DC Circuits: Circuit Theorems · EENG223: CIRCUIT THEORY I A large complex circuits Simplify...

EENG223: CIRCUIT THEORY I

DC Circuits:

Circuit Theorems Hasan Demirel


• Introduction • Linearity Property • Superposition • Source Transformations • Thevenin’s Theorem • Norton’s Theorem • Maximum Power Transfer

Circuit Theorems


A large complex circuits

Simplify circuit analysis

Circuit Theorems

‧Thevenin’s theorem ‧ Norton theorem ‧Circuit linearity ‧ Superposition ‧Source transformation ‧ Max. power transfer

Introduction


• Homogeneity property (Scaling)

• Additivity property

iRvi

kiRkvki

Rivi 222

Rivi 111

21212121 )( vvRiRiRiiii

Linearity Property


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

v V0

I0

i

Linearity Property


• A linear circuit consist of :

• linear elements (i.e. R=5 W)

• linear dependent sources (i.e. vs=6Io V)

• independent sources (i.e. vs=12 V)

mA1mV5

A2.0V1

A2V10

iv

iv

iv

s

s

s

nonlinearR

vRip :

22

Linearity Property


Linearity Property Example 4.1: For the circuit in below find Io when vs=12V

and vs =24V.



and vs =24V.

KVL

0412 21 svii

03164 21 sx vvii

12ivx

01610 21 svii

(1)

(2)

(3)

2121 60122 iiii

Using Eqs(1) and (3) we get

Eq(2) becomes



and vs =24V.

From Eq(1) we get:

A76

1220 iI

V12sv

A76

2420 iI

V24sv

when

when

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

76076 22

ss

vivi


Linearity Property Example 4.2: Assume Io = 1A and use linearity to find the actual value of Io in the following circuit.


Linearity Property Example 4.2: Assume Io = 1A and use linearity to find the actual value of Io in the following circuit.

A,24/

V8)53(thenA,1If

11

010

vI

IvI

A3012 III

A27

,V14682 23212

VIIVV

A5234 III A5SI

A510 SIAI

A15A30 SII


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, kill, inactive source(s):

• independent voltage source: 0 V (short circuit)

• independent current source: 0 A (open circuit)

• Dependent sources are left intact.


Superposition • 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.


Superposition

• How 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.


Superposition

• Example 4.3: Use the superposition theorem to find v in the circuit below.


Superposition

• Example 4.3: Use the superposition theorem to find v in the circuit below.

Since there are two sources,

Let

Voltage division to get

Current division, to get

Hence

And we find

A2)3(84

83

i

V84 32 iv

V108221 vvv

21 vvv

V2)6(84

41

v


Superposition • Example 4.4: Find Io in the circuit below using

superposition.



superposition.

Turn off 20V voltage source:



superposition.

Turn off 4A current source:


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

R

viRiv ssss or


Source Transformation • Equivalent Circuits:

R

v

R

vi

viRv

s

s

i i

+ +

- -

v v

v

i

vs -is


Source Transformation • Source transformation: Important points

1. Arrow of the current source is directed toward the positive terminal of the voltage source.

2. Transformation is not possible when:

♦ ideal voltage source (R = 0)

♦ ideal current source (R=)


Source Transformation • Example 4.6: Use source transformation to find vo in the

circuit below.



circuit below.

Fig.4.18



circuit below.

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

and

A4.0)2(82

2

i

V2.3)4.0(88 ivo


Source Transformation • Example 4.7: Find vx in the circuit below using source

transformation.



transformation.

KVL around the loop in Fig (b)

01853 xvi

ivvi xx 3013

(1)

Appling KVL to the loop containing only the 3V voltage source, the 1W resistor, and vx yields:

(2)



transformation.

Substituting Eq.(2) into Eq.(1), we obtain

Alternatively

thus

A5.403515 ii

A5.40184 iviv xx

V5.73 ivx


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 sources are turned off.


Thevenin’s Theorem • Property of Linear Circuits

i

v

v

i

Any two-terminal

Linear Circuits

+

-

Vth

Isc

Slope=1/RTh


Thevenin’s Theorem • Replacing a linear two-terminal circuit (a) by its Thevenin equivalent

circuit (b).


Thevenin’s Theorem

• How to find Thevenin Voltage? Equivalent circuit: same voltage-current relation at the

terminals.

:Th ocvV ba atltagecircuit voopen



• How to find Thevenin Resistance?

:inTh RR b.a atcircuitdeadtheofresistanceinput

circuitedopenba

sourcestindependenalloffTurn




There are two cases in finding Thevenin Resistance RTh.

CASE 1

• If the network has no dependent sources:

● Turn off all independent sources.

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




There are two cases in finding Thevenin Resistance RTh.

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

i

vR Th

o

o

i

vR Th




The Thevenin resistance, RTh , may be negative, indicating that the circuit has ability providing (supplying) power.


Thevenin’s Theorem • After the Thevenin Equivalent is obtained, the simplified

circuit can be used to calculate IL and VL easily.

L

LRR

VI

Th

Th

Th

Th

VRR

RIRV

L

LLLL

Simplified circuit

Voltage divider


Thevenin’s Theorem • Example 4.8: Find the Thevenin equivalent circuit of the circuit

shown below, to the left of the terminals a-b. Then find the current through RL = 6,16, and 36 W.




shortsourcevoltageV32:Th R

opensourcecurrentA2

W

4116

124112||4ThR

Find RTh:




Find VTh:

analysisMesh)1(

:ThV

A2,0)(12432 2211 iiii

A5.01 i

V30)0.25.0(12)(12 21Th iiV

AnalysisNodal ely,Alternativ)2(

12/24/)32( ThTh VV

V30Th V




Find VTh: transformsource ely,Alternativ)3(

V3024396

122

4

32

THTHTH

THTH

VVV

VV

Thevenin Equivalent circuit:




Calculate IL:

LL

LRRR

Vi

4

30

Th

Th

6LR A310/30 LI

16LR A5.120/30 LI

A75.040/30 LI36LR


Thevenin’s Theorem • Example 4.9: Find the Thevenin equivalent of the circuit below at

terminals a-b.



terminals a-b.

