电路原理课后答案18章

Problem18.1TogetzMandz2l,considerthecircuitinFig.(a).12/13/1999

z”=1+611(4+2)=4。

Z2I

Togetz22andz12,considerthecircuitinFig.(b).

z22=-=2II(4+6)=1.667Q

21

I。V,6IO=I2

Z]2=/=lC

Hence,

41

[z]=

11.667

Problem18.2ConsiderthecircuitinFig.(a)togetzHandz2I.

ICI。，1C1Q1Q[2=()

1

1Q1Qinv2

IQIC1QIQ

(a)

V,

1

zH=-=2+lll[2+lll(2+l)]

L

(3、(1)(11/4)11

z„-2+1112+1-2+；-2+,-2.733

k471+11/415

11.

I°\+3l°-J

14

I=---------T=—T

°-1+11/41151

141

【。=屋百

1

匕=1。=由

V21…

z21=——=-=z1?=0.06667

I115

Togetz22,considerthecircuitinFig.(b).

h=01CIQIC1。

++

inv2c

V1IQIQ

IQIQ1Q1Q

(b)

z22==2+1II(2+1II3)=zH=2.733

Thus,

-2.7330.06667"

[z]=C

L」0.066672.733

Problem18.3

(a)TofindzHandz21,considerthecircuitinFig.(a).

z“=¥=jii(i-岑Ri+j

LJ+i-J

Bycurrentdivision,

I”启…

V2=I0=jI,

V2.

%=丁=J

Togetz22andzl2,considerthecircuitinFig.(b).

v,=ji2

V..

Z”=M=J

l2

Thus,

1+jj

[z]=

io

(b)

Z”=「•=j+l+lll(-j)=l+j+/；=1.5+jo.5

LJ

V,=(1.5-j().5)I,

z2I=-=1.5—jO.5

Togetz22andzp,considerthecircuitinFig.(d).

z22=^-=-j+l+HI(-j)=1.5-jl.5

V,=(1.5-j0.5)I2

V.

z=——=1.5—j().5

p12

Thus,

1.5+j0.51.5-j0.5

[z]=

_1.5-j0.51.5-jl.5.

Problem18.4TransformthennetworktoaTnetwork.

ZiZ3

(12)(jl0)j120

12+jlO-j512+j5

-j60

Z

212+j5

50

Z3

12+j5

Thezparametersare

_(-j60)(12-j5)

z=z

i22i=Z2=144+25=-L775-j4.26

-z「L775+j4.26

z1i=Z]+Ng

(5())j5)

z22=Z3+z2,=^~+z2l=1.7758-j5.739

Thus,

1.775+j4.26-1.775-j4.26

[z]=

--1.775-j4.261.775-j5.739_

Problem18.5ConsiderthecircuitinFig.(a).

S2+S+1

11=s3+2s2+3s+l

11s

111-

_________ss+1Ts+1

11

1II-+1+S+-+1+S+-------+s2+S+1

sss+1ss+1

s3+2s2+3s+l1

s?+2s-+3s+1

zJ=1

21Ijs3+2s2+3s+l

ConsiderthecircuitinFig.(b).

(l[l+s+T)1+s+L

Is人s+17s+1

Z22nr"='s"

-+1+S+----71+S+S-+----;

SS+lS+l

s~+2s+2

22=s3+2s2+3s+l

Hence,

S24-S+l

s3+2s2+3s+ls3+2s2+3s+l

s2+2s+2

-s3+2s2+3s+ls3+2s2+3s+l-

Problem18.6TofindzHandz21,connectavoltagesourceV〕totheinputand

leavetheoutputopenasinFig.(a).

3()3

where*=20+30,=-V

50

V,-Vo3.2

I=FF=032V。

4.2V。

=13.125。

0.32Vo

v2_0.6Vo

Z21=1.875C

I."0.32Vo

Toobtainz22andz12,usethecircuitinFig.(b).

I2=().5V2+^-=0.5333V2

V2_1

Z22=1.875C

I2-0.5333

V,=V2-(20)(0.5V2)=-9V2

_Y_-9V2

Z12=-16.875。

~I2~0.5333V2

Thus,

13.125-16.875

[z]=

1.8751.875

Problem18.7TogetzNandz2I,refertothecircuitinFig.(a).

Atnode1,

KT

⑴

L=T+4

Atnode2,

V-V

VxV2V2+2VX

c-----Vx__V，⑵

42

Substituting(2)into(1),

2V,

v--=-1.5V,

42

JL_s3

V2-2

-2

Z2I-j-=y=-0.6667Q

25

Y=VX+L=-V2+I、L+L=#

V,5

zH=-^=-=1.66670

1]J

Togetz22andz12,refertothecircuitinFig.(b).

