Solution Irwin 10th - ch05

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Irwin, Basic Engineeling Circuit Analysis, 1DIE 
5.1.	 Use linearity and the assumption that Vo = 1 V to find 
lhe actual value of V in Fig. P5.1. o 
+2k.l1 : 2kO
.. 
-	
: 2kO V2k.l1 
• o 2k!l~ 1,=12 rnA 
-
Figure PS.l . 
SOLUTION: 
'I 
I :LX 
i 
l 
I 
,i
 
,l 
\
 
i 
I 
I 
I 
.....
·····1
. ­
iV 
--0 
t­liz. Ie.. 
2K 
2./(. 
-to. 
2J( . 3V 
'} 
'2..-lL 
+ 
3V 
kCl
 
I 
~ .', 
I 
.. j 
"·1 
! 
i 
I
 
I 
I
 
1 
Chapter 5: Additional Analysis Techniques	 Problem 5.1 
Irwin, Basic Engineering Circuit Analysis, 10/E 1 
5.2 Using linearity and the assumption that 1 = 1 rnA, find0 
the actual value of I. in the network use Fig. P5.2. 
eb 6~ 
± V =24 Vs 
41<0 
. 
6kfi 
3~. 
;. 2~4~ 
Figure PS.2 
SOLUTION: 
9v 
1" -
Gl:: 312.1::.+­
::~~~ 
" ... 
4-k. 110 
4-k 
2.k 
II 
tk !h
 
3J.L.. 
"ikf-k 
I+.G ::.
.lLt 
­4- 4- +f 
.!! = iikVL 2­
~ +It ss. 
'- 2­
+
 
I 
J.'k 
2'1/ 
. 
,'.' 
"-', 
~,.. 
.1.
-k 
:='. 
Chapter 5: Additional Analysis Techniques Prnblem5.2 
c 
Iiwin, Basic Engineering Circuit Analysis, 10/E2 
i
-/( L 
- 24- ! j 
f ; 
1;,­
t: 
" ~ 
I 
I 
i 
i 
I 
I 
..; 
! 
l­
! 
Problem 5.2 Chapter 5: Addnional Analysis Techniques 
I 
+
-
6k
T'-0
K
'
"
I 
Irwin,Basic Engineering CircuitAnalysis, 1DIE 
5.9 Find V in the network ill Fig. P.5.9 using superposition. o 
akO41dl 
+ Vo -t O 
61dl 
6mA t + 4mA 41dl21dl j 
, 
Figure P5.9 
SOLUTION: 
-
Gk. 
.2.k 4-k. 
'----------1'"4-----------
G (Gk) (Gk)v.D ' ­ k G.K t l'3k 
(J;(Gx6 0 
- IV ~; 
.".' 
!
! 
i 
! 
i 
l f' 
F 
Chapter5: Additional Analysis Techniques Problem 5.9 
Irwin, Basic Engineering Circuit Analysis, 1OlE2 
4-kj
:":j
.....·1
I
 
!
 4­i 2l! 4-K-k 
Vo\I = -±.. [8ki-l.jl()(6k) = !i. {~\l6~= ILV k: l 2-4 K k 2-'-1Kj 
"4 I 1" Vo II:: qt I'2...= ~ I V 
Problem 5.9 Chapter 5: Additional Analysis Techniques 
"win, BasicEngineering CircuitAnalysis, 10tE 
5.10 Find \'. in thenetworkin Fig. P5.10using superposition. 
..j 
1, Figure P5.10 
6k!l6mA 
6kO6k.n 
12V 
6k.n 
6kU 
0+ Va 
SOLUTION: 
V. I 
o 
61<. 6/< 
+-
L-_-'--- ~:----}------ _ 
4vGk. ) = )%lc­
3 
Gt Gk.. 
j. 
6k 
1 
, 
••• 1 
Chapter5: AdditionaiAnalysisTechniques Problem 5.10 
NavjeetK
Pencil
2 Irwin, Basic Engineering Circuit Analysis, 10/E 
Problem 5.10 Chapter 5: Additional Analysis Techniques 
 
 
-. ··1 
Irwin, Basic Engineering Circuit Analysis. 10fE 1 
5.11 Find 10 in the network in Fig. P5.11 using superposition. 
12V 2kfi 
4kn 4kfi 
2kfi 
Figure P5.11 
SOLUTION:
 
12- T
- 21<:. 
:~
 
2.1:. 
/1­
12­
-
-"---­
2.k.1" ~k 
5 
-ro= 
, 
-I 4-k.. 
= (~;~ )(k:)/ 10k... 
I:;. A
-
 /lk. 
Chapter 5: Additional Analysis Techniques Problem 5.11 
.c 
---
i 
· ~ 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
It> II 
+k 
2t 
+1<:. 
· .' ~ 
:·i 
2.t 
:C;I 
I.I.­~ r.: 
T :
r: 
I' 
~. 
G 
k 
4--k. 
:3 
4 k. 
- GLo I~ - T 
i 
· .'1 
_ 12.- 4g1 
-lik. Jlk.. 
Problem 5.11 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.12 Find Vo in the circuit in Fig. P5.12 using superposition. 
o + Vo -,~ 
21<0 
12V SmA 
21<0
 2kfl2 kO 
Figure P5.12 
SOLUTION: 
2.1:: 
V. I0_ 
2./( 
/2­
G----{-t 
2k 
2k 
.~. 
10 1 
/2V
 :;: 2.t. 
I.
 
Chapter 5: Additional Analysis Techniques Problem 5.12 
i 
--
Irwin,Basic Engineering CircuitAnalysis, 1DIE2 
= .±. Ie.
-3 
... 
-36 
_ - 18
 
10(. sit:.. 
-)~v,/ = 5'k. 
(-t) -
V. IIT D_ 
.---MN\J1\.--.------, 
21::.. 
I.- rr-. 
I-' J'D 
>~ 
,2.1c. !; 2.1:: . ~ G/t ~ 2.1:,. 
= -­
+ "11) , I 
-
2.K 
Glk. 
.3k..
 
Problem5,12 Chapter5: Additional AnalysisTechniques 
Irwin, Basic Engineering Circuit Analysis, 10/E 3 
Vo "-
-
b
k ( 3t) (:~k.) 5k.. 
- ~V
 
5" 
V, II'./' 
..,.. 0VD = 0 
- /;;..
- -t~
-g s: 
-
- '3J:v 
S" 
f, . 
! 
I 
i 
Chapter 5: Additional Analysis Techniques Problem 5.12 
Irwin, Basic Engineering Circuit Anaiysis, 1DIE 
5.13 FindVo in the circuitin Fig. P5.J3 using superposi lion. 
41<{} 
21<{} 2kO 
4kO 
Figure P5.13 
SOLUTION:
 
)~V 
-611:. 
2k. 
+ V. " o 
V II4­ o == - sv 
k 
,/1 ,/ 11 _ /2-8 _+V
'1 :: V o -r Vo ­0 
Chapter 5: Additional Anaiysis Techniques Problem 5.13 
Irwin,BasicEngineering CircuitAnalysis, 1DIE 
5.14 Find 10 in the circuit in Fig. P5.14 using superposition. 
12V 6ka 
6ka 6k!l 
6kO 
Figure P5.14 
SOLUTION:
 
Yo' +I ' o 
IlK.. 
(I s ) ( I~f) ~ tt = 
_~4-15" _ -£'1+ 
ct. 51::. 
i 
I 
i 
6/(.xi tI 
G1<. 
Chapter5: Additionai AnalysisTechniques Problem 5.14 
Irwin,BasicEngineering CircuitAnalysis, 1DIE2 
. It 
1 i 
i 
, 
3(" )'J­1- 6 :; -T 1 
--­( )~) -15A:. ..9c.'"'" - Ie.. 
II £Ac10 -- 1 "/z. Sic.­
.-.. III'() ..,.. :Lv
.1;; = 
r"o 
Gk. 
j 
...
··"1
,
,
, 
Problem 5.14 Chapter5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 10tE 
5.15 Find V. in the circuit in Fig. P5.15 using superposition. 
12V 
.' 
L----.....----.....---;O 
Figure P5.15 
+ 
6kO 
6k!l 
6k!l 
6kn 
SOLUTION:
 