• (independent + dependent sources case)

0sourcetindependen

intactsourcedependent

,V1ovoo

o

ii

vR

1Th

Find RTh: Use Fig (a):



terminals a-b.

21 3ii

Find RTh: For Loop 1:

2121 or0)(22 iiviiv xx

2124But iiivx



terminals a-b.

Find RTh: For Loops 2 and 3:

0)(6)(24 32122 iiiii

012)(6 323 iii

gives equations theseSolving

.A6/13 i

A6

1But 3 iio

W 61

Th

oi

VR



terminals a-b.

Find VTh: analysisMesh :(b) Fig Use

18/602 i

V206 2Th ivV oc


Loop 1:

Loop 2:

Loop 3:

A51 i

20212 32 ii

06)(2)(4 23212 iiiii

4026 32 ii

0)(2))(4(2

0)(22

2321

23

iiii

iivx


Thevenin’s Theorem • Example 4.10: Determine the Thevenin equivalent circuit in Fig (a).

0Th V

o

o

i

vR Th

:anaysisNodal

4/2 oxxo viii

Find RTh:

(dependent source only case)

Apply a current source io at a-b

Use Fig (b):

Find VTh:



Find RTh:

22

0 oox

vvi

But

4424oooo

xo

vvvvii oo iv 4or

W 4Thus Th

o

o

i

vR power)(Supplying

Use Fig (b):



Input circuit:


• When RTh is negative, you must evaluate (check) if the two circuits are equivalent or not!



Input circuit:


Check with RL (9W) and voltage source (10V)




Input circuit:




)1(3,062

,0)(248

2121

12211

iiii

iiiiiii xx

A2105

0109)(2

22

212

ii

iii

A2105

01094

ii

ii

Load current in both circuits are equal.

So the Thevenin Equivalent circuit is OK.


Norton’s Theorem • Norton’s theorem states that a linear two-terminal

circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN .

• 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.


Norton’s Theorem

v

i

Vth

-IN

Slope=1/RN

• Property of Linear Circuits


Norton’s Theorem • How to find Norton Current

Thevenin and Norton resistances are equal:

ThRRN

Short circuit current from a to b gives the Norton current:

Th

Th

R

ViI scN


Norton’s Theorem • 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. V

N

N

ThTh

scN

ocTh

RI

VR

iI

vV


Norton’s Theorem • Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.


Norton’s Theorem

Find RN: Use Fig (a):

W

425

52020||5

)848(||5NR

• Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.


Norton’s Theorem

Find IN: Use Fig (b):

:Mesh

Nsc Iii A12

(short circuit terminals a and b)

012420,2 121 iiAi

(ignore 5W. Because it is short circuit)



Norton’s Theorem

Find IN: Use Fig (c): Alternative Method Th

ThN

R

VI

:ThV

:analysisMesh

012425,2 343 iiAi

A8.04 i

V45 4 iVv Thoc

(open circuit voltage accross terminals a and b)

,Hence

A14/4 Th

ThN

R

VI



Norton’s Theorem

Input circuit:

Norton’s Equivalent circuit:



Norton’s Theorem • Example 4.12: Using Norton’s theorem, find RN and IN of the circuit in Fig

4.43 at terminals a-b.


Norton’s Theorem

shortedresistor4WParallel:2||||5 xo ivW

Hence,

W 52.0

1

o

oN

i

vR

Find RN: Use Fig (a):

A2.05

V1

5,A0 0

0 W

W

v

iix

• Example 4.12: Using Norton’s theorem, find RN and IN of the circuit in Fig 4.43 at terminals a-b.


Norton’s Theorem

Find IN:

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

.5A,24

010

xi

A72(2.5)5

102 xxsc iii

7A NI



Norton’s Theorem

Input circuit:

Norton’s Equivalent circuit:



Maximum Power Transfer • The Thevenin equivalent is useful in finding the

maximum power a linear circuit can deliver to a load.

LL RRR

VRip

2

LTH

TH2


Maximum Power Transfer • Maximum power is transferred to the load when the

load resistance equals the Thevenin resistance as seen from the load (RL = RTH).

LL RRR

VRip

2

LTH

TH2


Maximum Power Transfer • Pmax can be obtained when: 0

LdR

dp

THL

LTHLLTH

LTH

LLTHTH

LTH

LTHLLTHTH

LTH

LTHLLTHTH

L

RR

RRRRR

RR

RRRV

RR

RRRRRV

RR

RRRRRV

dR

dp

0)(0)2(

0)(

)2(

)(

)(2)(

0))((

)(2)(

3

2

4

22

22

22

TH

TH

R

Vp

4

2

max


Maximum Power Transfer • Example 4.13: Find the value of RL for maximum power transfer in the

circuit of Fig. 4.50. Find the maximum power.




W

918

126512632THR

Find RTh:




VVVii

AiAiii

THTH 220)0(23612

3

2 ,2 ,121812

21

2221

Find VTh:

W 9THL RR W44.1394

22

4

22

max

L

TH

R

Vp

KVL around the outer loop:

DC Circuits: Circuit Theorems · EENG223: CIRCUIT THEORY I A large complex circuits Simplify...

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Transcript of DC Circuits: Circuit Theorems · EENG223: CIRCUIT THEORY I A large complex circuits Simplify...