chap4

Circuit TheoremsCircuit Theorems Eastern Mediterranean UniversityEastern Mediterranean University 11

Circuit Theorems

Mustafa Kemal Uyguroğlu

Chap. 4 Circuit TheoremsChap. 4 Circuit Theorems

Introduction

Linearity property

Superposition

Source transformations

Thevenin’s theorem

Norton’s theorem

Maximum power transfer

Circuit TheoremsCircuit Theorems 22EEastern astern MMediterranean editerranean UUniversityniversity

4.1 Introduction4.1 Introduction

Circuit TheoremsCircuit Theorems 33

A largecomplex circuits

A largecomplex circuits

Simplifycircuit analysis

Simplifycircuit analysis

Circuit TheoremsCircuit Theorems

‧Thevenin’s theorem Norton theorem‧‧Circuit linearity Superposition‧‧source transformation max. power transfer‧

‧Thevenin’s theorem Norton theorem‧‧Circuit linearity Superposition‧‧source transformation max. power transfer‧

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4.2 Linearity 4.2 Linearity PPropertyroperty


Homogeneity property (Scaling)

iRvi kiRkvki

Additivity property

Rivi 222 Rivi 111

21212121 )( vvRiRiRiiii


A linear circuit is one whose output is linearly related (or directly proportional) to its input

Fig. 4.1


vV0

I0

i


Linear circuit consist of

● linear elements

● linear dependent sources

● independent sources


mA1mV5

A2.0V1

A2V10

iv

iv

iv

s

s

s

nonlinearRv

Rip :2

2


Example 4.1Example 4.1

For the circuit in fig 4.2 find I0 when vs=12V and vs=24V.



KVL

Eqs(4.1.1) and (4.1.3) we get


0412 21 svii

03164 21 sx vvii

12ivx becomes)2.1.4(

01610 21 svii

(4.1.1)(4.1.2)

(4.1.3)

2121 60122 iiii



Eq(4.1.1), we get

When

When

Showing that when the source value is doubled, I0 doubles.


76 076 22

ss

vivi

A7612

20 iIV12sv

A7624

20 iI

V24sv



Assume I0 = 1 A and use linearity to find the actual value of I0 in the circuit in fig 4.4.




A,24/

V8)53(thenA,1If

11

010

vI

IvI

A3012 III

A27

,V14682 23212 VIIVV

A5234 III A5SI

A510 SIAI

A15A30 SII


4.3 Superposition4.3 Superposition

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 currents through) that element due to each independent source acting alone.

Turn off, killed, inactive source:

● independent voltage source: 0 V (short circuit)

● independent current source: 0 A (open circuit)

Dependent sources are left intact.


Steps to apply superposition principle:

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

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.


How to turn off independent sourcesHow to turn off independent sources

Turn off voltages sources = short voltage sources; make it equal to zero voltage

Turn off current sources = open current sources; make it equal to zero current


Superposition involves more work but simpler circuits.

Superposition is not applicable to the effect on power.



Use the superposition theorem to find in the circuit in Fig.4.6.



Since there are two sources,

let

Voltage division to get

Current division, to get

Hence

And we findCircuit TheoremsCircuit Theorems 1717

21 VVV

V2)6(84

41

V

A2)3(84

83

i

V84 32 iv

V108221 vvvEEastern astern MMediterranean editerranean UUniversityniversity


Find I0 in the circuit in Fig.4.9 using superposition.




Fig. 4.10




Fig. 4.10


4.5 Source Transformation4.5 Source Transformation

A 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


Fig. 4.15 & 4.16Fig. 4.15 & 4.16

Rv

iRiv ssss or


Equivalent CircuitsEquivalent Circuits

R

v

R

vi

viRv

s

s


i i

++

--

vv

v

i

vs-is


Arrow of the current source positive terminal of voltage source

Impossible source Transformation

● ideal voltage source (R = 0)

● ideal current source (R=)



Use source transformation to find vo in the circuit in Fig 4.17.




Fig 4.18



we use current division in Fig.4.18(c) to get

and


A4.0)2(82

2

i

V2.3)4.0(88 ivo



Find vx in Fig.4.20 using source transformation



Applying KVL around the loop in Fig 4.21(b) gives

(4.7.1)

Appling KVL to the loop containing only the 3V voltage source, the resistor, and vx yields

(4.7.2)


01853 xvi

1

ivvi xx 3013



Substituting this into Eq.(4.7.1), we obtain

Alternatively

thus


A5.403515 ii

A5.40184 iviv xx

V5.73 ivx


4.5 Thevenin’s Theorem4.5 Thevenin’s Theorem

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 source are turn off.


Property of Linear CircuitsProperty of Linear Circuits


i

v

v

i

Any two-terminalLinear Circuits

+

-Vth

Isc

Slope=1/Rth


Fig. 4.23Fig. 4.23


How to Find Thevenin’s Voltage How to Find Thevenin’s Voltage

Equivalent circuit: same voltage-current relation at the terminals.


:Th ocvV ba atltagecircuit voopen


How to Find Thevenin’s ResistanceHow to Find Thevenin’s Resistance


:inTh RR b.a atcircuitdeadtheofresistanceinput

circuitedopenba sourcestindependenalloffTurn


CASE 1

If the network has no dependent sources:

● Turn off all independent source.

● RTH: can be obtained via simplification of either parallel or series connection seen from a-b


Fig. 4.25Fig. 4.25

CASE 2

If the network has dependent sources

● Turn off all independent sources.