Atnode2,

V2+2VV2-Vx

⑶

Atnode1,

V-VV

Vor(4)

=广一＞2=5VXVX=0.2V2

Substituting(4)into(3),

IV2+0.4V2+匕-0.2匕03v

V,1

z,,=j=—=1.1111。

22

I20.9

111()2

Y="三§i2=/

Y2c-

z1p2=—=77=0-2222Q

I29

Thus,

-1.66670.2222-

[z]=C

i」-0.66671.1111

Problem18.8ItisevidentfromFig.18.5thataTnetworkisappropriatefor

realizingthezparameters.

R,=zH—z12=10—4=6Q

R2=z22—z12=6—4=2c

R3=zI2=z21=4C

Problem18.9

(a)Thisisanon-reciprocalcircuitsothatthetwo-portlooksliketheone

showninFigs,(a)and(b).

1125Q10Q

I2

1_____

++

Vi20I5IiV

2>2

(b)

(b)Thisisareciprocalnetworkandthetwo-portlookliketheoneshownin

Figs,(c)and(d).

Kll-Z12Z22-Z12

IlI2

-1

++

Z12

Viv2

(c)

21

z”-Ng=1+—=14

S0.5s

Z22-Z12=2s

1

Z|2

h1610.5F2Hh

++

IF

Viv2

(d)

Problem18.10Thisisareciprocaltwo-portsothatitcanberepresentedbythe

circuitinFigs,(a)and(b).

FromFig.(b),

V1=(8+4114)1,=101.

Bycurrentdivision,

Problem18.11Thisisareciprocaltwo-portsothat12/13/1999

thecircuitcanberepresentedbythecircuitbelow.

Weapplymeshanalysis.

Formesh1,

-120+901,+10I2=0——>12=91,+I2(1)

Formesh2,

3011+12012=0——>I,=-4I2(2)

Substituting(2)into(1),

12=-361.+1.=-351.——>

222235

ll|2if12?

P=5lUR=J(100)=5.877W

Problem18.12TofindZTh,considerthecircuitinFig.(a).

Y=ZuL+ZI2h⑴

V2=z2|I|+z2212⑵

But

v2=1,V.=-ZSI,

-Z19

Hence,O=(Z]1+Zs)L+z1212>1.=——

Zll+4ZIs-I2

Z2IZ%

1+Z22

Z11+Zs2

rjV21Z21Z12

TofindVTh,considerthecircuitinFig.(b).

Substitutingtheseinto(1)and(2),

Problem18.13Asareciprocaltwo-port,thegivencircuitcanberepresentedas

showninFig.(a).

5Q10-j6Q4-j6c

15Z0°Vj4C

Atterminalsa-b,

ZR=(4-j6)+j6ll(5+10-j6)

Jj6(L15-j6)

Zn=4-j6+15^4-j6+2.4+j6

ZR=6.4C

i6

V(15ZO>j6=

^j6+5ilO-j6°^I

TheTheveninequivalentcircuitisshowninFig.(b).

Fromthis,

K=j&4(j6)=3.18/148。

vo(t)=3.18cos(2t+148°)V

Problem18.14ToobtainzHandz2I,considerthecircuitinFig.(a).

(a)

Inthiscase,the4-Qand8-Qresistorsareinseries,sincethesamecurrent,Io,passes

throughthem.Similarly,the2-Qand6-Qresistorsareinseries,sincethesamecurrent,

Io,passesthroughthem.

V.(12)⑻

zn=-^=(4+8)||(2+6)=12118=y-^=4.80

JL14U

82,3

=51=-I

°=8+12°5

But-V2-4IO+2IO=0

-86,-2,

V=-4I+2I+5I[=T11

2OO51

V,-2

z=~=—=-0.4Q

21L5

(b)

Togetz92andz12,considerthecircuitinFig.(b).

V,,(6)(14)

z22—=(4+2)II(8+6)=6II14=———=4.2Q

122°

z12=z?1=-0.4Q

Thus,

4.8・0.4

[zl=

・0.44.2

WemaytakeadvantageofTable18.1toget[y]from[z].

