Download solution Chp8 Basic engineering circuit analysis and more Exercises Electronic Circuits Design in PDF only on Docsity! ) A Download any solution manual for free All Departments , All Subjects www.myUET.net.tc CREATE YOUR OWN WEB PAGE VISIT www.myUET.net.tc Download any solution manual for free All Departments , All Subjects www.myUET.net.tc Irwin, Basic Engineering Circuit Analysis, 9/E
8.12 Find the impedance, Z, shown in Fig. P8.12 at a
frequency of 60 Hz.
10 mH 20
> 10 uF
Figure P8.12
SOLUTION:
10 mH 20
= 10 F
2.5 j(341)(}0m) = j3 772
2c] 5 265-25 2
30377) (10H)
j 277 Dlr.
-
at,
nN
an
oO
NS
S
Z2 = [ 2 -j265.25 | + j 3.77
Chapter 8: AC Steady-State Analysis Problem 8.12
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8.25 Find the frequency at which the circuit shown in
Fig. P8.25 is purely resistive.
Z—> $in 5 mH > 1 mF
Figure P8.25
SOLUTION:
= 1 mF
= 447-2) xaklfZ
Jce = JS5m(im)
Fo ww = 447-21
IL 2m
Fs Th2 Aa
Chapter 8: AC Steady-State Analysis Problem 8.25
Irwin, Basic Engineering Circuit Analysis, 9/E
T= 06 £53-26°A
L(t) = 06 Cox (10004 53°26") A
Ve = (0°6 £-53-26°) (-j 14-93)
<
"
Ve = 8°96 /-143:26° V
Ve = (0:6 453- 26°) (20)
v
Vp = Ite 5326°V
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q0°
mes >
7 —56-26 Re
Ve
143.26
7
Vv
Vs
Problem 8.29 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.33 If v,(t) = 20 cos St volts, find p(t) in the network in
Fig, P8.33.
30 05H
re
v(t) @) tHE ©1093 0.02F Vo(t)
Figure P8.33
SOLUTION:
Zi, = j5(05) = j 2-52
2.7 JCS = p52
Zz = a) Tj Jo
JC5)(0-02)
WMWA- Tana"
3 INS A |
. toe
20 Lo"V jv 3 jon yar
z= (-jlo lite) His
= [ ett) Ilys
lo- J 10
J = Pot Ztyse nl
Chapter 8: AC Steady-State Analysis Problem 8.33
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8.38 Find é,(t) and i2(t) in the circuit in Fig. P8.38
10.cos (25f + 20°) V ®
50 0.01 F
\W\——+ t
in(t) i(t)
80 mH 40 f 200 mH sy 20 cos 25¢V
Figure P8.38
SOLUTION:
Zy= j 29 (8om) =j2-2
Zu= jQs)(vom) =j52
Ze= | = 7) 4
so 4 bet ond
WW JN _
t, “JUD T,
Ip £20°V @) jn Yn j5D *) 20 LV
50
AWW it
_ |; =
102 20°V Ti y 34? 71,
©) j22 un f j52
T/=0A, Short across U2:
T’ 52
104 20°v
Chapter 8: AC Steady-State Analysis
Problem 8.38
Irwin, Basic Engineering Circuit Analysis, 9/E
a °
T's 102o = 156 [-1466A
(j2 l|-]4) +5
tq! - 56
* ap" aa “)
Ty’ = 3:12 1-16. 66°A
Wt iy
= Zz" “/** 3)
40 js © 20]o°v
Problem 8.38 Chapter 8: AC Steady-State Analysis
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8.46 Find the frequency-domain voltage V,, as shown in Fig.
P8.46.
+
-[§ 2 >R ions V5
0.525°A OD jen
US
Figure P8.46
SOLUTION:
To
~
05 L25°A zl 18-2 Mo
y= j2u-J5 = jr (-V5)
-j5 tyr
Z= 322 2A? 2
iz 3-33 290° Coser .
