networks theorypart 2

7/30/2019 NetWorks TheoryPart 2

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II. THE NETWORK THEORMS :-

The need of Theorms:-

In a complicated Network with several sources, Nodes and Meshs, If the response in asingle element is desired then the Network Theorms are used.

1. The superposition Theorm (SPT):-

Def:- In a linear Network With several independent sources, the response in a Particular

branch When all the sources ar acting simultaneously is equal to the linear sum of

Individual responses calculated by taking one independent source at a time.

All the ideal voltage sources are eliminated from the Network by shorting the sources,

All the ideal current sources are eliminated from the Network by opening the sources

and donot disturb the dependent sources present in the Network.

Properties:-

1. This therm is applicable to only for the linear Networks i.e. the Networks with

R,L,C and Transrformer, linear controlled soureces as elcments.

2. The Presence of the dependent sources makes the Network an active and hence the

SPT is used for both active as well as for passive Networks.

The Homogenity Principle:-

This is the principle obcyed by all the lincar Networks

Def> In a Linear Network If the excitation is multiplied with a constant k then the

responses in all the other branches of the Network are also Multiplied with the same

constant k.

4.THE MAXIMUM POWER TRANSFER THEOREM (MPTT):-

Properties: same as SPT

This theorm is applicable to only When the load is a Variable, otherwise choosethe minimum internal impedance of the souce, wich results a Maximum current

through the fixed load and hence a Maximum Power dissipation across the load.

Under the Variable Load Conditions:-

(1) Rs and Rl :-

I=ls

s

RR

V

+P=I2-RL W

P= )W()RR(

RV2

Ls

L

2

s

+

18

+_VS I

RS

RL


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[ ]22Ls

LsL

2

s

2

s

L )RR(

)]RR[2.R.1.)]RR[(V

dR

dpL

+

++

=

For MPT 0dR

dp

L

=

RL=RsPmax = sL RRp =

= WR4

V

L

2

s

= WR4

V

s

2

s

Pdel = pabs

=I2.Rs+I2RL

=W

R2

VW

R2

V

s

2

s

s

L

s

=

%=50% so the efficiency of the Maximum Power Transfer Theorem is Utmost

50%.

2. Zs and ZL:-

Zs =[Rs+jXs]

I= )XX(j`RR

V

lsLs

s

+++

2

s

2

Ls

s

)XLX()RR(

VI

+++=

`P= )W(R.I L2

P =2

Ls

2

Ls

L

2

s

)XX()RR(

RV

+++

2(i) Only RL is a variable:

22

Ls

2

Ls

LsL

2

Ls

2

Ls

2

s

L ])XX()RR[(

)]RR(2.R1.)XX()RR[(V

dR

dp

+++

++++

=

19

+_VS

I

ZL=(R

L+J

XL)

ZS=(R

S+J

XS)


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For MPT

0dR

dp

L

=

])XX()RR[(V 2Ls2

Ls

2

s +++ = 2RL(Rs+RL)

++= 2Ls2

sL )XX(RR

Pmax =2

Ls2sL )XX(RRp ++=

2(ii) Only XL is valuable:-

22

Ls

2

Ls

LsL

2

Ls

2

ls

2

s

L ])XX()RR[(

)XX(2.R0].)xX()RR[(V

dX

dP

+++

++++=

For MPT 0dx

dp

L

=

0xx Ls =+

Pmax= 0xxp

Ls =+

2(iii). Both RL& XL are varced simultancovsly:-

In this case both the conditions

RL=22 )( Lss xxR ++ & Xs +XL=0 are valid at a time

sL RR =

XL= -XsSo ZL =RL+jxL=Rs-jxs

= *sL ZZ

Pmax =sL RR

p=

XL=-Xs

)(4

)(4

22

WR

VW

R

V

s

s

L

s ==

% %50


4/21

Applied to Linear, Passive and bilateral Network i.e. R,L,CS& Transformer.

Def: In a linear, Passive and Bilateral Network,the rartio of response to the excitation is

constant even though th source is interchanged from the input terminals to the output

terminals.

i.e ttanconsV

I=

2

2

1

1

V

I

V

I=

Proof:

Reg= 4+3||4

= 4+7

43

=7

40

I = AR

V

eg 2

7

740

20

==

AI 2

34

4.2

7

1 =

+

=

Properties:-

This theorm is applicable to only for the linear, passive and bilateral Networks, i.e. the

Networks with R,L,C and Transformer as the dependent sources makes the Network an

active and hence the dependent sources makes the Network an active and thence the

reciprocity theorm is not applicable, so called the Non-Reciprocal Network.