△z=(4.8)(4.2)—(0.4)2=20

z224.2-zI20.4八

y”=0.21y,,=--=—=0.02

△z20力2Az20

-Z2]0.4z4.8

­=0.02y=—N=——=0.24

丫21=’22A20

Az20z

Thus,

0.210.02

[y]=

”」0.020.24

Problem18.15Togetynandy2I,considerthecircuitinFig.(a).

V、=(6+6113)L=8L

T-6-2V,-y

I-i

V->i=—2=—

J2iV,12

Togety22andy12,considerthecircuitinFig.(b).

I2111

312

V2311(3+6116)31162

Io31

।2'°3+6123

(b)

2s

V=(sll2)I=

22S4-2

12s+21

0.5+-

丫22=工=工s

-s-ss+2-V2

1广打2=帝卞丫2=丁

Yi2=9=。5

Thus,

Fs+0.5-0.5

[yl=L-0.50.5+l/s」S

Problem18.17TogetyHandy2I,refertoFig.(a).

L

L=g+O.2Y=o.4Y—>y11=^=o.4

I2=-0.2V,―>丫2]=*一0.2

Togety22andy12,refertothecircuitinFig.(b).

SinceVj=0,thedependentcurrentsourcecanbereplacedwithanopencircuit.

2i

v=1()I“y22=—=0.1

221()

y小田=o

Thus,

0.40

S

-0.20.1

Consequently,th>yparameterequivalentcircuitisshowninFiq.(c).

Problem18.18

(a)TogetyHandy21refertothecircuitinFig.(a).

Atnode1,

i"Y-v,»I1=1.5Y—0.5V。(1)

Atnode2,

]工X」>1.2V,=V⑵

F23O

Substituting(2)into(1)gives,

L=1.5M-0.6V1=().9Y——>%1=5=().9

Vi

12=母=-().4丫|---->丫21=装=。4

Togety22andyl2refertothecircuitinFig.(b).

Vx=V,=0sothatthedependentcurrentsourcecanbereplacedbyan

opencircuit.

I1

=2==

V2=(3+2+0)I2=5I2------>y225

1!==-0.2V2------>yl2=^-=-0.2

Thus,

-0.9-0.2'

(b)Togetynandy21refertoFig.(c).

z=jll(l+lll-j)=jll1+4=

injll(1.5-j0.5)

v1-J/

j(1.5-j0.5)

=0.6+j0.8

1.5+j0.5

I.11

V.=Zin->y..=—=----=-------------=0.6—j0.8

V,Zin0.6+j0.8J

j.1.5-j0.5

I=-----1i=------i

01.5+j0.515°1.5+j0.5

..-j________I,_______

I。=k1。=(7)(1.5+也5)=西

(1.5-J0.5)22+i

12=I0+1°~L51+

-I2=(1.2-j().4)(0.6-j0.8)V,=(0.4-jl.2)V1

丫21=茎=-04+只-2=丫|2

vi

Togety22refertothecircuitinFig.(d).

ZoUt=jH(l+lll-j)=0.6+j0.8

1

y22=7-=0.6-j0.8

■oul

Thus,

「0.6-j0・8・0・4+jl.2

[y]=L-0.4+jl.20.6-j0.8

Problem18.19Sincethisisareciprocalnetwork,aFInetworkisappropriate,as

shownbelow.

(a)

1/4S

1/4S1/8S

(b)

S,Z1=4Q

S，

4Z2

311

S,

Y3=y22+y21=r48

Problem18.20Togetynandy21,considerthecircuitinFig.(a).

Atnode1,

V

-^+2V2V.=-V⑴

2X14s

V,-VV,+2V.I,

xL

But==1-5V1—>y„=^=i.5

Also,I2+^-=2VX——>I2=1.75VX=-3.5V,

y2i=£=-3.5

Togety22andy12,considerthecircuitinFig.(b).

Atnode2,

V-V

Atnode1,

Substituting(3)into(2)gives

I2=2VX-^VX=1.5VX=-1.5V2

丫22=^=-L5

-VxV2I,

>y『=o.5

Thus,

Problem18.21ConsiderthecircuitinFig.(a).

025

vi=4ii—>y”=*=今=

I2=20I,=5V,—>y2T=5

ConsiderthecircuitinFig.(b).

I.0.1

4I,=0.1V2>y,2=v-=—=0.025

V,I,

I2=20I,+^-=0.5V2+0.1V2=0.6V2——>y22=^-=0.6

Thus,

0.250.025'

Fvl=_S

Alternatively,fromthegivencircuit,

VI=41,-0.1V2

I2=201,+0.1V2

Comparingthesewiththeequationsforthehparametersshowthat

h”=4,h12=-().1,h21=20,h22=0.1

UsingTable18.1,

11-h120.1

y12-卜=-=0.025

Yu=7=0.25,h

-h,!4n4

△h

20nh0.4+2久

yA

y2\=7=5,22-h-,—u.o

h”11114

asabove.