3-332 90° +10
Tr = onsge 7658 A
Voz Io (01582 96°58")
Vo = 158 29658 V
Chapter 8: AC Steady-State Analysis Problem 8.46
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Zr = 2}Iy) = 2 ()
2-J)
Zz O84 2763-4292
j2n
a mn
oO
’ ~64- t
lo 73 2-63-43v an BVp
Vo = 2. (10-73 2-63" 43°)
22+ OP4L-G3 43° .
Vo = &L-90:)°V
em Tne,
$ 0
Problem 8.54 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.62 In the network in Fig. P8.62 1, = 4 {0° A, find I,.
193
-j1.0
2/2 A
19s
Iy
wfrv 310
Figure P8.62
SOLUTION:
kcL at G) : I,= I,+ T,
Vi-Vy = V2 =
1 JI -JI
JI (W-9)) = V,-V3 7 Vit Va
“jiy, +j1W+V, 7 Wy = 0
Chapter 8: AC Steady-State Analysis Problem 8.62
Irwin, Basic Engineering Circuit Analysis, 9/E
KL at@: TT, +2l0° =I,
Y-Y +2L0° = —Vy
!
A- Vy JIC Vs-Wy ) = SW
Yo -jlV2 + (-1452) Vy =9
Vu = 1( 420°) = 420°V
My t+ C-14 §2) Wy = 12240°
Vv) = 12£40° - Ylo® (-I4jU
Vv, = 366 £ 45° Y
Ui - Vy, = 2L0°
Problem 8.62 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
3V,-V2 = 102 180°
JIG, +(-INDV = 4290"
VM = 822 L-179%3P Vv
VW= O55 L-563°V
To = 3-234-1773°
{
To = 3-22 21177:2°A
Problem 8.66 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.68 Find V, in the network in Fig. P8.68 using nodal analysis.
L yi
WW t JW
20 ito 10
j2a
12°V i 2/0? A ) 2 ¢ 4fov
ee V5
Figure P8.68
SOLUTION:
J,
1
VW iv
20 -j10
12/0°V © 2/0? A
KcLat O: I,+T, = 240°4+T,
220 - Vy + Y2ZO™ V, =2L04+ My
2-jI ) rts
Vs 4+ M +YW = 122o° +4z20°-220°
2452 2) 2-))
D> ar Zba-uy®
65 LZA-7M®
Chapter 8: AC Steady-State Analysis Problem 8.68
Irwin, Basic Engineering Circuit Analysis, 9/E
VY, = 437 2212°V
T= YW = YzZapee
2432
2452
T= 155 2 -23°8°A
Vo = 20155 2-23-7%)
Vo = 212-2328 VY
Problem 8.68
Chapter 8: AC Steady-State Analysis
2 Irwin, Basic Engineering Circuit Analysis, 9/E
-}!I, + jrT, = 620°
-j2 7, +(2432)T, = 12245°
T, = 1039 2115-26°A
T= 325 L2157R5°A
Vor 2(3 25 L157°5°)
Vo = 65 2 157:5°
MRAM S AGC
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Problem 8.77 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.82 Using loop analysis, determine V, in the network in
Fig. P8.82.
