CIRCUIT ANALYSIS CIRCUIT ANALYSIS METHODMETHOD

CIRCUIT ANALYSIS CIRCUIT ANALYSIS METHODMETHOD

TOPICTOPIC•Node-Voltage Method•Mesh-current Method•Source of embodiment

principle•Thevenin’s Circuit•Norton’s Circuit•Maximum Power Transfer•Superposition Principle

INTRODUCTION TO NODE-VOLTAGE

METHOD

•Base on Kirchhoff’s Current Law

•Important step: select one node as a reference.

Example:Node-Voltage method

•Previous circuit, set node 3 as a reference. By using Kirchhoff’s Current law at node 1,

251

100 2111 VVVV

Node-voltage equation at node2

2102

0 212

VVV

•Solve previous equation

VV

VV

91.1011

120

09.911

100

2

1

NODE-VOLTAN METHOD THAT CONTAIN

DEPENDENT SOURCE

•If the circuit contains dependent source, the node-voltage equation imposed by the presence of the dependent source.

Find power that absorb by 5Ω Resistor using node-voltagemethod.

•Circuit have 3 node. •Need 2 node-voltage

equations.•Summing the currents away

from node 1 generates the equation,

05202

20 2111

VVVV

•Summing the currents away from node 2 yields

02

8

1052212

iVVVV

•These two node-voltage equations contain three unknowns, namely, V1, V2 and iø . To eliminate iø, we must express this controlling current in terms of the node-voltage,

521 VV

i

•Substituting this relationship into the node 2 equation simplifies the two node-voltage equations

06.1

102.075.0

21

21

VV

VV

•Solving for V1 and V2 gives,

V1 =16V V2 = 10V

•Then,

•Power absorb by 5Ω resistor

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 are three essential nodes in this circuit, which means that two simultaneous equations are needed.

•There is only one unknown node-voltage V2, but V1 =100V.

•Solution of this circuit thus involves only a single node-voltage equation at node 2.

055010

212 VVV

Have V1 =100V, and solvedV2 =125V.

SUPERNODESUPERNODE

•A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.

Example:supernode

•Select node:

•Node voltage equation at node 2 and node 3

0505

212

iVVV

04100

3 iV

•Add previous equation

04100505

3212 VVVV

•Previous equation could get when use supernode concept at node 2 and node 3.

Supernode

•From 5Ω resistor

04100505

3212 VVVV

•Pink equation was equal to green equation.

•Using supernode at node 2 and 3 make it simple to analyse the circuit.

•Have V1 =50V and V3 can be describe with V2,

iVV 1023

5

502 V

i

•Replace V1 =50, V3 and iø, pink equation become

1410500

10

100

1

5

1

50

12

V

VV

V

60

15)25.0(

2

2

•Insert V2

Ai 25

5060

VV 8020603

INTRODUCTION TO INTRODUCTION TO MESH-CURRENT MESH-CURRENT

METHODMETHOD

•One mesh mean a loop that no others loop inside.

•This mesh-current method used Kirchhoff’s voltage law to find current each mesh.

Example:Mesh-current

•From Kirchhoff’s law

(1)

(2)

321 iii

33222

33111

RiRiV

RiRiV

•Use i3 from equation (1) and insert to equation (2)

323111 )( RiRRiV )( 322312 RRiRiV

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

•Using KVL at those two mesh

311 RiiRiV baa

232 RiRiiV bab

•After ia and ib known, then we can calculate voltage at power at each resistor.

MESH-CURRENT METHOD THAT HAVE DEPENDENT SOURCE

•When circuit have dependent source, mesh-current equation will have constant value related to dependent source.

Example:mesh method with dependent source

• Find power that obserb by 4Ω resistor using mesh-current method.

•From Kirchhoff’s voltage law

iiiii

iiiii

iiii

154200

4150

20550

2313

32212

3121

•Have

•Insert equation iø to related equation,

31 iii

321

321

321

9450

41050

2052550

iii

iii

iii

•By using Cramer law, i2 and i3 can be calculated as below,

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 that absorb by 4Ω resistor

W

Rip

16

4)2628( 2

2

SPECIAL CASE (SUPERMESH)

•When a branch of current source can be remove and use supermesh concept (current source assume as open circuit)

• Assume that current source as open circuit

•Supermesh equation

06450

23100

ac

bcba

ii

iiii

cba iii 65950

•Mesh-current equation for mesh 2

cb

bab

ii

iii

2

1030

•Known

ic –ia= 5A By using Cramer law at those three equation, value for those three mesh current could be calculated.

SOURCE SOURCE TRANSFORMATIONSTRANSFORMATIONS

•Source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.