0 0 
+ 
Glt t GK.. Gk Vol 
3l 
-
--0 
.£( (6k.) (9t) I 
J: I~k- J 
GX(;"Xr:r 3 
= 
fC5 
~ lOB V
',')JI s: 
, 
,
 
i,
 
1 
I 
! 
Chapter 5: Additional Analysis Techniques Problem 5,15 
Irwin, Basic Engineering Circuit Analysis, 1DIE2 
v. 
i GK. 
I, 
. ::1 
+ Gll.. 
"VI VOGILGr-
-
II
.'d t
:-1 " 
IZ." 
( 4-1:.)
'I 
-Il. -4-8V =' 10I {ok.. = 
Vo 1/ = -2-
1/
-
- /0 
1 V., -]Lj. V 
V' V.V-... a -r 0 II o 
- 108 - /;..
-
S" -S­
-
J.§..V
s: 
Problem 5.15 Chapter 5: Additional Analysis Techniques 
.
Irwin, Basic Engineering Circuit Analysis, 10JE 
5.16 Find 1 in the circuit in Fig. P5.16 using superposition. :"0 ..' 
6k!l 
12V 6V 
i 
1,
, 
I 
10 
6kO 
Figure P5.16 
SOLUTION: 
~_.: 
,---'vVv1,~---'----, 
61=. 
12. 
6K.. 
----'\II.AA-~. -----1 
I'I_ (;
 
o - /2.. t 
:1 
z.1:: 
Chapter 5: Additional Analysis Techniques Problem 5.16 
I 
Irwin,BasicEngineering CircuitAnalysis,10/E2 
}1 i'1,," 
6t ~I:: 
WI' t.G( 
1/ bItT~ = - ..J.A 
2­ /C.. 
3 ..,.. 1- ::..L 41/) 
­
~ 
:2Jc ::1.1:k-
I· 
Problem 5.16 Chapter5: Additional Analysis Techniques 
i 
Irwin, Basic Engineertng Circuit Analysis, 10/E 
5.17 Use superposition to find 1 in the circuit in Fig. P5.l7. 0 
12 mA 
12­
G/c. 
~~ 
f"', 
i 
!
I 
i 
i 
Chapter 5: Additional Analysis Techniques Problem 5.17 
:.:: 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
I" = o 
: .j ~ " 
.;.1 
.j 
- .2 A 
T 
,.
, 
Problem 5.17 Chapter 5: Additional Analysis Techniques 
Irwin. Basic Engineering Circuit Analysis, 10/E 
5.18 Use superposition to find 10 in the network in Fig. P5.l8. 
6kU 
10 
6kfi 
6kU 
6kfi 
I. 
I Figure PS.18 
SOLUTION: 
GI:.. 
k; 10': 2 mA 
k~k. 
<:;t.
6K
..i 
i 
,i 
I,
;
 
;
 
Gk.. :r; II 
J, 0 
G/<. r Gt 
~k.. 
10 1'= -.±r:(.1.2=.):: tIII l~.· 
r::
..(- .a L
.3 i 
f 
I 
! 
Chapter 5: AdditionalAnalysisTechniques Problem5.1B 
Irwin, Basic Engineering Circuit Analysis, 1DIE
5.19 Use superposition to find Vo in the circuit in Fig. P5.19. 
3kO 
12V 6V
 
3kO
 3kO
+ 
3kO Vo , 
.;. j f: 
i\1
'·'-1, :: 
Figure P5.19 ! 
I.• 
1 
SOLUTION: 
0 
I').V +
-
3k..3k. 
3t T 
<.3(. 
'Ie> 
, 
_ /2.. (3/,.K: ) $ I,). ( ~ ) ""-
31c T 3/;.'(. 
+ 
V. II 
D 
-+ G 
Chapter 5: Additional Analysis Techniques Problem 5.19 
Irwin, BasicEngineering CircuitAnalysis, 1DIE2 
!
 
t 
! 
··1 
.:'".'l 
-,:=/j 
i
 
i
 
- 4- -t..z ::: -2Y 
.~ . 
Problem 5.19 Chapter5: Additional Analysis Techniques 
hwin, Basic Engineering Circuit Analysis, 10/E 
5.20 Use superposition to find V. in the networkin Fig. P5.20. 
+ 
6V 
6kn 6kn 
12V 
j f·~: 
.,
... 6kn 6mA 6kn r-:'1 t 1-­! 
Figure P5.20 
SOLUTION:
 
I~
 
t; k.. 
-----------" 
1')-( at. ) 
$Ic. of ~ It 
Chapter 5: Additional Analysis Techniques Problem 5.20 
I .. 
i 
2 Irwin,BasicEngineering CircuitAnalysis, 10/E 
\1 IIVo :: 
.,. 
V. 1/1 
o 
+ 
v: ,1/() ::. -/~ V 
Problem 5.20 Chapter5: Additional AnalysisTechniques 
Irwin, Basic Engineering Circuit Analysis, 10/E 3 
-
t ... ~-J2.
-
i -10'1 i,
:1 k' t.:
.. ! ] c· 
r" 
..~' 
Chapter 5: Additional Analysis Techniques Problem 5.20 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.21 Use superposition to find 10 in the circuit in Fig. PS.21. 
6kO + 12V 6k!l 
! [ 6 k!l 6 k.ll! 
! 
....
/1;.; 
12 k.ll 
Figure P5.21 
SOLUTION:
 
t: '= -imA!. /2­
---0 
:4>' 
'--------'V\IVl------+~~ 
61c.Glc. 
,
-k.. 
0.',' 
..~ 
Chapter 5: Additional Analysis Techniques Problem 5.21 
i 
I 
! 
I 
,J! 
1 
I
 
I
 
! 
Irwin,Basic Engineering CircuitAnalysis, 10/E2 
:t I'
 
/2.L
 
Gk....
 
,--V\l\A~--,I 
i 6K .:4J "' 
10 
i 
i 
I
',:.1 
1 j
 
j
 
i
 
I
 
I 
I 
l'
 
Z II..,.. .z III
- ]rl'-r 
--1'i"0-2­
::: ' 3m A 
Problem5.21 Chapter5: Additional AnalysisTechniques 
Irwin,Basic Engineering CircuitAnalysis, 1OlE 
5.22 Use superposition to find 10 in the network in Fig. P5.22. 
12V t SmA 
10 4k!l 
i 3kfi
'00'01 k,: 
.. 1
, SV Sill 
Figure PS.22 
SOLUTION:
 
G 
~'
 
Chapter 5: AdditionalAnalysisTechniques Problem 5.22 
Irwin, BasicEngineering CircuitAnalysis, 1DIE2 
til,; =: 4­
-r. 
i 
1· 
II ~:: Jol-r 1v :: 
- (4-- ;J./!;) LA
- /L 
::c 
.!-Q.A­
.?> /L 
-. 
,.. 
~' 
V 
Problem 5.22 Chapter5: Additional Ariaiysis Techniques 
Irwin, Basic Engineering CircuitAnalysis, 10/E 
5.23 Use superposition to lind V. in the circuit in Fig. P5.23.
+
....----....,.----_--:0
12V t 6mA 
3kil 
3k.n 
3k.n 3k.n 
Figure P5.23 
SOLUTION: 
-
3L 
Vo': -)2- ( ~~ ) 
-~V 
5 
Chapter5: Additional Analysis Techniques Problem 5,23 
.
2 Irwin, Basic Engineering Circuit Analysis, 10/E 
+ 
31::. 
31<.. 
G
-k. 
V II 
D 
.~ 
t
 
t t. ~ 3l \/, 
0 
\I 
\/:. lL 
It. 
-
'D 
Problem5.23 Chapter5: Additional Analysis Techniques 
3 Irwin, BasicEngineering Circuit Analysis, 1DIE 
,I 
J
 
.-----.--------,---0 
+ 
H. 31<.. 
II<. 
v, 11_ 
() -
-
6 ( ~)(3~)it... 
-
-
)g v' 
S 
1 I'
"1 -;- VO0Vb -= 
/8
-
-
-3(, 1" 
-f -s: 
==­ -~V 
5 
Chapter 5: Additional Analysis Techniques Problem 5.23 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.24 Find VA inFig. P5.24usingsuperposition. 
2kn 4kO 2kn 
5kfi 
10V 2kn 
--
1: Figure P5.24 
SOLUTION: 
-
2.t.Q.. 4-1::: ..ll- Z/::....n.... 
tW 
j 5/::....12.­
~ ,
, 
~ 
! 
,i 
~ 
I 
! 
i 
! 
10\1 +­ .j. 
_ 
'Ih IA 
2/::'...n... 
V II 
1--
V' = A 8V 
+ 
VAil 
Chapter 5: Additional Analysis Techniques Problem 5.24 
2 Irwin, Basic Engineering Circuit Analysis, 10/E 
-
2mA 
\ I II
 
VPr ::::
 
1f:J' 2 V 
V III 
-!-- A-Vk AAA MA 
.~ sl.:--'L 
" ",+ 1 J., )D fl') II- ~ .. PI 
-
/ 
...... 
I'"A 
Problem 5.24 Chapter 5: Additional Anaiysis Techniques 
3Irwin, Basic Engineenng Circuit Analysis, 10/E 
;.. 
~
-A 
T ~I 10m ( 41:-) 
-
4-mA
4-t +6k 
VA II, 
_(+m) (~L) 
i
,
-
-
i i
·'·1 
-BV..: ...j V, 
II 11/\h: 
TV :::. VA '1" ~ 'Ir Pt 
-
gt- 1~12.-g
.­ ~. 
-
)a.2 V
-
-
, 
L,. 
~; 
f 
Chapter 5: Additional Analysis Techniques Problem 5.24 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.25 Find II in Fig. P5.25 using superposition. 
-;-. 
21<0 41<0 
:;l. t.n­ 4-t .a, CP r: 
t' 
12-" ~ 
-
6kn.. 
I 
1, 
.. 
'. 
il:..~ 
.. 
.. 
!
'>1
.,.,.! \2.V 
+-
VI +lJl.. 12. k:. si: 
i, 
( 
[ 
" 
r;::; 
"
,::,", 
:r,1 
Chapter 5: Additional Analysis Techniques Problem 5.25 
---
Irwin, Basic Engineering Circuit Analysis, 1DIE2 
i 
-z-. 
I
 
I
 
I
 
I
 
-
g
 
!
! I. ';. 
,I '-k..j 
I
i 
! 1,'1
_. i
 
I, " 
Problem 5.25 Chapter 5: Additional Analysis Techniques 
.
Irwin,Basic Engineering CircuitAnalysis, 1DIE 3 
I /. :+1 k.-U.i i .,. - ­
-
- J~k... ~~R"i­
i 
.J- -tJ- --t.-L- ;2.t.
- - - ~L4-'­~ 
t.J2­
-
I· o~
-~ 1, '+I k....Il.- .. 
-c. 
..Jvv\A. 
. 
8V~ 
_\111fT
 
I
----Ajt/I!I.<-------l
 
I' o4t.J1­
/. 0'1 Ie- _ . ) =
 
}. D1/t -t-). ~I k.
 