● Apply a voltage source vo at a-b

● Alternatively, apply a current source io at a-b


o

o

iv

R Th

o

oTh i

vR


The Thevenin’s resistance may be negative, indicating that the circuit has ability providing power


Fig. 4.26Fig. 4.26

Simplified circuit

Voltage divider


LL RR

VI

Th

Th

ThTh

VRR

RIRV

L

LLLL



Find the Thevenin’s equivalent circuit of the circuit shown in Fig 4.27, to the left of the terminals a-b. Then find the current through RL = 6,16,and 36 .


Find RFind Rthth


shortsourcevoltageV32:Th R

opensourcecurrentA2

4116

124112||4ThR


Find VFind Vthth


analysisMesh)1(

:ThV

A2,0)(12432 2211 iiiiA5.01 i

V30)0.25.0(12)(12 21Th iiV

AnalysisNodal ely,Alternativ)2(12/24/)32( ThTh VV

V30Th V



Circuit TheoremsCircuit Theorems 4343Fig. 4.29

transformsource ely,Alternativ)3(

V302439612

24

32

THTHTH

THTH

VVV

VV




:getTo Li

LLL RRR

Vi

430

Th

Th

6LR A310/30 LI16LR A5.120/30 LI

A75.040/30 LI36LR



Find the Thevenin’s equivalent of the circuit in Fig. 4.31 at terminals a-b.



(independent + dependent source case)


Fig(a) :findTo ThR

0sourcetindependen intactsourcedependent

,V1ovoo

o

iiv

R1

Th



For loop 1,


2121 or0)(22 iiviiv xx

214But iivi x

21 3ii




:3and2Loop

0)(6)(24 32122 iiiii

012)(6 323 iii

gives equations theseSolving

.A6/13 i

A61

But 3 iio

61

ThoiV

R




0)(22 23 iivx

51 i

Fig(b):getTo ThV

23 iivx

analysisMesh

06)(2)(4 21212 iiiii 02412 312 iii

.3/102 i

V206 2Th ivV oc

xvii )(4But 21



Determine the Thevenin’s equivalent circuit in Fig.4.35(a).

Solution


)caseonlysourcedependent(

o

o

iv

R Th0Th V:anaysisNodal

4/2 oxxo viii




220 oo

x

vvi But

4424oooo

xo

vvvvii

oo iv 4or

:4Thus Th o

o

iv

R powerSupplying






4.6 4.6 Norton’s TheoremNorton’s Theorem

Norton’s theorem states that a linear two-terminal circuit can be replaced by equivalent circuit consisting of a current source IN in parallel with a resistor RN where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent source are turn off.


Fig. 4.37Fig. 4.37


v

i

Vth

-IN

Slope=1/RN


How to Find Norton CurrentHow to Find Norton Current

Thevenin and Norton resistances are equal:

Short circuit current from a to b :


ThRRN

Th

Th

RV

iI scN


Thevenin or Norton equivalent circuit :Thevenin or Norton equivalent circuit :

The open circuit voltage voc across terminals a and b

The short circuit current isc at terminals a and b

The equivalent or input resistance Rin at terminals a and b when all independent source are turn off.


ocTh vV

NI

ThTh N

Th

VR R

R

sci



Find the Norton equivalent circuit of the circuit in Fig 4.39.




:)(40.4Fig a

425

52020||5

)848(||5NRNRfindTo




NifindTo

.andterminalscircuitshort ba

))(40.4.Fig( b

:Mesh 0420,A2 2121 iiii

Nsc Iii A12




NIformethodeAlternativ

Th

ThN

R

VI

voltagecircuitopen: ThV ba and

:))(40.4( cFig

:analysisMesh

012425,2 343 iiAi

A8.04 i

terminalsacross

V45 4 iVv Thoc




,HenceA14/4

Th

ThN

R

VI



Using Norton’s theorem, find RN and IN of the circuit in Fig 4.43 at terminals a-b.




NRfindTo )(44.4. aFig

shortedresistor4Parallel:2||||5 xo iv

Hence, 2.05/15/ ox vi

52.0

1

o

oN

iv

R




NIfindTo )(44.4. bFig

xiv 2||5||10||4 Parallel:

.5A,24

010 xi

A72(2.5)5

102 xxsc iii

7A NI


4.8 Maximum Power Trandfer4.8 Maximum Power Trandfer


LL RRR

VRip

2

LTH

TH2

Fig 4.48


Fig. 4.49Fig. 4.49

Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen the load (RL = RTH).



TH

TH

THL

LTHLLTH

LTH

LLTHTH

LTH

LTHLLTHTH

L

RV

p

RR

RRRRR

RRRRR

V

RRRRRRR

VdRdp

4

)()2(0

0)(

)2(

)()(2)(

2

max

32

4

22



Find the value of RL for maximum power transfer in the circuit of Fig. 4.50. Find the maximum power.




918

126512632THR




WRV

p

RR

VVVii

Aiii

L

TH

THL

THTHi

44.1394

224

9

220)0(231612

2 ,121812

22

max

2

221


Homework ProblemsHomework Problems

Problems 6, 10, 21, 28, 33, 40, 47, 52, 71


chap4

Documents

introductioncircuit

currentcircuit theorems

getcircuit theorems

independent voltage

voltages sources

open current sources

short voltage sources

source value