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Irwin, B asic En gin e e rfn g Circu it An alysis, 10tE
,:_ g/< .rz_
&= 2r<ÿ
Vÿ,= gv
SOLUTION:
.... - J _
1''o j
Z
...J
i
€
w
&÷aS_ "
-- , , , ,
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
/
goooÿ-20oo¢ ;oco
Problem 5,
Chapter 5: Addttloÿfÿl Ahaÿysis Techniques
lrryin, HaEic f;rqine€ring Sircedt
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lo
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lrwin, Basic Engineering CircuitAnalysis, 10ÿ 1
5.32 Use Th4vetÿin's theorem to A Vo in [her network in
Fig. P5.32.
b
SOLUTION'
Vÿ 2...
t
÷
iÿ'r ÿ P' //Pz
_ 4x9ÿ;_ 4V
,,--..
,.....--
)ÿ,. , vÿ :
t
/ \
R/R2_
Chapter 5: Additional Analysis Techniques
Problem 5.
L
0
Iÿr ÿ; AÿIÿ Analÿis Technk:lueÿ
Pmÿem 5.ÿ
CA-
_ J,
Pÿbÿ 5A0 Chapter 5: ÿitÿorÿ Analysis Technÿqueÿ
Irwin, Basic Engineering Circuit Analysis, 10!E
in Fig. I>5,54.
.... gl ÿ ÿ,,
4--
b
Rze - 2Ko.._
..................... ,#,, TI i
" '". "t'.".':: '" .::'it'tttz::zÿ ......
o "ÿo,
5
:z
....------..-.
/ÿ'2._
Chaplet 5: Additional Arÿaly.ÿis Techrÿ1quss Prolÿlem ÿ;,ÿ4 '
9
9
In, Arm, Bÿsic Engbeeffn9 Grcui: Analys!,ÿ, lOtE ÿ
lg,3 ÿ .2. /ÿ_ ÿ-L_
W "" ÿ:LVs, c.
2.. ""
SOLUTION:
,1 R2..
'Vv L
Vr :ÿV
'VÿI = Vvÿ = /2 V
--.ÿC R z ti'z.
2 !ÿinÿ Basic Engineÿirlg Cirri .Aÿlysis, 10/E
.+.ÿLt _, .3¢o ,
Iÿ: Ik
½ = 25o m
Problem 5,63 Ch;ÿer 5; Additional Analysis Technkÿes