Chapter5: Additional AnalysisTechniques Problem 5.25 
Irwin, Basic Engineering Circuit Analysis, 1DIE 4 
l' "_ 0,5""18,,3 If)f+4./ .. 
.: 
,VIAi\.. 
~lc..~?-k..4­
',,1.0 , •• Ai 
~ ~ ~ k:...S).. ( 
~ 1/'" 
I 
i 
,i 
I ~~ .. -L -+.J- -to J.. R-t-t
. I V 2t. zs: , k. T 
3l:....Q.. 
-
I~V 
V /I' 
, 
i 
,•
: j.~ Slt4 I 
F'
"" !~ 
r. 
D 
+ /6V
-+- ( 
Y1\1 _ I - -If, 0, 'b ~=l-II:: )( 0' ~ 5""+1 ;t-t-3 t. 
Y \1'_ - 3, ~6 VI -
V, I"1'" - :._ b,~q.a3YY\A 
-"1 
Problem 5.25 Chapter 5: Additional Analysis Techniques 
1 Irwin. Basic Engineering Circuit Analysis, 1DIE 
5.26 Use superposition to calculate Ix in Fig. P5.26. 
12 kG 
6kO t, 
24V 
4kO 
12V 
Figure P5.26 
SOLUTION: 
i 1 1 I 
i -! ­, 
1
 
1
 
I
 
I
 
1 
1 1')..:1 
1 
~ ; 
i 
1
 
I
 
! 
1 II 
I 
~
 
-·····-1
I 
I 
i 
I
 
!
 
I 
I 
-+­
1 ::
-x 
x'J( 
= 
- })..
-
/01:... 
Gk. ,.l).. 
<­
1"
1,
-
.2.t-V 
Chapter 5: Additional Analysis Techniques Problem 5.26 
Irwin, Basic Engineering Circuil Analysis, 10/E2 
1 "'
 ~
 
~ k..ll. :"..k. Sl/ AAAA 
:1. 111 
~mA l' 
lIZ II, 
2mA 
:'.­
i 
I 
i 
Problem 5.26 Chapter 5: Additionai Analysis Techniques 
3Irwin, Basic Engineering CircuitAnalysis, 1DIE 
l' 'II 
-~ .=. 
0, %mA
 
:: 
-
-1.2m A ­
-
.'.' 
..; :-~ 
Chapter5: Additional AnalysisTechniques Problem 5.26 
--
Irwin,BasicEngineering CircuitAnalysis, 1OlE 
5.27 Calculate Vo in Fig. P5.27 using superposition. 
12V 
6kO 
2kO 
4kn 
6mA 
10 kn 
10 kn 
18V 
Figure P5.27 
SOLUTION: 
loiz.. ..n. 
+ 18V 
.. 
',\ 
YD I )Gt....Q.. ;- VI) 
-
lot.. J).. 
MM MAit ~ 
'. 'J t +'-.Q...r )..~.f(... ~ 
$p." G lo~J2-~ 
(10):,) (lo':L =- 5k-SL 
IOIc. + lolL 
Chapter5: Additional Analysis Techniques Problem 5.27 
-.--l 
2 Irwin, Basic Engineering Circuit Analysis, 10/E 
V. I'+ 11_ 
'I
V. II+ D _ 
1,
 
Problem 5.27 Chapter 5: Additional Analysis Techniques 
3Irwin, Basic Engineering Circuit Analysis, 10/E 
, j 
. I III+ VD _ 
l~v
 
Iol::. A. 
-
; 
1
..
 
c 
~: 
;:: 
~. 
-
Chapter 5: Additional Analysis Techniques Problem 5.27 
4 Irwin, Basic Engineering Circuit Analysis, 10/E 
v, =­
,J 11 1_ 
Vb -
-'1/I 
\I. Ii I
'It;
Yr;'T V.II "...,.. ~ o 
-
j, 2--T ~. G.- ~,'1 
::. 10· t.., V 
----'
-
i 
I 
i 
L 
c: 
,t .
j 
Problem 5.27 Chapter 5: Additional Analysis Techniques 
A^
eq/
eq/
-
0'
'''
+-
Irwin, Basic Engineering CircuitAnalysis, 1OlE 
5.33 Use Thevenin's theorem to find 1.in thecircuit using 
Fig. P5.33. 
12V 2kll 
4kfi 4kfi 
Figure P5.33 
2kfi 
SOLUTION:
 
/1
1/ 6,21./:t.- -l­
-k 
4-1::. 4-1:.. 
Voc 
4t I::::. }2­
-t I(1, - i) 
b t 1/:::. .2 4-
r - _4-A~J - k: 
Yo{: :;; -4KI, - 2- 4­
s: _ J6 -.2-4-- s: - +0 V 
Chapter5: Additional Analysis Techniques Problem 5.33 
I 
i 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
4-L 
, 
t· 
(111:; 4-t-r .2tJ)4-L 
- 4-t T 4- L - J5:... L .: 
d 3 
"j 
I 
I, 
! 
I 
I 
-t!, 
i 40v 
10 -- - 4-0 "- - I?- b 
~).:t !it ~£ 
..3 
Problem 5.33 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.34 Use Thevenin's theorem to find V.in the circuit using 
Fig. P5.34. 
+ 
2k.n 
6mA 
12V 
2k.n2k.n 2kCl 
Figu re P5.34 
SOLUTION: 
'Yac 
I 
I 
l ; 
12­
f----{-+}----+-{ ~ 1--------. 
bit 
Chapter 5: Additional Analysis Techniques Problem 5.34 
i 
I 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
':
.,· 
\'2­
v
 
'{j
····1 
I
 
. 
II 
j 
",". 
Problem 5.34 Chapter 5: Additional Analysis Techniques 
'
Irwin,BasicEngineering CircuitAnalysis, 1DIE 
5.35 UseThevenin's theorem to find V. in thecircuit in 
Fig. P5.35. 
3 kfl 
4k!l 
+ 
Vo 
-
, 3k!l 
6V + 
6k!l 
-
6k!l 
[' 
~:~ 
6mA 
Figure P5.35 
SOLUTION: 
3l:. 
~~.4-)k­
-I-
'VDC 
-+­-
I~ 3k 
?-)1:.. I~+ 
Gk.. 6l 
~ 
G It 
\foe ~ - GV .. '. 
,.... 
~~: 
e7f-( ~ 
> ;:
2>"­ ~ ~~ 
-. ,-. A .. 
E31:... r:t 
~~ 
~ 
Chapter5: Additional AnalysisTechniques Problem 5.35 
.~. 
.
--
Irwin, Basic Engineering Circuit Analysis, 1DIE2 
6
 
-
-
6 (4-J:)
-
4-t~ q/;).(. 
,~
~. 
-
-
;..4­
-
-/r/). 
- -
+~ II.;"_1 
!-:::19'" 
Problem 5.35 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering CircuitAnalysis, 10JE 
5.36 Use Thevenin's theorem to find 10 in the network in 
Fig. P5.36. 
4mA 
,-------{__1--------, 
2k!l 
1 kO 1 k!l 
Figure P5.36 
SOLUTION:
 
6) 4-1~
 
)oj/(. )0/1:. 
2-c.. 
V -.b k3;.­ +­/L Ik.. ac 
~ 
VOC 
-
- ~~(J~) -r "!{J 
::=- 2~ V 
3lc..JL I ~ '711 :- I ~ -
Chapter5: Additional Analysis Techniques Problem 5.36 
Irwin, Basic Engineering Circuit Analysis, 10lE 2 
!,
 
i 
I 
······.1..... 
It.
 
