Irwin, B asic En gin e e rfn g Circu it An alysis, 10tE

Figure PS.6

,:_ g/< .rz_

&= 2r<ÿ

& = ,/<.c7_

Vÿ,= gv

SOLUTION:

.... - J _

1''o j - 3ooa = 4.5 ÿ'-ÿ

Z

...J

i

€

w

&÷aS_ "

-- , , , ,

ge>oÿ,, , o,t,ÿ co3_

g,oOO -t 2.0 oo

-'Vÿ- i ....

I_ _ Rÿ

'ÿ 6ooo

--- / ÿtO

Chapter 5: Additional Analysis Techniques Problem 5.6 '

2 Irwin, Bask En gin e e tin g Ci rcu it Aÿ a lysis, 1 (Jÿ

+ P.s't- Rz/

/goooÿ-20oo¢ ;oco

Problem 5,6 Chapter 5: Addttloÿfÿl Ahaÿysis Techniques

Iÿwÿ Basic Engineering Circuit ÿaÿysis, t0LE I

5,:1/4 Find Io in the circuit in Hg. P5.14 using supeÿosition1

Kk !

Figure P5.ÿ4

&= 6Kÿ-

V¢l "ÿ 12_V

SOLUTION;/- /('3 +Re -- COOÿ 4- ÿooo

Vo+ÿ-ÿ

Chapter 5: Addÿtfonÿ Arÿlys[s TeÿhniqLÿ Problem 5,14

lrryin, HaEic f;rqine€ring Sircedt Arraly*ir, 1{FE

To"* I" lo +

g; E Iu'+ To'

rdr

T*D#

-o-8 xr{3 f'_t. L ;,{3

o,4 ,hH '

-7z.hr)o-tttt4t:-z

Ttt Rax Jr,

: 6oeo X 6xtd36o0o r 6ooo f 6D oo f6oq{ooo tboo

7" = 2'4 -A

= l'Z qnA

flrLB

R1 ar,

Rz

#hapla S: *ddtiDruiAndb€k TedffiquesPrffi!€rfl 5.'14

lrwin, Basic Engineering CircuitAnalysis, 10ÿ 1

5.32 Use Th4vetÿin's theorem to A Vo in [her network in

Fig. P5.32.

bFigure P5.32

Rt- ÿ2L

SOLUTION'

Vÿ 2...

t

÷ iÿ'r ÿ P' //Pz

_ 4x9ÿ;_ 4V

,,--..,.....--

R,r cÿ

)ÿ, . , vÿ :t

/ \

R/R2_

R-re ÿ

Chapter 5: Additional Analysis TechniquesProblem 5.32

2 Irwin, Basic Engineering Circuit Analysis, II:)/E

'I'ÿ = t-3ÿ K_o- ' \

Problem 5,32Chaplet 5: Additional Analysis Techniques

Figure P5.4o• i iiiiirT rill ......................................"S-6f.UÿON:

L0

'V',::,c "- .... ÿc=V'

Iÿr ÿ; AÿIÿ Analÿis Technk:lueÿPmÿem 5.ÿ

CA-

_ J,

Pÿbÿ 5A0 Chapter 5: ÿitÿorÿ Analysis Technÿqueÿ

Irwin, Basic Enflineeiinfl CircuitAnalysis, 10/E !

5.49 Given the linear cimoii in Fig. P5.49, it is knovÿ thatwhen a 2-!d2 load is ooÿmecÿed to the terminals A-B, theload cut"rent is I 0 mA. If,ÿ lO-kÿ load is connected [othe Eerminals, the load cllrtent is 6 i1ÿk. Find lhe cunrenlin a 20ÿLQ load,

J'ÿ, = to ÿ 5 ÿLI=2-ÿ

Figure P5.49

"ZL. .... I .............................. ..==..,,,.l= __ ,,Nmÿm ......................................................................................................... ii u HI/IIIIIIIIÿIIIII .....................

SOLUTION"

(-I- .... I

I i

If !II I

_ÿ__:_ _'t

_ _Voo : ..ÿ, (/,-, ÷ ÿ',ÿ)

__ .4ÿD

.................... i,.

Chapter 5: Additional Arrÿlysls Techniques Problem 5.4g

Irwin, Basic Engineering Circuit Analysis, 10!E

in Fig. I>5,54.

.... gl ÿ ÿ,,

4--

b

Rze - 2Ko.._

V<r= :zzÿv

Figure P5.54

..................... ,#,, TI iSOL.ON: ...... " '" . "t'.".':: '" .::'it'tttz::zÿ ......

(

o "ÿo,

(

dÿ

+ ,

5

:z4....------..-.

/ÿ'2._

Chaplet 5: Additional Arÿaly.ÿis Techrÿ1quss Prolÿlem ÿ;,ÿ4 '

99

ll'v/In, ÿaSlC Enÿneenng Cÿro.W ÿ,nal'!sÿs, IOlE

Fi-":ÿrrrÿ Ps.ÿ',o

SOLUTION:

+Vÿ

,,....---

_ P_.ÿ,---iÿsqÿÿ --/- ÿ

__- Y,,

=©

iO3 X

In, Arm, Bÿsic Engbeeffn9 Grcui: Analys!,ÿ, lOtE ÿ1

Fÿÿ_ / /z ÿ-&

lg,3 ÿ .2. /ÿ_ ÿ-L_

VvI -: / ;z V

W "" ÿ:LVs, c.2.. ""

SOLUTION:

,1 R2..

÷ V/- -t-

'Vv L

Vr :ÿV

'VÿI = Vvÿ = /2 V

/I 1/

--.ÿC R z ti'z.

2 !ÿinÿ Basic Engineÿirlg Cirri .Aÿlysis, 10/E

.+.ÿLt _, .3¢o ,4

Iÿ: Ik

½ = 25o _m_

Problem 5,63 Ch;ÿer 5; Additional Analysis Technkÿes