4f°A ad 102 = 10
‘ 12forv
ia
rn €>)
TS +
310 2fea S20V,
Figure P8.82
SOLUTION:
KCL I= It 7,
T= 34-
Kee: p= T/T,
I JI,7 J,
kyet 220° + | (-T) SIC T)=9
4{3,-T,)-J! I, = 122190"
Chapter 8: AC Steady-State Analysis Problem 8.82
2 Irwin, Basic Engineering Circuit Analysis, 9/E
-T,t (FD T= [22180°
I, = 420°A
(I-j) T, = ladive +42o0°
TL = 566 Z-!35° A
kve: I(-T) - JI, +27 t (I) til =o
{L-T,)- JIT, +2Ty =F T, 4 jl(T,-T, =o
(-I-S) + (IT + Citi) Ty +2 Ty=O
Oi) 420°) FC Ij) (5 66 £1359) t( It) +20,:0
(tj) Ty t 2Ty = 12°65 Liga
KeL: Ty t2lo=T,
yh- du = 2Lo°
(I4j!) T, +27, = 12°69 218° 43°
T,- Ty = 2L0°
T, =5222-4-4° A
T= 3:23 L713 A
Problem 8.82 Chapter 8: AC Steady-State Analysis
2 Irwin, Basic Engineering Circuit Analysis, 9/E
KeLat@: T= ith
WM = Vi + Mo
“JI | J
—
GR = TCW-W)-\”
V, ti!" JIG =o
Vo —2V, = YL1F°
UV, #+(-1 +52) Va ~J2V3 = 4490°
Vi + JIG, - JIM =O
V2 — 203 = YL190°
V, = 2:53 2 161 579V
V, = 537 2-63-43°v
V3 = Y 2-368 TY
Voz Vv 6
Ver YWL- 3677
Problem 8.87 Chapter 8: AC Ste:
Hee OS AMEKSZ- SSS
Irwin, Basic Engineering Circuit Analysis, 9/E
8.88 Find V, in the circuit in Fig. P8.88
al,
ji n=
Figure P8.88
SOLUTION:
kecat@: 2%, +1,t+ti=0
= Me
-j\
aly y-Vy +tY¥=0
J! | -jl
rv, = ji U,- V2) + Vv; st)
QjuvVut jin, =o
Chapter 8: AC Steady-State Analysis Problem 8.88
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Vi= 12Loev
(2-5SDY, tJICi2Lo°) =o
VY = 39 L£71-69V
kc. ot @: JT, +25, +2L0°=9
V-V +2 [we +2Lo=0
2 =i
ndor- Vy +2 PEbas") 4ole° =o
2 “J
Y= 308 L8-96°V
Vo = Ve
Vz = 30728-96V
Problem 8.88 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.101 Find V, in Fig. P8.101 using Thévenin’s theorem.
na
tom.
12fev(4) 320
¢ 20g,
=j10
16frv o @ 2/oa
Figure P8.101
SOLUTION:
Mt (=! - 1)vt Voc = 210°
-jl J! 2
UV, = —16Lo°Vv
Vo. Y= 12L0°
Chapter 8: AC Steady-State Analysis Problem 8.101
Irwin, Basic Engineering Circuit Analysis, 9/E
Voc =
Vo.= UHZ1%0°V
12Z0° + 1621 80°
Zryr 0-2
2m jin
eG +
Voc= YZ1 80% 702 &V%
Vo = (Yzise)
sire i
Vp = 3:6 £15342 V
Problem 8.101
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.107 Find V, in the network in Fig. P8.107 using Thévenin’s
theorem.
1
a
OV
)
i
a
G
2Vx< | 19 iN=RV, $10 V3
Figure P8.107
SOLUTION:
-jrxVy —jiV, +M=O0
«jl%, +Cl-jrayR=zo
U,-Vy = 12L0°
Here OS AMOK SSS
Chapter 8: AC Steady-State Analysis
Problem 8.107
Irwin, Basic Engineering Circuit Analysis, 9/E 4
8.108 Find the Thévenin’s equivalent for the network in
Fig. P8.108 at terminals A-B.
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B
|
Figure P8.108
SOLUTION:
(1451) G -sINe = 42-90
KCL ot @: TtT= 2VYy
Chapter 8: AC Steady-State Analysis Problem 8.108
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Voc “WV + VV _ 2[Vo.-V J
=jl
Voe 7 “a JIC v,-) = ~pl Tod
-j\NV, +(-1-JDM% + (1452) Yor =O
kel ot @: T= YLo°
Von ~V2 = 4LO
I
—H t+ Voc = YL-90°
(451) Go JIN, = Y2-90°
~j1V, + (-1-JI) Me +( 42) oe = 0
—V, TW. =44°9e
WY = &L 190° V
Y= 794 ZNUGE OV
Voe = 5°66 ZI35°V
Problem 8.108 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
Ty ie YLO0° ~ The
Kve: ICI) -JIGI,) + 51CT,)=0
T-T, +il(420°-T) ti! =o
-T, +(4i)) Th -s Tse = 42-90"
KeL! 2Vxy +Ic= Th
= Sl C420%~ Fe)
-~j20 420° -Tye) + Tee = TZ
Chapter 8: AC Steady-State Analysis Problem 8.108
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Cm > a. * 354 22-12 “12°
i 120°
Qty = 354 LPB TD
Problem 8.114 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.115 Find the Thévenin equivalent for the network in
Fig. P8.115 at terminals A-B.