~
-
" [. 
Problem 5.36 Chapter 5: Additional Analysis Techniques 
.
.:'. 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.37 Find 10 in the network in Fig. PS.37 using Thevenin's ;', 
theorem. 
~:-, 
an ~ 
'. j 
1 kn 2kn 
+ 6V 
~ 
! 
i 
i 
I 
f'r:
r' 
Ffgure P5.37 
. 
SOLUTION: 
2-) k, 
IK 
.<.
- Voe. (L~ +~ (il:)~O ~ Le­
,Voe 
-
0 
o '" 0 
-»
1 s: 0 
r 
, 
:::: 
Chapter 5: Additional Analysis Techniques Problem 5.37 
,
Irwin, Basic Engineering Circuit Analysis, 1DIE 1 
5.38 Find Vo in the circuit in Fig. PS.38 using Thevenin's 
theorem. 
12V 
_--...(-+l----~ 
,~ 
1 kfl 2 kfl\j + f 
::::.: ~~ 
2 kfl 4mA 1 kfl Vo 
Figure PS.38 
•.. 
SOLUTION:
 
ilL 
81:.1, - ;ik. 12-;:' /2­
'_.i""1 :;:. - +/t.~2. 
31c J., -r 4-~ /2­
1', ~ .l­
--- ole.. 
X;~ 
,i
i
 
I 
Chapter 5: Additional Analysis Techniques . Problem 5.38 
Irwin, Basic Engineering Circuit Analysis, 1OlE 2 
ilL 2L 
AA" A"A 
1= .2l t::' 
~.:! ~ :? c:. 
4-- f1/f ::: 2.k.­
~ 
~ 
:lit. t 
il 'V0-.1 O 
Problem 5.38 Chapter 5: Additional Analysis Techniques 
hwin, Basic Engineering Circuit Analysis. 10tE 
5.39 Find V. in the circuit in Fig. P5.39 using Thevenin's 
theorem 
+ 
1 kO 
2kO 
t 2mA 
12V 11<0 
Figure P5.39 
SOLUTION: 
It 
\')... +
-
)..\1.. 
l' ?;:. ~ 
r 
l 
t 
Voe 
'0 
-
-
/+V 
r 
I 
~ 
i 
i 
,;. 
lie-
Ik.. 
Chapter 5: Additional Analysis Techniques Problem 5.39 
2 Irwin,Basic Engineering CircuitAnalysis, 10/E 
j4-V J! (~lt.) 
<3K.. 
-
- Ev 
-3 
k 
C' 
,i 
r' 
,
~.:
~ 
!, 
! 
Problem 5.39 Chapter5: Additional Analysis Techniques 
1 Irwin, Basic Engineering Circuit Analysis, 1OlE 
5.40 Find 10 in the circuit in Fig. PS.40 usingThevenin's 
theorem. 
12V 2kO 
1 kO 
2kO 
1 kG 
Figure PS.40 
SOLUTION:
 
\k..
 
- VOC - 11- - ). k. (1=) s: 0 
VOC -16V 
Chapter 5: Additional Analysis Techniques Problem 5.40 
2 Irwin, Basic Engineering Circuit Analysis, 1DIE 
.~. 
t ~lL 
~ 
;:--t­
1-\1­ s R1'1 
,t; 
,.:.,. 
-
-
i 
I [ 
r.,
r: 
Problem 5.40 Chapter 5: Additional Analysis Techniques 
--
1 
Irwin, Basic Engineering Circuit Analysis. 10/E 
5.41 Find v, in the network in Fig. PS.41 using Thevenin's 
theorem. 
'I 4mA 
no 12V 
t 
2kn 
1kn 21<0 
+
 
Figure P5.41 
SOLUTION: 
~ Il I')...>~ 
• 
k t 
+lL \l VOC 
-
lit. err - -tJ -t 2 k.3{ - -I'J.. 
.3l~ 1.r ~ ~ g 
L -- -~ i i3.l i I 
:<1 1:Vo G ; 2.)(. C-! )~ 4- ~ 
1 
I 
i 
! ~ - )6 ,.,.. 4- -::: - It. -tJ1:" .::: 
"7f ::5.B 
Chapter 5: AddiUonal Analysis Techniques Problem 5.41 
:
.1 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
--------10
 
.------~~-~-v--
IJI. 
..:1
.":-:j 
IlL. 
i 
I
 
1 
!
!
i 
I 
I 
I 
i 
! 
.
I
.co:j 
.: ! 
: .:.~ 
;j
 
I
 
i 
i 
I 
~I-!> k..
 
'If) =- - +/3 
c(L-r!"(~t 
;:. - g/~ !IV 
co< 1" .r/3 
~ 
i 
i 
[;
r 
s :
.. 
t 
:.: 
t 
v 
_ 0 
(:J.. k) 
,,: 
-
-g 
.. 
'. 
< 
-
Problem 5.41 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.42 Find 10 in the networkin Fig. P5.42 usingThevenin's 
theorem. 
1 kO 2kO 
12V 
+ 6V 
21<0 1 k!l 
Figure P5.42 
SOLUTION: 
4-}l 
0 
\K.. 
').~ 
e­
h­
.ill. 
z.ll.. 
-I­
'!oC 
-
4- -tg ~ Voe :: o 
VOC ;:; - 4-v 
Chapter 5: Addifional Analysis Techniques Problem 5.42 
- -
hwin, Basic Engineering CircuitAnaiysis, 1DIE2 
)k. 
l-. 
t. 
, 
,
, 
, 
.. 
~~ G-tr
-
41st 
ot 
..:,.1 ..­
".,] ~IB.L 
I 
I J- (fl A
-... 6I ~ :t.­~f 
i 
I 
Problem 5.42 Chapter5: Additional Analysis Techniques 
---
Irwin, Basic Engineering Circuit Analysis, 10/E 
5.43 Use Thevenin's theorem to find 10 in Fig. P5,43. 
2mA 4kfl 
10 
2kfi
-·1'
-- '.
"0"_) 2 kfi 4 kfi 
+ 12V + BV 16V 
Figure P5,43 
SOLUTION:
 
+ gv t 16V 
I'}..:J 
VDG-
-
- 87 J~-8 
-
)6V 
--.
.­
i: 
Chapter 5: Additional Analysis Techniques Problem 5.43 
2 Irwin,Basic Engineering CircuitAnalysis, 1DIE 
L-_--J'''V\"A,V'-­__....!
'. \:-2. I:.. .l2,... 
Problem 5.43 Chapter5: Additional Analysis Techniques 
ii 
,. 
1Irwin, Basic Engineering Circuit Analysis, 1DIE 
~l5.44 Calculate Ix in Fig. P5.44 usingThevenin's theorem. ::: 
d 
1 
i 
12V 
12 k!l 
4k!l 
t 2mA 
Figure P5.44 
SOLUTION: 
Voe 
.12 k.!l. 
i, 
~: 
,.~ 
c: 
r
t~ 
.~ 
I" 
! 
24V 
.~.B A 2.,.,.,A t k ..Q.. 
,.
, 
gv­
It-v r 
2,.,., A + 2.+" 
Yoe ,. 
\f0(;,. = 
8-2.+- 12­
-2.'6V 
-
.... ~ 
~ !
f--f(-v
1J1
 
~AAA i2 k.l2. 
Q----II/VI/I/V------I 
Chapter 5: Additional Analysis Techniques Problem 5.44 
.
2 Irwin,Basic Engineering CircuitAnalysis, 10/E 
A
t 
I'· 
\ 
-28 
,~ 
. ! 
.,'.1 
j 
j 
! 
Problem 5.44 Chapter 5: Additional Analysis Techniques 
I 
NavjeetK
Pencil
-
T-
"
+-
+-
+-
+–
=
Irwin, Basic Engineering Circuit Analysis, 10/E 
5.63 Find 10 in the circuit in Fig. PS.63 using Thevenin's 
theorem. 
1 kfi 
Vx ­ 1 kO 
2V"x12V 2kO 
.·.i
I 
.' '. ~ 
.' .;! 
Figure P5.63 
SOLUTION: 
r 
+1/.'­)I.. 
+ 
+ 12.. 
ik. 
I 
, 
2Vx 
J~- .2- \Ix1 ­
~lL 
'Vf' :. ir:l 
:= 12. 
~I 12.- 3 - qvVOC = 
Chapter 5: Additionai Analysis Techniques Problem 5.63 
Irwin,Basic Engineering CircuitAnaiysis, 1DIE2 
lk
 
2V~ 
··.·-1 
..... 
)2- +3:..=L
-lie lit 
.. } 
,
 
7
1, 
i 
i·
 
Problem 5.63 Chapter5: Additional Analysis Techniques 
+
-
+-
+
-
+-
1Irwin, Basic Engineering Circuit Analysis, 1DIE 
~;~ 
5.67 Find Vo in the network in Fig. P5.67 using Thevenin's 
c:
/; 
:;;: 
theorem 
i 
i 
i 
fr 
l 
I 
I'. 
Figure P5.67 
SOLUTION:
 
i-/kIZ it 'Ioe 
!~ 
ik. 21~ 
-
-r.2 _1~ t 2IxI 
--
0 
k.. 
,~. 
-3 I I 
-
2-
X -
-It:. 
, 
-,2... ~ ~, ~ i It .:: 
',~, ~,1 3Jc. 
)= )2- -4-' -- 3hVVoe - 12. -t J.k. (2. It . :3 
-3 
Chapter 5: Additional Analysis Techniques Problem 5.67 
L 
· ;", 
·.1 
Irwin, Basic Engineering CircuitAnalysis, 10/E2 
+­
'-- j \~ 1;'
::.:J 
­ Z~ l:~ ;~::""'1 
:1.k..ik 
Z4
 
