Electronics and Communications for Scientists and Engineers, Second Edition [2nd Ed] (Instructor's Solution Manual) (Solutions)

40 Pages·2020·3.09 MB·English

by  Martin Plonus

These proofs may contain colour figures. Those figures may print black and white in the final printed book if a colour print product has not been planned. The colour
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B978-0-12-817008-3.09989-7,09989
Instructor’s Solutions Manual
c49945
Solutions to All Exercises to
Accompany
Electronics and Communications
for Scientists and Engineers
Second Edition
p5005
by Martin Plonus
p5010
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
Date:19/11/19 Time:19:14:03 PageNumber:1
typesetter SPi. It is not allowed to publish this proof online or in print. This proof copy is the copyright property of the publisher and is confidential until formal publication.
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B978-0-12-817008-3.09989-7,09989
2 Instructor’s solutions manual:Solutions to allexercises to accompany
Chapter 1: Solutions to problems
s0010
Ð
o0010 1.1 W:m‘2, F:m‘, F⋅d‘¼m‘⋅‘¼m‘2¼W
t2 t2 Ð t2 t2
p0020 V :W, E:F, V ¼(cid:2) E⋅d‘¼F ⋅‘¼m‘ ⋅‘¼m‘2¼V; can replace Qby It
Q Q Q Qt2 Qt2
o0015 1.2 E¼5V/0.001 m¼5000 V/m, F¼QE¼(cid:2)1.6⋅10(cid:2)19⋅5000 V/m¼8⋅10(cid:2)16
Newtons qffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o0020 1.3 IntegratingF¼matwice,weobtaint¼ 2ms¼ 2⋅9:11⋅10(cid:2)31⋅0:05¼6:9⋅10(cid:2)8 s,where
F 1:9⋅10(cid:2)17
F¼EQ¼VQ/‘¼12 V⋅1.6⋅10(cid:2)19/10 cm¼1.9⋅10(cid:2)17 N.
o0025 1.4 V¼RI, I¼V/R, R¼V/I
o0030 1.5 P¼VI, P¼I2R, P¼V2/R
o0035 1.6 V¼RI¼4Ω⋅1.5A¼6 volts
o0040 1.7 Cross-sectionalareaof wireis A¼πr2¼3.14⋅(6.5⋅10(cid:2)4 m)2¼1.33⋅10(cid:2)6m2.
Using ð1:6Þ ‘¼RA¼4Ω⋅1:33⋅10(cid:2)6m2¼5:32 m:
ρ 10(cid:2)6Ω(cid:2)m
o0045 1.8 From(1.6) R/‘¼ρ/A¼1.7⋅10(cid:2)8Ω-m/2.081⋅10(cid:2)6m2¼8.17⋅10(cid:2)3Ω/m
o0050 1.9 P¼VI¼300⋅220¼66,000 watts (W).
o0055 1.10 P¼I2R¼(5)2⋅10¼250W
o0060 1.11 V¼IR¼5⋅10¼50V. P¼V2/R¼(50)2/10¼250W.
o0065 1.12 W0¼I2RT¼(4)2⋅5⋅10¼800 joules (J).
o0070 1.13 1 hp¼1000/1.341¼746 watts; 1BTU/s¼1000/0.984 ¼1055 W; 1 cal/s¼1000/
239¼ 4.18 W.
o0075 1.14 1 kWh¼3,600,000 W-s.Since 1 W-s¼0.738 ft-lb, then
1 kWh¼3.6⋅106⋅0.738¼2.66⋅106 ft-lbs.
o0080 1.15 Heatrequiredtoraise250gofwater90°CisH¼90⋅250¼22,500calories.Alsofrom
(1.9) H¼V2T/RW-s ¼0.239V2T/Rcalories.Therefore, 22, 500¼0.239(110)2T/15.
SolvingT¼22,500⋅15/0.239⋅(110)2¼116.7s¼1.945 minutes.
o0085 1.16 RatingofheaterisP¼IV¼(120/10)120¼1.44kW(cid:3)8(cid:3)24(cid:3)30¼8294cents/month.
o0090 1.17 12V(cid:2)9V¼3 V¼V .
R1
o0095 1.18 1 A(cid:2)0.5 A¼0.5 A¼I .
R1
o0105o0100 1.19 (a) W¼Pt¼VIt¼120⋅120/10⋅5¼7200J
o0110 (b) (169.7)2⋅5/2⋅10¼7199 J.
o0115 1.20 ADCvoltageof120VisequivalentindeliveringpowertoaresistorasanACvoltage
with peak valueof V ¼169.7 V.
P
o0125o0120 1.21 (a) p(t)¼vÐ2/R(cid:3)¼VP2c(cid:4)os2 10t/R; (cid:5) (cid:6)
p0140 P ¼1 T V2=R cos2 10tdt¼1Vp2 1 110T+1sin20T ¼V2=2R.
ave T 0 P T R 10 2 4 p
o0130 (b) Power flowis always from sourceto resistor.
o0140o0135 1.22 (a) Using formulain textfollowing(1.15), spacingbetweenplates is
‘¼εA/C¼6⋅8.85⋅10(cid:2)12⋅10(cid:2)2m2/0.05⋅10(cid:2)6¼1.06⋅10(cid:2)5m.
o0145 (b) Electricfieldstrengthbetweenplates.V/‘¼100 V/1.06⋅10(cid:2)5m¼9.43⋅106V/m
