ac network theorems

8/13/2019 AC Network Theorems

1/22

AC Network

TheoremsPrepared by:

Karthik Chandran PillaiIV Sem, EEE (B)


2/22

Superposition Theorem

The voltage across (or current through) an

element is determined by summing the

voltage (or current) due to each independentsource.

All sources other than the one being

considered are eliminated.

Replace current sources with opens.

Replace voltage sources with shorts.


3/22


A circuit may operate at more than onefrequency at a time.

Diode and transistor circuits will have both dc

and ac sources. Superposition can still be applied.


4/22


The superposition theorem can be appliedonly to voltage and current.

It cannot be used to solve for the total power

dissipated by an element. This is because power is not a linear quantity,

but instead follows a square-law relationship.


5/22

ET 242 Circuit Analysis II Network Theorems for AC Circuits Boylestad 4

Independent Sources

Ex. 18-1 Using the superposition theorem, find the current Ithrough the 4

resistance (XL2) in Fig. 18.1.

Figure 18.1 Example 18.1. Figure 18.2 Assigning the subscripted impedances to the network in Fig.18.1.

j3jXZ

j4jXZ

j4jXZ

,(Fig.18.2)circuitredrawntheFor

C3

L2

L1

2

1

9025.1908

010

412

010

412

010

90121212

34

)3)(4(

13//2

1

32

323//2

1

AV

jj

V

jj

V

ZZ

EI

jj

jj

jj

ZZ

ZZZ

havewe,(Fig.18.3)Esource

voltagetheofeffectsthegConsiderin

s

1


6/22

Figure 18.3 Determining the effect

of the voltage source E1on the

current Iof the network in Fig. 18.1.

9075.31

75.3)25.1)(3(

)(

3

1

Aj

A

j3-j4

Ajj

ruledividercurrentZZ

IZI'and

2

s3

Figure 18.4 Determining the

effect of the voltage source E2

on the current Iof the network in

Fig. 18.1.



7/22

Superposition for Dependent

Sources

If the controlling element is external to the

circuit under consideration, the method is the

same as for independent sources. Simply remove the sources one at a time and

solve for the desired voltage or current.

Combine the results.


8/22

Superposition for Dependent

Sources

If the dependent source is controlled by an

element located in the circuit, the analysis is

different. The dependent source cannot be eliminated.

The circuit must be analyzed by considering

all effects simultaneously.


9/22

Dependent Sources For dependent sources in which the contr oll ing variableis not determined by the network to which the superposition is to be appli ed, the

application of the theorem is basically the same as for independent sources.

Ex. 18-5 Using the superposition, determine the current I2for the network inFig.18.18. The quantities and hare constants.

Figure 18.18 Example 18.5. Figure 18.19 Assigning the subscripted impedances to the network in Fig.18.18.

66.38/078.066.388.12810864

),20.18.(

864

),19.18.(

21

2211

VV

j

V

j

V

ZZ

VI

FigsourcevoltagetheFor

jjXRZRZ

FigsystemtheofportionaWith

L



10/22

66.38312.066.38)078.0(4

66.388.12

))(4()(

),21.18.(

21

1 hIhIhI

ZZ

hIZI

FigsourcecurrenttheFor

66.3822.1666.3862.066.3860.15

66.38)020)(100(312.066.38/)010)(20(078.0

,100,20,010

66.38312.066.38/078.0

2

2

2

AAA

mAVI

handVVFor

hIVIIIisIcurrenttheFor

Figure 18.20 Determining the effect of the voltage-controlled

voltage source on the current I2for the network in Fig.18.18.

Figure 18.21 Determining the effect of the current-controlled

current source on the current I2for the network in Fig.18.18.



11/22

Thvenins Theorem Thvenins theorem converts an ac circuit into

a single ac voltage source in series with an

equivalent impedance.

First, remove the element or elements across

which the equivalent circuit is to be found.

Label the two terminals.

Set all sources to zero - replace voltagesources with shorts, current sources with

opens.


12/22

Thvenins Theorem

Calculate the Thvenin equivalent impedance.

Replace the sources and determine the open-

circuit voltage.

If more than one source is involved, use

superposition.

Draw the resulting Thvenin equivalent circuit,including the portion removed.


13/22ET 242 Circuit Analysis II Sinusoidal Alternating Waveforms Boylestad 12

Ex. 18-7 Find the Thevenin equivalent circuit for the network external to resistor R

in Fig. 18.24.

Figure 18.24

Example 18.7.

18033.3620

28)10)(2(

)(

:)27.18.(4

9067.2906

16616

28

)2)(8(

:)26.18.(3

28:)25.18.(21

21

2

2

21

21

21

Vj

Vjjj

Vj

ruledividervoltageZZ

EZE

FigStep

jj

jj

jj

ZZ

ZZZ

FigStep

jjXZjjXZFigandSteps

Th

Th

LL

Figure 18.26

Determine the Thevenin

impedance for the

network in Fig.18.24.