.1/t 
I~ 
i· 
,
 
Problem 5.67 Chapter5: Additional Analysis Techniques 
I
Irwin, Basic Engineering Circuit Analysis, 10/E 3 
:-: 
.. 1 
I'). 6/3 ~A f_.":OJ
':-.-:·1 1sc/' r,3k.. i~:::~:-:! 
:"/3 ~
3'1~ 
,
'. 
.. 
. . 
.. 
.>:j
'I 
i 
,i 
R111 = 'lac. 
i; 
-
.32./3 
3;;.13k. 
-
.:i.k. 
" 
~, 
.. 
:J.k.. 
ik. Jj..v
3 
•• "1 
::·::i 
i,
l)
f-" 
Chapter 5: Additional Analysis Techniques Problem 5.67 
c 
i 
Irwin,Basic Engineering CircuitAnalysis, 10/E 
5.68 Use TMvenin·s theorem to find V in the circuit ino 
Fig. P5.68. 
!, 
.j 
, 
I
 
! 
i 
k 
+
-k 
4-2.VDC ~ Voe. - 6"1" Voc = 
\IDe-': 16 V 
-, 
Chapter5: Additional Analysis Techniques Problem 5.68 
--
Irwin,BasicEngineering CircuitAnalysis, 1DIE2 
":= 
~' . 
t 
,------(--7 t-------, 
G :. 
<j--J'V\/IiI--.>---_(_+J--_ ! 
1, 
e .... /£ 
l/~k 
T I - -k. 
c. 
-
If. 
-
JtJL 
I~( I<.. '~: 
-cl1c " 
Ik. it. 
H.. 
+ 
V = If. (lk.) 
f) "3k 
~!£V 
3 
Problem 5.68 Chapter5: Addijional Analysis Techniques 
.
. .'.' ~ ;:~
,-'.' '.-.
,..' 
Irwin, Basic Engineering Circuit Analysis, 10/E 
~-.:5.69 Use Thevenin's theorem to find Vo in the circuit in 
Fig. P5.69. 
4mA 
,-----(-)---------, 
12V 
1 k!l 1 k!l1 k!l 
'" 
.< 
+ ~ 
+ f;
2mA Vx+ ZVx 1 k!l Vo i' 
'-------<1_----..---_---_----(;:1 
.±Voc - (V, + I).) -T 
- k.
, ,,""'..-­
k... 
" +I'}..-VOC = 2­
v,_lVoe-"" V, -r V, 
'l - \),-12. -t-Voc:;. 4­oe 
3\)1 - 3\loc =- 2- 12-= -10 
Chapter 5: Additional Analysis Techniques Problem 5.69 
--
i 
Irwin, Basic Engineering Circuit Analysis, 10tE 2 
- \II -+:;. VO~ = 16
 
3'./oc == - I 0
2\/1- i. 
2.\/oe - ­
-"I -r I~
 
A=- ~- 30:=. 3
 
i 3 ] [-/ 0 ]
,3 -+16Cio'~ J~ 3 [~ 
v: -- t [-)0 -r +0: ~ V 
4-lt 
F-----~}----'---------, 
~ 2-Va.. 
kT 
',:-: 
_/0 VV~ :;. -:3 
;- (1"/~ 
t 
:3 v').:= - 10 
-r~ 
.3 
2--b --t-.E::. 
- 3k. 3/c.. 
l·', 
1~· 
~: 
Problem 5,69 Chapter 5: Additional Analysis Techniques 
I 
3Irwin, BasicEngineering CircuitAnalysis, 10/E 
;: II<.. 
.----fvW.---O--.JVV1,Ik...~~--_<O 
ik.. 
+­
, 
-,".-
I 
.ik. 
1---~OI----------~...__-f) 
,
, 
, 
Chapter5: Additional Analysis Techniques Problem 5.69 
Irwin, Basic Engineering Circuit Analysis, 10/E 1 
5.70 Use Thevenin's theorem to find Vo in the network in 
Fig. P5.70, 
Vx' - 1.2 - (Vo~ -2.V'A' ) -+ V;.../ T V"/..J~ VPC = ~ :J.lk. 
- t- t 
! 
r.\joc - 2. Vx' - \/'1- I -r 1'2- -T \foe - Vj..,' -'i 'Joe:: 4- f.' 
rV"f-' - 1'2.- 'Joe" -r 1-\f~ ~ V./ ..>.; V~ - Voc:: - 2-
Chapter 5: Additional Analysis Techniques Problem 5.70 
Irwin, Basic Engineering Circuil Analysis, 10lE 2 
3 \joe - 4-vx' =- - 8 
-'1-Voc -r 5v./: 10 l-
A... W'- 8 = 1­
;.1 
I
!
! 
[~<c J-i [: ;J[-,UI 
~ . 
i
: 
l"-
I 
L ~, 
~. 
f 
~ 
~ I 
f 
,
, 
! 
Problem 5.70 Chapter 5; Add~ional Analysis Techniques 
-
+
+
–
–+
–
+
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.76	 Use Thevenin's theorem to find 12 in the circuit in 
Fig. P5.76. 
30 
".' 
lA. 
VA 120 +
-
ZVA 
+ 90 
12V 
60 
70 
12 
Figure PS.76 
SOLUTION: 
v: 
)\11 
' ..:.'. 
I 
! 
I 
I 
..
! 
+ 
~t~~~~:~=~A CD 
. 13 =­ +Ip,' :: 4-1./ 0 
ax/'t 2-VA1 -;-G(~-:r3)+/:l.z;= 0 
31} 1'" 24-1/ -r btl - 613 "1-/2.1, :. D 
+011 - G13 := 0 @ 
12..0.. 
b~ 
SA 
Chapter 5: Additional Analysis Techniques	 Problem 5.76 
:.~ 
Irwin, BasicEngineering CircuitAnaiysis, 10lE2 
+6 7j::;; C;; 13 
13 ::: 'to S X/ 
~::. 13 :=. 0 
Voc .:: 12. V 
® 
i 
r-:, 
~t 
1~ ~ 3.I2.­
- h) 
\ 2....cL + 2.V/' )ll
-
-1 5A 
'.' 
i" 
, . 
6.n. ~.sl. VA -, 121:1 
p..v ]f
:3 
A 
'kc. 
e ]4 -1 4-1/1 " 
VI)
" 
J4.;, 4-~ 'I ~ 4-1/ 0.'-..," "~", 
-8 X,-I- .2 VA' " + G (1; - 13) f" s: 0J2:It 
31, f2+1, Tb.1./-bIJ ;-12.:IJ ;; n 
.
/!i 4-5"1, - (;13 := 0 ® ! 
~l13 -1,).=. /2- (j) 
Probiem 5.76 Chapter5: Additional Analysis Techniques 
3 Irwin,Basic Engineering CincuitAnalysis, 10/E 
I, ;;: ..2. r 
- .....!l)~ .;.J 
(;(13- 1?13 ) ;;: J2 
~.31 A
 
XI::; ~ (~,3' ) : 
ts: 
+to. -30g) :. /.;).. 32.A,J4;: 4-1, ~ 
-=- ~. g ~ -I, 2.3 1. =- J. D.g.g A 
Chapter5: Additional Analysis Techniques Problem 5.76 
Irwin, BasicEngineering CircuitAnalysis, 10lE 
5.77 Use Thevenin's theorem to find Vo in the circuit in 
Fig. PS.77. 
20 40 
so 50 
+ 
L..­ ..... .J----<;) 
, 
I c~ 
p
,', 
i 
Figure P5.n 
SOLUTION: 
. i 
I 
VOG 
I
-
4-SL 
i 
2.A 
I 
Ip, ')ZY" I2. 
~v 
VA}'" &(X/-13 ) 
~:'-r .!2f[ v..p..;t"~ : 12-;:' - 21+ ([) 
73 z: - 31ft -; -3(1.:;..-4) 
3:11 - 312. - 13 =:=- 0 ® 
Chapter5: Additional Analysis Techniques Problem 5.77 
Irwin,Basic Engineering CircuitAnalysis, 1DIE2 
3 XI -t ,-VA I -I- 8( X, - 13 ) s: 0 
3X j -t z. [BUt/-1S)] T [5 (~_13]= () 
32/ T 2.4-( Z-~):; 0 
'I eX'+X/ - <X 4- ; = 0 ® 
;.:': ~ 
Z=- at., 1,3 
~0 Jt~ :It ~ -3.2,4 
13 :=. ~ 3-bA 
'Ioc .: (1). -t s (i; - 13) t 1 
::: 5 (-~) T & t- 3 ' 2- - (.3. 6)) T 7 
V :: ~,2- Voc 
-;-. 
I.ll. .'. 
-c-: 
Problem5.77 Chapter5: Addltional AnalysisTechniques 
Irwin, Basic Engineering Circuit Analysis, 10lE 3 
~-1~: Z;.=-2.A (j) 
"" 1fi) ~ 't !'Jf"\:.I+ - 3': -:5 4t =.:> CT:. - 'l, ) 
3XI - j X.. - 13'" ~::. 0 0 
3:E 11" .2. VA II + B( XI -13 ) ,: CJ 
3X, 'T 2[8(1/-13)J1- is' (:£,-13 ) :; 0 
3X,'T f2it (.i,-:!j)-"' 0 
2".1-1- - ~t,1j:o1 
XI - Bier 13:, D ® 
Be X3 - 1,) +s( ~- :t).).= 1 
- g ~ - 51). 1" B13 7 ~--.I,::"o ~ @ 
0 
-.2­0 110 
o 
::1...I 0 -&j ~ 0 o I~3 -3 -I / '1S,r 1'1
-1$ -~ s 
r» 
I;> " ~J 11e104 :L'1 s: 0' 19 2.~-A -
-' .....: 6 I 1.19 ')..f' Ir'" ~:.$.e z: ".--....-_..,....",.-- c: 
'IDe. ,;).2- YI 4[.} ,SJ..~fr/1 :; 
- - ­
- '" l!.c, t)'41;..J'" 
,
I:.f: 
Chapter 5: Additional Analysis
Techniques Problem 5.77 
i 
4 Irwin, Basic Engineering Circuit Analysis, 10/E 
Problem 5.77 Chapter 5: Additional Analysis Techniques 
 