LAPLACE TRANSFORMS

I(t) I(s)

21

V N I

Fig (a)

VNI

Fig (b)

20V

4

2A

Fig (c)

34A

4 I1

4

20V

Fig (d)

3 I

4

Req


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V(t)V(s)R RL SLC

SC

1

8(t) 1U(t)

2

1

s

R(t) 3

1

s

P(t) as+

1

astue at

+

1)(

k.u(t) s

k

= )s(Z)s(I

)s(V

V(s)= Z(s).I(s)

)s(Z

)s(V)s(I =

)s(V).s(Y)s(I =

)s(Y

)s(I)s(I =

)s(Z

1)s(Y =

Z(s) Y(s)

R

SL

SC

1

R

1

SL

1

SC

The impedances in series and the admittances in parallel we can add.

22


6/21


7/21

The inductor acts as a constant current source in the steady state.

Since VL= oVdt

dilL L . cSL .

Similarly2

2

1cCV = max & constant.

vc= max &ConstantSo the capacitor will act as a constant voltage source in the steady state.

Since ic=C. cocidt

dvc

c .0

2. The inductor current and the capacitor voltages at t = 0 - and at t = 0+instants:

L: dt).t(VL1)t(i

t

LL =

dt).t(VL

1dt).t(V

L

1t

0

L

0

L

+=

At t=0+ += + )0(i)0(i

LL

dt).t(VL

10

0L

+

For all the

Practical excitation

)0(i)0(i LL+ =

So the inductor current cannot change instantancous i.e With Zero time for all the practical

inputs stmilarly the energy.

If VL(t) = 8(t) them

IL(0+

) = iL(0-

)+dtt

L

t

.)(8

1

0)(

1)0()0( A

Lii LL +=

+

))(0()0( JEE LL+ >

So the inductor current can change instantaneously i.e With in Zero time,

impulse voltage across it similarly the energy.

More Specifically for impulse:-

24

dt).t(VL

1)0(i

t

0

LL

+=

)0(i)0(i LL+ =

-

0- 0+

0

t


8/21

0tfor0)t(8)t(VL


9/21

Vc = dtic

t

c.

1

oc

t

ccc VVdtiC

VV =+= )0(.1

)0(0

By KVL 0.1

0

= t

coc dtic

VV

Transients:-

dc ac

d.c: RL 0IL1.Source free circuits

Without independent source. RC 0VC

RLC 0IL OR C 0VOr both

Property:- In all the source free circuits the energies are maximum at t= 0 and as a function

of time This stored energies will be delivered to the resistances in an exponentially decayed

manner, Hence the energy Present in the Network in steady state is Zero.

2.With sources the initial (t = 0+) and final (t = ) conditions :-c.oL0t == +

c.sc =

c.oc =At these two instants, the L&C elements will loses their significance and hence the nature of

the circuit is the resistive.

3.With sources the laplace transforms approch of solving the transient problems for t0 source free circuits:-

1.Source free RL circuit

By KVL

0VV LR =

26

+_ I0V

L

i

L

0V

VR

R

+

_

c.sLt ==


10/21


11/21

dt

)t(di.L)t(VL = By ohms law

)J()t(i.l.2

1)t(E 2L =

2T

t

0 )eI(i.l.2

1i

=

)J(Ecoe ETt

= For t0

5

05

0e

IeI

T5t)t(i

===

= 0.0067 I0~ 0.0

R2

L

2

TT iE == sec

So the energy decay is 2 times faster than the current decay.

2. The Source free RC circuit:-

28

i(t)

I0

0.368 I0=

0.0067 I0

t = T t = ST t

Tr. periodSS. period

EL(t)

E

0 t

+

_V

0

V iC

C

0V

(V > 0)

R

iR


12/21

By KCL 0Ii CR =+

0dtdv.C

RV =+

0R

V.

c

1

R

V=+

0V)RC

1D( =+

VKe)t(V Rct

= for t0

For t 0

At t = 0+ ; A resistive circuit

0V)0(V =+

So t = 0+

0Ke)0(V

+ =

VeVK RCt

0

= for t0

VeVK RCt

0

= for t0

T = RC = Sec = The time constant of the circuit

dt

)t(dv.Cic = By ohms law

29

V

+

_ C

+_ V0

0V

R

V(0+)

+

_

V0

0V

R+_

v(t)

V0

0.368 V0=

0.0067 V0

t = T t = ST t

Tr. period SS. period

0


13/21

)t(V.C.2

1)t(E 2

c=

=

ETt

0ev.C.2

1(J) for t0

2

RC

2

T

Tv

E == secSo the energy decay is 2 times faster than the Voltage dacay.