Problem18.22WeobtainyHandy2IbyconsideringthecircuitinFig.(a).

4Q

-------•-------—.-----•—

++

V16Q

p2QV2=0

---•---------------------------•-----

(a)

葭・6

y21=V；=34=-0-1765

Togety22andy12,considerthecircuitinFig.(b).

工=2”(4+6h)=2仙+金⑵(34/7)=%=工

y222+(34/7)24I2

24

y22=—=0.7059

-62147

---------：—L=-1=—V

i2+(34/7)'48723492

-6I,-6八

.=~—VV?=34='0-1765

,342y.2

Thus,

0.2941-0.1765

lyl=S

-0.17650.7059

TheequivalentcircuitisshowninFig.(c).Aftertransformingthecurrentsourcetoa

voltagesource,wehavethecircuitinFig.(d).

2c

8.5V2Q

(2111.889)(8.5)(0.9714)(8.5)

V==0.5454

2111.889+8.5+5.667-0.9714+14.167

V2(0.5454)2

=0.1487W

2

Problem18.23

(a)Transformingthe△subnetworktoYgivesthecircuitinFig.(a).

Itiseasytogetthezparameters

zl2=z21=2,zN=1+2=3,z22=3

△z=zHz22-Z|2z21=9—4=5

z993-z|2-2

九一%一5一丫22'y"y2i一△工-5

Thus,theequivalentcircuitisasshowninFig.(b).

32

I,=1()=-V,--V,——>50=3V,-2V(1)

55?

-23

I2=-4=-y¥,+-V2——>-20=-2V,+3V,

10=V,-1.5V2——>V,=10+1.5V2(2)

Substituting(2)into(1),

50=30+4.5V2-2V,——>V2=8V

V,=10+1.5V2=22V

(b)Fordirectcircuitanalysis,considerthecircuitinFig.(a).

Forthemainnon-referencenode,

V

10-4=———>V=12

2°

V-V

10=」『..——>V,=10+Vo=22V

Problem18.24

(a)Converttozparameters;then,converttohparametersusingTable18.1.

zn=zi2=z2i

=60Q,z22=100Q

△z=zn，22-zi2z2i=6000-3600=2400

Az240060

h”-__—,hn=--=0.6

Z2210()Z2210()

_一1

112]=-0.6,h22=0.01

Z22Z22

Thus,

24Q0.6

|hl=

L」-0.60.01S

(b)Similarly,

zN=30flZ12=z2I=z22=20Q

=600-400=200

20020

h=一=10=一=1

1n1201220

1

^21=-l1122=——=0.05

20

Thus,

0.05S

12/13/1999

Problem18.31Wereplacethetwo-portbyitsequivalentcircuitasshownbelow.

50kQ

Zin=-p,2001150=40kC

V,=-50I](4()xl()3)=(_2xl()6)L

Fortheleftloop,

V-IO-4V,

—s--------=--I

1000,

\-10-4(-2X1061,)=10001,

Vs=10001,-2001,=8001,

V

Zin=-^=800Q

1I----------------------

Alternatively,

(1()4)(5())(50xKT)

Z=200+800-=800Q

in1+(0.5x10-5)(50x1(")

Problem18.32Togetg”andg21,considerthecircuitinFig.(a).

Y=(3j6)L_V1卷=0Q667+J。。333s

121,2

=0.8+j0.4

g2i(12-J6)1,

Togetgpandg22,considerthecircuitinFig.(b).

-12>gn4=有/=力21=O8-j().4

，2I'-JO

V2=(jlO+12ll-j6)I2

8-=iV72=jl0+i(12)r(-jk6)=2-4+j5-2fi

0.0667+j0.0333S-0.8-j0.4

0.8+j0.42.4+j5.2Q

Problem18.33Forthegparameters

L=gnv、+g12i2⑴

V1=§21V]+§2212⑵

ButV,=VS-I,ZSand

V2=-I2ZL=§21V1+§2212

V++ZI

0-g211(g22L)2

Substitutingthisinto(1),

।_(g22gli+Z-g||~~g21gl2)

-§21

or

giZ+Ag

Also,v2=g21(Vs-I,Zs)+g22I2

=§21Vs-g21ZsI]+g22I2

=§2iVs+Zs(g“Z