+- +-
2 x
x
-
–+
Irwin, Basic Engineering Circuit Analysis, 10/E 
5.82 Find the Thevenin equivalent circuit of the network in 
Fig. P5.82 at the terminals A-B. 
~; 
: 
i 
! 
i j 
i,,
 
!
 
I 
1 
\:,j 
, 
0-
A B 
Figure PS.82 
SOLUTION:
 
SA 5:fL1A 
+ ;2 Vf>t i212.S2.. ~JVA 
6Sl- 9'SL1"" 
:+-SL13 J ]'14-A I AB
e@e 
+ Vi-. -
VA :=.1~8:, 
~~ , 
'X ,,4i A- CD ~:: 4Xp ., 4:L, ® 3 
;2.VA ~ 5~ -t '1( 1).- 14-)
 
2.4-.:L I z: P·rI'). - ~ 1+-
Chapter 5: Additional Analysis Techniques Problem 5.82 
2 Irwin, Basic Engineering Circuit Analysis, 10lE 
~4-LI - 14XJ.,. -t-q~ ~ IJ ® 
3I, 1- 2.Vp. 1- ~ L~~ 13) -1-121 z: 0 
3J: + 2.4.1, -t(;'1, - GI3 t 12'It :; V 1 
4-5 X 1 - bX-3;' 0 ® 
~~ j;~ ~ e, 2.1)s:-A~ 4=- 81 KA 
V;..:; k,C13- s, ) r 'f{}3-I.,) 
:: ~ ( I - 2-1 IS-) t '*(I - 8 JI~.) 
,'. 
:-' 
::: 1>. Lt'fV
 
Rrrw :: Vx $ 8. 'il.f. ::- g·4i-Sl..
 
1"1 T I 
Problem 5-82 Chapter 5: Additional Analysis Techniques 
.
Irwin, Basic Engineering Circuit Analysis, 10/E 
5.84 Find I, in the network in Fig. PS.84 using Thevenin's 
theorem. 
, 
i 
!. 
4Ix 1 kO av, i 
+ Vx ­ t,i 
·'·'·1
.-..-"' 
':"1 1kO 1 kO 
1 kO 
6V 
10 
Figure PS.84 
SOLUTION:
 
, 
2.V'jo...'Ar'}.,. 
"\}.I
+- '1---
Ii 
"'3
 
b J,. ~I k 
i 4-1'x. 
V)<\ b-V2-
T; :::: V~-V2­
k 
~~ 
,,
, 
~V')/ 
Chapter 5: Additional Analysis Techniques Problem 5.84 
Irwin,BasicEngineering CircuitAnalysis, 1DIE2 
y -v. 
-t .2- ~ ... O 
It.. 
\/,- y2-r V3- '1/.2- ~ - 2. T +V.; -4-'1/... 
V1_Yj = I,';\.- ~Y').. 
V2-"" V:l-- V, or V,-- V6 ~ ~ 
V1-V2- 1" V3- V2,. -4\/,3 -;- +V). = - '2.. 
VI" '";. Y...- Vo:: 12 
\/:.- 't \j2,.- V, -r V2..- V3 :o ,I 
I VI 1" 2.'12.. - '6\13, :: -.2.I 
-' 
! VI -r ')..V2- - V3:: 12­
- \J I t 3 V ')... - \l~ ::. to 
'\j I ~ - 2.- ')...\1)-1""3\13 
- '2- - y;::r 3V3 1";>ri:- \j3 = l),. 
~ 't 2-'12- - 3'13;- 3V;,.- V,3:: b 
~v3~ 14 --. 
" ?J t:- 1-V 
5\1).. - 4- V&.:: 4­
~"\11.- - 'l.&:: 4 
va.:. 3,./.r- V 
\1\:. - 2- - ')..v... 1" &V~ 
Problem 5.84 Chapter5: Additional AnalysisTechniques 
Irwin, Basic Engineering CircuitAnalysis, 10/E 3 
= - 2.. - 2. (hi SOJot ::J.I 
:. - .2- - ~A·I ~ t 2.) 
-;: -\O-"~'"\"lOr-
.311 s:v= ---"'----~~~ 5 
I~~-
,,..\ '5 
aIr21~ ~~JV@ 
10/(''i)\~ 
..... 
::: 
V, 
-
Chapter5: Additional AnalysisTechniques Problem 5.84 
Irwin, BasicEngineering CircuitAnalysis, 10tE 4 
-:-. 
+1 2-II 
;{ - ­ k. 
", - 12­
k. 
VI TV1- I~ :::. 4", - +8- 2­
S'5.. J-V, 
/4v., '" !. 
.~ : 
'V ~ = VI - 11. s: 19 - , l. & i-v 
®
 
II' \ 1"1 " 
..... ..... 
"£ i­
, 
.\ ~ 32.­ , II 
II 
10 ::: 3).!ri I/s~t 
I l~-r 'ls:k.!, 
.ik.i ~\,S' .. 
.3.. 
:,:::. GIt.'1 To 
- I~ rr,AT 
,
i·
 
Problem 5.84 Chapter5: Additional Analysis Techniques 
–
–
+
+
+
+–
–
2
Irwin, BasicEngineering Circuit Analysis, 1DIE 
5.92 Use source exchange to find /0 in thenetwork in 
Fig. P5.92. 
6V 
,-----,.----(- +.-----. 
20 rnA t 61<.0 
31<.0 
4kO 
12kO 
_'::'~._.J 6 ill 
-: i 
10 s xo 
BV 
Figure P5.92 
SOLUTION: 
.,
, G 
, ~---'\rvv"'-----/"'" 'l":J.... , .--~-? )-----,
, 
-) y31<. 
+k. -=->
 6k. 
Chapter5: Additional Analysis Techniques Problem 5.92 
2 Irwin,BasicEngineering CircuitAnaiysis, 1DIE 
+ 4-1<... 
,...-----\ - +r---II/WJ,-----l..-----; 
ok.2-1lL. .6k.. 
r o 
-+-~ 
~ 
:;.J k. (3k)10= 
-
.2­
-9L "3k- A 
.'; 
Problem 5.92 Chapter5: Additionai Anaiysis Techniques 
i 
i 
Irwin,Basic Engineering CircuitAnalysis, 1DIE 1 
5.93 Use a combination of Y-a transformation source trans­
formation to find 10 in the circuit in Fig. P5.93. 
6V6k!l 6k!l 
-+ 
10 
6kf1 4kO 3kO 
" .,j 
.. j 4kf1
 
!
 
4k!l6kf1 2mA t 
6k!l 
Figure P5.93 
SOLUTION: 
6k.. 
~-Wv,.,..-----{ 
rJ-t. ~ 6k. ==) 
;~ 
1:-: 
C' 
i· 
1· 
i 
1. 
I 
t , 
I,' 
r 
i
 
~ .
, 
'-----\'" _.~-
I; 
t; ~ 
16l: 3lt6k. 6k.. 6k. 
'j 3k. 
.... ) ;) 
or 
+It. 40- ­
8' 
Chapter5: Additional AnalysisTechniques Problem 5.93 
, . 
i 
L 
Irwin, Basic Engineering Circuit Analysis, 1DIE2 
:..L ~ .. 
31<. ; 
...L 
3k. 
r~
I" 
i­!. 
i. 
i.
:;j 
"f­i . 
i 
~ .I 
I , 
i 
I 
i 
.. 
Problem 5.93 Chapter 5: Additional Analysis Techniques """ 
.1 
Irwin, Basic Engineering Circuit Analysis, 1D/E 
5.94 Find Vo in the network in Fig. PS.94 using source
 
exchange.
 