Figure 18.27 Determine

the open-circuit Thevenin

voltage for the network in

Fig.18.24.

Figure 18.25 Assigning the

subscripted impedances to

the network in Fig.18.24.


14/22ET 242 Circuit Analysis II Parallel ac circuits analysis Boylestad 13

Step 5: The Thevenin equivalent circuit is shown in Fig. 18.28.

Figure 18.28 The Thevenin

equivalent circuit for the

network in Fig.18.24.

Ex. 18-8 Find the Thevenin equivalent circuit for the network external to resistor

to branch a-a

in Fig. 18.24.

5

43

86:

:)30.18.(21

2

1

3

22

11

jjXZ

jjXRZ

jjXRZimpedancesdsubscriptewithcomplexity

reducedtheNoteFigandSteps

L

C

L

Figure 18.29

Example 18.8.

Figure 18.30 Assigning the

subscripted impedances forthe network in Fig.18.29.

:)3118(3 FigStep


15/22

09.7708.596.2385.9

13.5350

96.2385.9

)010)(13.535()(

:.0

,:)32.18.(4

12

2

23

VVV


EZE

ZacrossdropvoltagetheisEThenI

circuitopenanisaaSinceFigStep

Th

ThZ

36.3249.594.264.4

06.264.4596.2308.55

96.2385.9

0505

49

0505

)43()86(

)13.535)(13.5310(5

:)31.18.(3

21

213

j

jjj

jj

j

jj

j

ZZ

ZZZZ

FigStep

Th

ET 242 Circuit Analysis II Selected Network Theorems for AC Circuits Boylestad 14

Figure 18.26Determine the Thevenin

impedance for the network in Fig.18.29.

Figure 18.27 Determine the open-

circuit Thevenin voltage for thenetwork in Fig.18.24.

Step 5: The Thevenin equivalent circuit is shown in Fig. 18.33.

Figure 18.33 The Thevenin equivalent circuit for the network in Fig.18.29.


16/22


17/22

Nortons Theorem

Determine the Norton equivalent impedance.

Replace the sources and calculate the short-

circuit current.

Superposition may used for multiple sources. Draw the resulting Norton circuit with

elements which were removed replaced.


18/22

Maximum Power Transfer

Theorem

Maximum power will be delivered to a load

when the load impedance is the complexconjugate of the Thvenin or Norton

impedance.

ZTh= 3+ j4 ZL= 3- j4

ZTh= 10 30 ZL= 10 -30


19/22

Maximum Power Transfer

Theorem

If the Zis replaced by its complex conjugate,

the maximum power will be

N

NNmax

Th

Thmax

Th

Th

R

ZIP

R

EP

RR

REP

L

LL

4

4

22

2

2

2


20/22


Ex. 18-19 Find the load impedance in Fig. 18.83 for maximum power to the load,

and find the maximum power.

866.1087.3633.1306

87.3680

886

)908)(13.5310(

8

13.531086

21

21

2

1

j

jjZZ

ZZZ

jjXZ

jjXRZ

:(a)][Fig.18.84ZDetermine

Th

L

C

Th

W

V

R

EPThen

VVjj

V


EZE

f indmustwepowerimumthef indTo

jZand

Th

Th

L

38.364.42

144

)66.10(4

)12(

4

9012069072

868)09)(908(

)(

,max

866.1087.3633.13

22

max

12

2

Figure 18.84 Determining (a) ZThand (b) EThfor the

network external to the load in Fig. 18.83.

Figure 18.83 Example 18.19.


21/22


Ex. 18-19 Find the load impedance in Fig. 18.83 for maximum power to the load,

and find the maximum power.

866.1087.3633.1306

87.3680

886

)908)(13.5310(

8

13.531086

21

21

2

1

j

jjZZ

ZZZ

jjXZ

jjXRZ

:(a)][Fig.18.84ZDetermine

Th

L

C

Th

W

V

R

EPThen

VVjj

V


EZE

f indmustwepowerimumthef indTo

jZand

Th

Th

L

38.364.42

144

)66.10(4

)12(

4

9012069072

868)09)(908(

)(

,max

866.1087.3633.13

22

max

12

2

Figure 18.84 Determining (a) ZThand (b) EThfor the

network external to the load in Fig. 18.83.

Figure 18.83 Example 18.19.


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Relative Maximum Power

If it is not possible to adjust the reactance partof a load, then a relative maximum power will

be delivered.

The load resistance has a value determinedby

22ThTh

XXRRL

ac network theorems

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