secsec =

vi TT =

RCR

L =

=2CR

L! = Unit less

=L

CR2! = Unit less

HenryLCR2 ==

== CR

L2

Farad

== RRC

LOhm

== 2RC

LOhm2

== RC

LOhm

Case(iii) LTA

30

EC(t)

E

0 t

ECO

R

IR(s)

VR

iR

-

+

S-D

VR(s)

R

+

-

t-D


14/21


15/21

(Q) Det V0

Note:-

(i) When the Network consists of Several sources, Multiple resistors & inductors

(Seperable) or a single inductor then iL(t)=iL( Tt

e))

A for t0

secR

LT = )t(

dt

di.L)t(V LL =

secR

L

eq

= By ohms law

secR

L

eq

eq=

Obs:- For source free circuits iL( ) = OA

Ae).0(i)t(i Tt

LL

= for t0

(ii) In the above case if inductors are replaced by capacitors then

T =RC sec

=Req C sec

=Req Ceq sec

dt

)t(dv.c)t(i cc = By ohms law

Obs: for source free ckts Vc( ) =OV

Ve)0(V)t(VTt

cc

= for t0

32

S-D

S

2+

-

Ic(s)

Vc

+

-

Ic

+

-

F2

18

CV0

= 8 V0

= 8

V0

= 16V16


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17/21

V(t) VR RLjwlC jwl

i

Note:- The analysis of a.c in the steady state is generally carried out in the Phasor domain i.e.the KCL, KVL, Ohms law, Nodal, Mesh & the source transformation are written in the

phasor domain only.

A.C.TRANSIENTS

1. The series RL circuit

C.S = C.F + P.I

= Natural response + Forced response.

= Response up to 5T + response after S.T

= Transient response + Steady state response

i(t) = Itr(t) + iss (t)

i(t) = )t(isske Tt

+

)t(isske LtR

+

iss(t):

Method 1:- Phasor approach

Transform the above Net work into the phasor domain .

34

i(t)

L

R

t =0+

V(t)= Vmsin(wt+)

_

Li(t)

R

+

_

V(t)= Vmsin(wt+)

NW is in S.S

_

jwl

I

R

+

_

Vm

-90o

NW is in P.D

_

++


18/21

V= Z.I

By KVL in P-D

0I.jwlT.R90V om =

+

=

+

=

R

WLtan)WL(R

90V

jwlR

90VI

122

o

m

o

m

o1

22

m 90R

WLtan

)WL(R

VI

+

=

=

= jejwte.IP.R)t(i = A

= )Wtcos( +

= )90R

WLtanWtcos( o1+

= iss(t)

Method(2):- LTA

V(t) = Vmsin (wt + )

SLR

1

)S(Z

1)s(Y)s(H

)s(V

)s(I

+====

H(jw)=JwlR

1

+

R

WLtan

)WL(R

1 122

+

=

)s(V).s(H)s(I =

R

WLtanwtsin(V.

)WL(R

1)t(i 1m

22

++=

= iss(t) i(t) = itr(t) + iss(t)

)R

WLtanwtsin(

)WL(R

Vtke 1

22

mLR +

++=

I(0-) = OA = i(0+)

)R

WLtansin(

)WL(R

VKe0

1

22

m0 +

+=

+

=

R

WLtansin

)WL(R

Vk 1

22

m


19/21

suppose 0R

WLtan 1 = 0= k

A transient free response.

So the condition for the transient free response at t=0 isR

WLtan 1=

i.e.

=

=+

R

WLtan0t

)Wt( 1

Note:- So, if the total phase of the excitation at the time of switching is equal to

tan-1

R

WL

Than no transients will result in the system at the time of switching for the sinusoidal

excitation.

If the switch is closed at t =t0 =tan-1

R

WL

=+

R

WLtanwt 10

If the excitation is V(t) = Vmcos (wt+) then sin is replaced by cosine in the steady

state response and hence K is a function of cosine. suppose

2tan 1

=

R

WL

0= K 0)( = titr)()( titi ss=

A transient free response

2R

WLtan

1 +

= at t=0

2R

WLtanWt

1

0

+

=+ at t=t0

In thae above cases if L is replaced by C then =R

Lis repaced by =

RC

2.The Parallel RL Circuit :-

ILss(t) by LTA:

36

L

iL

t= 0

R

SI(t) = I

msinWt+


20/21

)(

)(

sI

sIL =H(s) = ?

SLR

RsIsI

L +=

).()(

SLR

RsH

sI

sIL

+== )()()(

H(jw)=

+

=+

R

WLtan

R

WL1

1

R

jwL1

1 12

)()( titi LssL +

+

+

+=

R

WLtanWtsin

R

WL1

IKe 1

2

mt

L

R


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38

networks theorypart 2

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