205

25

2

222

84

32

sc

ThTh I

VR

AI sc 44

16

Short-circuit current:

Thevenin resistance:

Thevenin equivalent circuit

V32

8a

b

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

NORTON EQUIVALENT NORTON EQUIVALENT CIRCUITCIRCUIT

• A Norton equivalent circuit consists of an independent current source in parallel with the Norton equivalent resistance.

• Can be derive from a Thevenin equivalent circuit simply by making a source transformation.

• Norton current, IN = the short-circuit current at the terminal of interest.

• Norton resistance, RN = Thevenin resistance, RTh

Example

V25

5a

b

20

4

A3

Step 1: Source transformation

A5 5

a

b

20

4

A3

Step 2: Parallel sources and parallel resistors combined

a

b

4

A8 4

Step 3: Source transformation, series resistors combined, producing the Thevenin equivalent circuit

V32

8a

b

THEVENINEQUIVALENT

CIRCUIT

Step 4: Source transformation, producing the Norton equivalent circuit

a

b

A4 8

NORTONEQUIVALENT

CIRCUIT

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

MAXIMUM POWER MAXIMUM POWER TRANSFERTRANSFER

• Two basic types of system:– Emphasizes the efficiency of the power

transfer– Emphasizes the amount of power

transferred.

• Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL.

• Maximum power transfer occurs when RL=RTh.

Example…Example…

ThV

ThR a

b

LRi

• Power dissipated by resistor RL

LLTh

Th

L

RRR

V

Rip2

2

• Derivative of p with repect to RL

4

22 2

LTh

LThLLThTh

L RR

RRRRRV

dR

dp

• Derivative is zero and p is maximum when

)(22LThLLTh RRRRR

LTh RR

• The maximum power transfer occurs when the load resistance, RL = RTh

• Maximum pwer transfer delivered to RL:

L

Th

L

LTh

R

V

R

RVp

42

2

2

2

CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS

• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle

PRINSIP PRINSIP SUPERPOSISISUPERPOSISIPRINSIP PRINSIP SUPERPOSISISUPERPOSISI

• In a circuit with multiple independent sources, superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are activate.

Step of Superposition principle

1. Deactivated all the sources and only remain one source at one time. Do circuit analysis to find voltages or currents.

2. Repeat step 1 for each independent sources.

3. Sum the resulting voltages or currents.

1. Independent voltage source will become short-circuit with 0Ω resistance.

2. Independent current source will become open-circuit.

3. Dependent sources are never deactivated when applying superposition.

REMEMBER!!!

Example…Example…

• Step 1: deactivated all sources except

voltage source

• V0 is calculated using voltage divider:

Vk

kV 54

1020

• Step 2: Deactivated all sources except

current source

• V0 is calculated by using current divider:

VkmV

mAmk

ki

2)2)(1(

1)2(4

2

0

0

V0 =2+5=7V.

• Step 3: Sum all the resulting voltages:

Question 1 (node-voltage)

• Calculate the value of Io

Solution

• Node 1:

622

3

4221

21

211

VV

VVV

• Node 2:

44

5

2

44

1

2

1

2

1

2

4422

21

21

2212

VV

VV

VVVV

846.1625.1

36

4

6

45

21

21

23

21

23

2

V

AI 923.02

846.10

Question 2 (mesh-current)

• Determine the value of currents, I1, I2 and I3.

• Supermesh:

• Mesh 3:

0)(510 321 III

125510

012555

23

233

II

III

• Dependent current source

• Vo

021 2VII

)(5 320 IIV

• Substitute V0

01011

)(10

321

3221

III

IIII

• Use Cramer rule

10111

1050

5510

10110

105125

550

1I

A1625

625

111

505

101

1005

1011

10510

1011

55125

• Current I2:

A

I

21625

13125625

101

510125

625

1001

101250

5010

2

• Current I3:

A

I

23625

14375625

101

510125

625

0111

12550

0510

3

Question 3 (thevenin)

• Open-circuit voltage, Voc:

• Node-voltage equation for Voc

VV

V

VV

VV

oc

oc

ococ

ococ

10

202

0424

0222

24

• Thevenin resistance, RTh:

5422THR

• Thevenin equivalent circuit:

VV 88.6)10(16

110

Question 4 (norton)

• Open-circuit current, Isc:

AI sc 64

123

• Norton resistance, RN:

RN = 4Ω

• Norton equivalent circuit:

VV 18)3(612460

Question 5 (superposition)

• Use superposition principle to determine the voltage Vo.

• Deactivated current source

V412

224V0

• Deactivated voltage source

V4212

46)2(iV oo

• Summing the voltage V0

V8VV 00

Question 6 (node-voltage)

• Determine the value of Vo.

node-voltage equation:

020

80

10

5

2003 000

ViVV

20

800 V

iCurrent iΔ:

• Thus:

V0 =50V