6k.ll
 
12V
 
.'.:'-"'1 
2k.ll
 
1 k.ll
 
6kfi 
+ 
2kO Vo 
-8 
2kO 
2kO 
2k.ll 
.''-
Figure P5.94
 
SOLUTION:
 
Gk.. 
<;; 
'-----it - t---­
\2. + It 
I~ 
t 
=-> 
12.. 
Chapter 5: Additional Analysis Techniques Problem 5.94 
·: ­
• <. 
2 Irwin, Basic Engineering Circuit Analysis, 10lE 
).j
 
-12. 
/2­
:1 
:,-·:·l
-'-1 
Problem 5.94 Chapter 5: Additional Analysis Techniques 
i 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.95 Use source exchange to find 10 in thecircuit in Fig. PS.9S. 
2k!l 
2k!l1 k!l 
6V 
6kO 
Figure P5.95 
SOLUTION:
 
zIt t 2./t s
 
~ 
~k.. 
.j 
j 
.: ~ 
I
 
I
 
,i
, 
1 
Chapter 5: Additional Analysis Techniques Problem 5.95 
.
Irwin, Basic Engineering Circuit Analysis, 10/E2 
, 
I
-I 
.. : 
. -.,,1 
2.K. 
t It 
2.(.Ik. 
1:: tQ - ..2=­
-k (~) k. 
f~~
',1 . i. 
.J.. 1 ..1rr,A :;:"~ - = e 2-
Problem 5,95 Chapter 5: Additional Analysis Techniques 
1 
· ·1 
Irwin, BasicEngineering CircuitAnalysis, 1DIE 
5.96 Use source exchange to find /0 in the network in 
Fig. P5.96. 
- /2.--?
=) 
3!:­
zk.. 
=) 
Chapter5: Additional AnalysisTechniques Problem 5.96 
Irwin, Basic Engineering Circuit Analysis, 10/E 
5.97 Use source exchange to find I. in the network in 
Fig. P5.97. 
4kO 3kO 
! 
i 
i 
3kO 4mA 6kO 4kO 1" 
.. 1 
;-:.)
.,
·'·1 10 2kO 
2kO 
12V 6kO + 6V2kO 
2kO 
3kO 3kO 
4kO 6kll 2kO 
. ;.' 
Figure P5.97 
SOLUTION: 
+l 
\~ 
=)
 
I 3h./~ 1'2­
.­
~'t. 
'k. 
! 
i 
! \'). ~ 
Chapter 5: Additional Analysis Techniques Probiem 5.97 
--
2 Irwin, BasicEngineering Circuit Analysis, 10JE 
,";.-i 
",-"j 
I 
i 
1 
i
-j 
i
 
I
 
! 
J]:::. I;J. ,(;)-/3) ( /8/ J~) .J-A
= 
:II 
1>1- ::O--t ­~ 
-
~;;..< 13k
. ­
".::.' 
-­34+1~ ~-r 1& /l~ 
-n -­
i 
-I 
! 
Problem 5.97 Chapter 5: Additional Analysis Techniques 
I 
-+
--
Irwin, Basic Engineering Circuit Analysis, 10lE 
5.1°9 Calculate the maximum power thatcan be transferred 
to R£in Fig. P5.109. 
6V 
6kO 
2kil 
3mA 
10 k!l 
10kO 
9V 
Figure P5.109 
SOLUTION: 
\jDe
-1~Vt&nA A B 
0-----r---Mi'vL--~ 
-t"1/ ­6k..JL.. iDl:...Q....
VDC- t-
21.Sl..GV t 10 k ,n, 
4·sv 
3mA 
YOe, .:: \81'"6- 4·0'= 14.5V 
~11 
! 
, 
---­
ek:s: A B
"H 
·v t 11 
21:..52 f e11 ~ s: 
lot. n­
,,, 
loJ::.5l_ 
->: 
I 
,,'col
.;",1 
:::>~l 
i 
I;: 
r" 
Ril-j ::. Gt. 1"" ( IO~)(IDl:.) . 
-/Ol+ I Dt 
-
II t.n.. 
.. 
Chapter 5: Additionai Analysis Techniques Problem 5.109 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
. i 
! A 
+I 
4.75 V II t.Q...I
I
-IS 
B. bYrnW 
-
Problem 5.109 Chapter 5: Additional Analysis Techniques 
~.- -
IrNin, Basic Engineering Circuit Analysis, 10/E 1 
5.110 Find the value of RL in Fig. P5.11 0 for maximum power 
transferand the maximum powerthat can be dissipated 
inRL· 
21<0 4k!l 
I
:<1 
i 
2k!l 
BV 
4kn 
RL 
+ 12V 6kl1 + 6V 
Figure P5.110 
... 
•• -_-.1 
SOLUTION: 
Voe. , "­
-
I 4-k..n.
'/4- \ 
J 
I I, A 
/?lV, 
4k...Q... 
\~..J 
By 
J 
~ ~e.v.l~~ 
~:. I2.V CD 
Vj = GV 0 
i 
i '1/+ - V1..:: 8V 0 j
:.:j
>!
, kCL@ ~~,k.: 
'12,.-1.2.. V'J- + V4--/2. 
.2.l -r -bit.. ..J. t. 
i', 
lr,
r:: 
+-Voe 
'13 t', 
bV 
Chapter5: Additional Analysis Techniques Problem 5.110 
2 Irwin, Basic Engineering Circuit Analysis, 10/E 
~~1 r V+ Jld..ol!: ~:; ILt·;tv 
;::. V4-~ ~ : It.,· 7 -6::: 8. q.VYOc 
HA AAA
.. 
b 
2lc-\2.. 4--L.JL. 
AAA Wvv 
~ 
~ Gt. n..,. 
i 
4-k.~ 
UA 
,"-~ 
tA~JC.. 
• 
.?l:.~ 
A A AA 
VVV 
O. '11l:.Jt.., 
e111 :; 4-. ~; I::)L.. 
Problem 5.110 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, iDlE 3 
A
 
4-.D:J-rnW 
:"-' 
i. 
l ; 
! 
j: 
f 
I 
~i;1j
, 
,.e:
i 
Chapter 5: Additional Analysis Techniques Problem 5.110 
---
Irwin,Basic Engineering CircuitAnalysis, 1OlE 
5.111	 Determine the valueof RL in Fig. PS.lll for maximum 
powertransfer. In addition. calculatethe powerdissipat­
ed in RL under these conditions. 
150 
30V 
200 
+ 20V 
Figure PS.111 
SOLUTION: 
Voc 
200 
Ioz, 
Vr- 30 z: - 4-
Is­
- eo #:- - /203'1/, 1"" ').."1/ 
i 
t	 
5V; ::.. -60 
il	 '1/, = 
-!2V "l~
····i V, - 2-0 = 
- /2- - 2.0 ::::. - !3.2. V	 
. 
1 
1	
­i 
, 
~ 
r,. 
I 
I. 
Chapter5: AdditionalAnalysis Techniques	 Problem 5.111 .c­
2 Irwin. Basic Engineering Circuit Analysis. 10/E 
) 
AA' 
v 
I!'-.J2.. 
JJ. I\~A 
·v 
:<'05l­
"<> 10 Jt. :~ ~ 
A[» Jt ~ l3 
". 
l~n. 
<' ~ ~ $. " ~O.J2,. ~ ~ ~ )05l­
R1'1 z: C'O XIS: )
• 
IO+Jr" 
t
 
-IbV 
-
Problem 5.111 Chapter 5: Additional Analysis Techniques 
-+
=
Irwin, Basic EngineeringCircuitAnalysis, 1OlE 
.> .., 5.115	 Find the value of RL in Fig. PS.llS for maximum power 
transfer and the maximum power that can be transferred 
toRLo 
en 40 
12V + 2A t 
:. ~ Figure P5.115 
" 
~'~SOLUTION: 
:=-c'VDC 
-	
- I2Vft
'5J1. 4->L A 
~ ~	 1" 
~	 -VA' + 
~ 
I').." Voe	 i 
,	 6 ('~XJ,);; )~VVA = 
T 12­t; - ~V/ 't l+) (:..) T V~ 
i -	 I
I 
i	 - .3 VA I -r 2.D 
, I 
,I l~,
.. ·l	 =' 
.:.: .~ 3 (16}t-:).O	 [c 
i 
-
68\/	 ~ 
-	 i 
~ ••
Chapter 5: AdditionalAnalysis Techniques	 Problem 5.115 
--
2 Irwin, Basic Engineering Circuit Analysis, 1OlE 
~c...
i 
I, 
--. - "l! 
.,'- ­
12.= I?-"Ix -
II 
'VA 
II 
;: -eIx 
12-1;( ­
-r 
.J-X 
::: /z.!x r 
=- ~@ Ix 
=- /2- _
-ZJi­
'.' 
:i~ 
, 
l·­r, 
f'·l' 2A 
r.
t-:
n­2. Vir :.:.
';. 
Z(- &I.J 
161)( 
~A 
9-
1,X. 't 2. ;:. rJ,LtBA 
Voc ,. 68 ::­
• 
~,'-t~:hr... 
Problem 5.115 Chapler 5: Additional Analysis Techniques 
Irwin, Basic Engineering CircuitAnalysis, 1OlE 3 
.------~AM__--~L----,
 
~~Jl.,. + 
.,34-V 
-
:;. 
-
-
Chapter5: Additional Analysis Techniques Problem 5,115 
Irwin,Basic Engineering CircuitAnalysis, 1OlE 
5.116	 Find the valueof R1 in Fig. P5.116 for maximum power 
transferand the maximum power that can be dissipated 
in RL• 
50	 50 
Ix 
100i 
I 
..,:::.	 u, + + 12VI 
RL 
Figure P5.116 
SOLUTION: 
VOc
- 551- 5S/... 1/ 
A 
is 
~ 
5 Ix I	 -+- si / )< T )2­
e~'~ -/2­
I	 - /2­I x.	 z: g 
, / 5:7 ' ...... 1:2- -Voc	 - -'"X I 
12V 
,
2IX	 z: D 
-I'~ A 
s(-" c}t/).. 
Chapter5: AdditionalAnalysis Techniques	 Problem 5.116 
.
I 
:': :; 
Irwin, Basic Engineering Circuit Analysis, 10/E2 
".: 
2-~. "t- p,. 
ID.Q..
1, 1yJ I 2.. V
! 
l·~· 
,-'·'1 
/) J 'kc.. f'
B 
Sl, -t }O (1, -1)..)~ 2Ix II:; zZ
 
)S,-z) - JO~ ~ ;2.I)..
 