Source Transformation

Example:Source transformation

•When resistor R=0, terminal a-b become close circuit. Beginning, close circuit current should be same. Therefore,

s

ss R

VI

•Close circuit current for second circuit was Is. Therefore,

ps R

VsI

•When resistor R = ∞, these circuit become open circuit.

•From first circuit, we have Vab =Vs . Therefore, it was voltage for open circuit.

psab RIV

•Vab for those two circuit should be same. Therefore, Vs = Is Rp .

•Replace Is

ps

ps

ss

RR

RR

VV

•Summarize for Source transformation

Tetapkan

s

ss R

VI

sp RR

MethodAfterBefore

Tetapkan

pss RIV

ps RR

Before After Method

THEVENIN EQUIVALENT CIRCUIT

•Introduced in 1883 by M. Leon Thevenin (1857-1926), a French telegraph engineer.

• Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh where VTh

is the open-circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent

sources are turned off.

•This theorem usually used to replace large sequence part (complex) with one simple equivalent circuit. This simple circuit makes voltage, current and circuit power could be calculated easily.

Thevenin equivalent circuit

•Thevenin voltage, VTh = open circuit voltage for origin circuit.

•When load decrease until zero, circuit become close circuit and current become:

sc

ThTh

Th

Thsc i

VR

R

Vi

Example

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

ThVVV

VV

32

03205

25

1

11

• Step 2: replace close circuit at a-b terminal

•Node voltage equation for close circuit:

VV

VVV

16

04

3205

25

2

222

84

32

sc

ThTh i

VR

Aisc 44

16

Close circuit current:

Thevenin resistance

Thevenin equivalent circuit

Norton equivalent circuit

•In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed Norton’s theorem.

•This equivalent circuit have one independent source that parallel

with one resistor.

•Norton equivalent circuit could have from Thevenin equivalent circuit by source transformation.

ExampleStep 1: Source transformation

Step 2: Combine source and parallel resistors

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

Step 4: Source transformation and Producing the Norton equivalent circuit

Norton equivalent circuit

TOPICTOPIC• Node-Voltage method• Mesh-current method• Source transformation

principle• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer

principle• Superposition principle

MAXIMUM POWER TRANSFER

•Power system designed to provide power to load at high-efficiency and decrease power loss when delivered to load. Therefore, we need to decrease source resistance and delivering resistance.

•Definition for Maximum power transfer tell that power that transfer from one source was represent by Thevenin equivalent circuit become max when load resistor RL and Thevenin resistor RTh was

ExampleExample

•Power absorb by resistor RL

LLTh

Th

L

RRR

V

Rip2

2

•Differentiate p with RL

4

22 2

LTh

LThLLThTh

L RR

RRRRRV

dR

dp

•Differential was zero and p become maximum

•Solve

)(22LThLLTh RRRRR

LTh RR

•Therefore, for maximum power transfer, RL must equal with RTH .

•Maximum power transfer equation:

L

Th

L

LTh

R

V

R

RVp

42

2

2

2

SUPERPOSITION PRINCIPLE

•The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or current through) that element due to each independent source acting alone.

Step to Apply Step to Apply Superposition PrincipleSuperposition Principle

1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active.

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to the independent sources.

1. Independent voltage source become close circuit with zero volt.

2. Independent current source become open circuit.

3. If dependent source exist, it should be active while superposition process.

REMEMBER!

ExampleExample

•Step 1: turn off current source

•Use voltage divider law to calculate V0 :

Vk

kV 54

1020

•Step 2: turn off voltage source

•Use current divider law to calculate V0 ,

VkmV

mAmk

ki

2)2)(1(

1)2(4

2

0

0

Total V0 :

V0 =2+5=7V.

Question 1

Answer•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

•Supermesh:

•Mesh 3:

0)(510 321 III

125510

012555

23

233

II

III

•Current source

•known

021 2VII

)(5 320 IIV

•Replace V0

01011

)(10

321

3221

III

IIII

•By using Cramer law

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

•Open circuit voltage:

•Node-voltage equation for Voc

VV

V

VV

VV

oc

oc

ococ

ococ

10

202

0424

0222

24

• Thevenin resistance:

5422THR

• Get:

VV 88.6)10(16

110

Question 4

• Close circuit current

AI sc 64

123

• Norton resistance

RTH = 4Ω

• Norton equivalent circuit:

VV 18)3(612460

Question 5

•Turn off current source

VV 412

2241

0

•Turn off voltage source

VV 4212

46110

•Total V0

VVV 0110

10

Question 6

Node-voltage equation:

020

80

10

5

2003 000

ViVV

20

800 V

iKnown:

•Get:

V0 =50V

•Finally we get:

V0 =50V