WW oO
10 I oA
10 =
aly <F
it af
B
oO
Figure P8.115
SOLUTION:
I,
-)) C201209-V, J -31 CZ) = Y
Chapter 8: AC Steady-State Analysis Problem 8.115
Irwin, Basic Engineering Circuit Analysis, 9/E
VY, (1s) = JN(2) C1L0% J (120°)
VY, = 212 2-45" V
Zep = M= 22/-45°
t 120°
Zeg= 2112 L-45° 2
Zey = ps5 -jhd 2
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Problem 8.115 Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E 3
—V, + Wy = 1220?
v, — JW +52 +(-142)Vy - flu, = 2 /- Gor
Ve i 21ya7°v
Vy = 35°38 Z- 137-39
V, = 174-727-137 3°v
Vy, = 12-04 f-94-7°V
Ve = 2554 <2-4q Rey
= 35-38 Z-137-3° — 17-7 £-137-3°
I
To = 11:7 2-187-3°A
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Chapter 8: AC Steady-State Analysis Problem 8.123
Irwin, Basic Engineering Circuit Analysis, 9/E
fia
KCL ° I5ti= Ty
ReL! 4, ta*= %
Ket: T,= 7, the
Ie = E,-L,
KCL > I’t T
re 5-G
KCL L+h= 15
yy = Ts ~Ie
Kkve: ICT) +1 CI) =2Nx
Problem 8.123
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
Vx = TCi) = Ty-Ts
T,-T tty -Is =2 (Ty-5s)
kvet = (-1) t IC T,) = 12 L0°
{iy -Ts )+ T5-K = lelo
[ae ee]
vet 12") 441 ( Tg) 41(-t =
7h tit -( I5-I)=9o
KCL + T,t 220° = Te
Ta = 2x
7, =2( 15-4)
I, -27, +2250
Nee OS aMEKS SES
Chapter 8: AC Steady-State Analysis
Problem 8.123
Irwin, Basic Engineering Circuit Analysis, 9/E
Te
Q ~ 2:0000;
6
0
12+ 0000
O -—2°0000i
Yu
—5-5000 +4: 5000;
— 26-0000 — 240000)
—13:0000 ~ |?’ 0000;
=150000 — 42-Q000%
16-5 000 — 19*5000/
Lo=
— {20660 — 12:0000/
Aer oe OS AMOK SESE
Problem 8.123
Chapter 8: AC Steady-State Analysis
Irwin, Basic Engineering Circuit Analysis, 9/E
8.126 The network in Fig. P8.126 operates at f = 60 Hz.
Use PSPICE to find the voltage V,
19 3 20 > jin
2/°A
20
Pot
*
j2a +) 12/ev 31 ay,
Figure P8.126
SOLUTION:
F= Go Hz ) lo= 217 pack Ik
ZU= jw
l= ZB = Je =5:3mH
Jw J(377)
Z2co= sethares
Swe | - _|
sa, Jam)
Chapter 8: AC Steady-State Analysis Problem 8.126
2 Irwin, Basic Engineering Circuit Analysis, 9/E
C= 2:65 mF
Re Sr c | 2-6 5mF
R) \
R
"OY _ wiv
ET VI 2
QS!
L 5’3MH
PSPICE Schomotc Diagram
ae: AC ANALYSIS
FREQ vMC$N_0004) vPC$N_0004)
6.000E+01 4.3686+00 1.018E+01
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Problem 8.126 Chapter 8: AC Steady-State Analysis