/};j I r;:; /1....1 2­
1t:::: 0, 81).. 
5Jl- -r 1'2-'" J 0 (Z~ ...:r,).:= 0 
-/oLI t Kl).. ;:'-/2­
-r ] J-
J I ::::­ o. ei; 
~t- :; 1, - 7~ 
:::: -r ], a':f -(-I· 9- /) 
(j . .sf-A 
Problem 5.116 Chapter 5: Additional Analysis Techniques 
-i 
Irwin, Basic Engineering CircuitAnalysis, 1OlE 3 
-
-
.-.! +­
.:·:-:1 
.....! 
B 
(~.~.r)~
_---or 
/3. 2-~ 
:::=- O·382.W 
• 
-.-.-.! , 
, .'~ 
,.. 
, 
i 
i 
i 
I 
I L 
1 
r,. 
~. 
! 
i, 
Chapter5: Additional AnalysisTechniques Problem 5.116 
--
Irwin, Basic Engineering Circuit Anaiysis, 10/E 1 
5.117	 Find the value of RL in Fig. P5.117 for maximum power 
transfer. In addition, calculate the power dissipated in 
RL under these conditions. 
4k.ll	 4kfi 
+ 
6kU 
12V 
t u, 
:,-"
Figure P5.117 
, 
" 
SOLUTION: 
\joe
12-v'	 t: 
A 
B 
+ 
I),V
/ 
'- '!')( 
/ 
- 2"""A1"-''f.. - , , , 
'.'. -Joc 
'0" 
!:'-: 
f' 
Chapter 5: Additional Anaiysis Techniques	 Problem 5.117 
--
Irwin, Basic Engineering Circuit Analysis, 1DIE2 
V, 
t 
1/ 2mA 
~ ­
2~ 'I 
1/'= V,
GE 
'If - /:2- -r VI ~ 2. J;..' + 2(">")
 
- +-1<.. Gk.
 
V (I -r J-) z: 2- (VI) t 2M -t,3 M 
I 4~ (,k. bj( 
V J. - J-) ;:- s: A I ( 4k.. 6 "-
V, _ {'('('I 1+ V, ~ 160 V
 
I')..k...
 
'h.~ - ~ - 2.ff) s: 
~l:. 
to 
; 
Problem 5.117 Chapter 5: Addijional Analysis Techniques 
I 
Irwin, Basic Engineering Circuit Analysis, 10/E 3 
J:" 
.... 
;A ,. 
k 
I: 
, ' .2.bL 
+ ~):.Jt.RI" -
IGv g" 
~ 
p­
-
i 
>j, I 
; I' 
, 
~:Chapter5: AdditionalAnalysisTechniques Problem 5.117 
--
Irwin, Basic Engineering Circuit Analysis, 10/E 
S.118	 Find the value of RL in Fig. P5.118 for maximum 
power transfer. In addition. calculate the power dissi­
pated in RL under these conditions. 
61<0 31<0 
+ 3mA t 
O.5Vx 
+ 
Vx 2kO12V 
Figure PS.118 
SOLUTION: 
\foe
-' 
~I 
'Ix 
t 3mA I 
,\{!
;( V; -J2-	 Vy. 7, IV8r0A 
­~T
-2-k... 9lc.. 
~	 
--
T '- 1),,- 1· ) o.r~4mA ~)( ­
-	 l 
i­q/c 
I	 o,::'i- I 
~v; ... -3k. 1x,1 -T Yx	 r, 
i:~: 
t·~3k.-l;< I -+ c.s v){ I r~' 
-r od:{:r· I) ~ ~ 3 l( D •~Lt Lt YYI) 
-
S-,/3V
-
Chapter 5: Additional Analysis Techniques	 Problem 5.118 
--
Irwin, Basic Engineering Circuit Analysis, 10/E2 
A
 
(3 
'2.. 
P:: ~.~q) 3. o8/Y1 L.0 
,), )~ Ie.. 
~ 
he..
- bk...Jl- :J./?) 3l:.. Jt.,
-
1t+ 
,~\j 
A 
-I­
lsc.. 1/;..11 
£5 t f\')A 
or V, h 
- D.l:' ;A 
-
4-.~\1 
V'f II 
1" -~~ 
/I 
V'A s: 
J,.(- L4.ri'jp'J."~ 
6k:... 
~ L!2.-­
"V - D'S-v'>(X 
3 Ie.. 
Problem 5.118 Chapter 5: Add.ional Analysis Techniques 
i 
Irwin, Basic Engineering Circuil Analysis, 10/E 3 
1tt.,.:Ie s: /. (;dl.(m -+0' 7.f~ 
-
s . B7JmA
-
.".'
·····1
;'.', 
-
'.~. 
." 
i 
i 
i 
..:.I 
Chapter 5: Additional Analysis Techniques . Problem 5.118 
l 
Irwin, Basic Engineering Circuit Analysis, 1DIE 
5.121	 Find the value of RL in Fig. P5.121 for maximum power 
transfer and the maximum power that can be dissipated 
in Rv 
20 40 
fA 
10 t·".;[. 
80 
2VA 
RL 
9V 20 
Figure P5.121 
SOLUTION: 
VOe. 
",," 
151., 
1Pr I 
+ 'J-VPr / 
8Jl... '}-1.,/ A­ '!> 
- Y/'r 't+- Voc- ­ -
6~1 4-\1 
+ 
V~;;. 8C2.J./f');;' i61/ir 
:?(1p.. I -r 2- ) -t 1(1.,/-1"').-)-/61'/-2'1'/-0 
.< 11>< I -r 4- -t II.pr) +.2. - 4-8 1ft I J:'. 0 
+f'~J ;;.- b 1p.'::. 6!+r'; Q,IB.E3A 
V ~ lC.1t.'-t 9+4-: /t,(O./3B?d}tJ'3::: 1S:·/3voc 
Chapter
5: Additional Analysis Techniques	 Problem 5.121 
2 Irwin,Basic Engineering Cincui! Analysis, 10/E 
~ 2.JJ..
 
1-I J ?-A 
I.Q.. 8~ 
-. i 
;-'. "Pr JI~ g C~ ,.2;)-: "1
- "" II ,.. 
I 1~ J 
~ ettl~' 
12,. ~ -:J-A (]) 
14­ - 13 : 3~ ~ = 3(1}..-1,) 3~ - 312­ - t 1" z., ~ 0 (3 
31, 't Z-vA," ..-r get:, Fr.) ~ 0 
: 2> ~ l' } c: l:t: -~) 1" ~ C~: .. 13) ~ 0 
! 
, 
I .3 1; r 
0<111 :; 
d{ It[11-!3); 0 
~ '-t J,3 t; :: cS/~ J; @ 
8(13-1,) -t.2 ~ ;. 9 
-gl) ~81.3 -t2-~~ 1 6 
~0 t: i, 7'~ 
X4-:;' 8 ,32,4 
~ 
I 
i 
I 
! 
r 
Problem5.121 Chapter5: AdditionalAnalysisTechniques 
1 
­
Irwin, Basic Engineering Circuit Analysis, 1O~E 3 
L, -1;). := S' 32. -e..-~) 
}o,SJ- A 
__.-n .... 
.....~ .....-.."'''''''' 
If'./3 
- "'Oe. - ~ I, ~75L:P­£:111 - ­ /0'.,3)....
-:HL.­
+ z-: r"o, 
-
-
..' 
: :.~ -" 
,
, 
~
1--
Chapter 5: Additional Analysis Techniques Problem 5.121 
i
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 5: Additional Analysis Techniques Problem 5.113 
 
SOLUTION:
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 5.113 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 5: Additional Analysis Techniques Problem 5.114 
 
SOLUTION:
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 5.114 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 5: Additional Analysis Techniques Problem 5.115 
 
SOLUTION:
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 5.115 Chapter 5: Additional Analysis Techniques 
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 5: Additional Analysis Techniques Problem 5.116 
 
SOLUTION:
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 5.116 Chapter 5: Additional Analysis Techniques