whichexceedsthebreakdownstrengthofmica,whichis6⋅106V/m.
o0150 (c) 6⋅106V/m⋅1.06⋅10(cid:2)5m¼63.6V.
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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typesetter SPi. It is not allowed to publish this proof online or in print. This proof copy is the copyright property of the publisher and is confidential until formal publication.
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B978-0-12-817008-3.09989-7,09989
Instructor’s solutionsmanual: Solutions to allexercises to accompany 3
Ð Ð
o0155 1.23 v¼0, t<0; v¼1 idt¼ 1 t20mAdt¼ 1 20⋅10(cid:2)3t¼4⋅103t, 0(cid:4)t(cid:4)3ms;
c 5μF 0 5⋅10(cid:2)6
v¼v(t¼3 ms)¼4⋅103t¼4⋅103⋅3⋅10(cid:2)3¼12V, t>3 ms.
o0160 1.24 Wc¼ÐCVp2sin22πt/2.Maxwhen sin2πt¼1, 5⋅10(cid:2)6⋅(200)2/2¼0.1 J¼Wcmax.
o0165 1.25 i¼1 t 2dt¼ 2V t¼103t A for 0(cid:4)t(cid:4)3ms; 1
L (cid:2)∞ 2⋅10(cid:2)3H Ð
ij ¼iðt¼3 msÞ+1 ∞0dt¼103tj +0¼3 Afor t>3ms.
t>3ms L 3 t¼3ms
o0175o0170 1.26 (a) I ¼.9V/3Ω¼.3A, R ¼(V (cid:2)V )/I ¼(1.5(cid:2).9)/.3¼.6/.3¼2Ω.
L i B L L
o0180 (b) (1.5+0.9)/2¼1.2 V.
o0185 (c) I ¼V /3Ω¼0.4A.
ave ave
o0195o0190 1.27 (a) P ¼I ⋅V ¼0.4⋅1.2¼0.48 W.
ave ave ave
o0200 (b) W-h¼0.48⋅6¼2.88 W-h.
o0205 (c) Batterycost in cents/kW-h¼120/2.88⋅10(cid:2)3¼41,667cents/kW-h. This is 41,
667/8¼5208 asexpense as energy supplied by electric utilities.
o0210 1.28 v ¼6V, R ¼v /i ¼6/2¼3Ω
s i oc sc
o0215 1.29 i ¼i ¼2A, R ¼v /i ¼3Ω
s sc i oc sc
o0220 1.30 Current flowing in the series circuitis12/(1+2+3)¼2A; V ¼2V, V ¼4V,
R R
1 2
V ¼6V.
R
3
o0225 1.31 1 ¼1+1+1, R ¼6=11Ω;V ¼IR ¼11⋅6=11¼6V;I ¼6A, I ¼3A, I ¼2A.
Req 1 2 3 e(cid:3)q (cid:4) eq R1 R2 R3
o0235o0230 1.32 (a) I ¼12= 10+ 100⋅1000 ¼0:12 A,
battery 100+1000
o0240 (b) I10Ω¼0:12A,I100Ω¼0:12⋅1001+0010000¼0:11A,I1000Ω¼0:12⋅1001+010000¼0:011 A
o0245 (c) V10Ω¼(0.12 A)(10Ω)¼1.2V, V100Ω¼V1000Ω¼12V(cid:2)1.2V¼10.8 V
10Ω
+
100Ω 1000Ω
12V
(cid:3) (cid:4) (cid:3) (cid:4) (cid:3) (cid:3) (cid:4)(cid:4)
o0255o0250 1.33 (a) I3Ω¼6= 3+11+⋅44 ¼1:58A, I2Ω¼1:58 22+⋅44 = 2+ 22+⋅44 ¼0:63 A,
I4Ω¼1:581+14¼0:32A
o0260 (b) V3Ω¼1.58⋅3¼4.7 V, V2Ω¼V4Ω¼6(cid:2)4.7¼1.3
o0265 (c) P¼V ⋅I ¼6 V⋅1.58 A¼9.47 W
bat. bat.
3Ω
+
2Ω 4Ω
6V 2Ω
o0270 1.34 6¼2i +10i (cid:2)10i but i ¼(cid:2)3A, i ¼(cid:2)24/12¼(cid:2)2A
1 1 2 2 1
o0275 (a) i10Ω¼i1(cid:2)i2¼(cid:2)2(cid:2)((cid:2)3)¼1 A, V10Ω¼i10⋅10¼1⋅10¼10 V
o0280 (b) i2Ω¼i1¼(cid:2)2 A, V2Ω¼(cid:2)2⋅2¼(cid:2)4 V
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
Date:19/11/19 Time:19:14:04 PageNumber:3
typesetter SPi. It is not allowed to publish this proof online or in print. This proof copy is the copyright property of the publisher and is confidential until formal publication.
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B978-0-12-817008-3.09989-7,09989
4 Instructor’s solutions manual:Solutions to allexercises to accompany
2Ω
+
10Ω 3A
6V
i i
1 2
o0285 1.35 Switching off the currentsource,we obtain i100Ω¼ð6=2+10Þ¼12 A and
V0 ¼6⋅ 10 ¼5V
10Ω 2+10
p0305 Switching off the voltage source,we obtain i1000Ω¼3⋅2+210¼12 A and
V00 ¼3⋅ 2⋅10¼5V
10Ω 2+10
o0290 (a) i10Ω¼i0+i00¼12+12¼1A, V10Ω¼V0+V00¼5+5¼10 V
o0295 (b) Similarly i20Ω¼6=ð2+10Þ¼12 A,V20Ω¼6⋅2/(2+10)¼1 V, i200Ω ¼3⋅10/(2+10)¼
2.5 A, V200Ω¼3⋅22+⋅1100¼5V, i2Ω¼i0+i00¼12(cid:2)2:5¼(cid:2)2A, V2Ω¼V0+V00 ¼
1(cid:2)5¼(cid:2)4 V
o0300 1.36 i2Ω¼(6(cid:2)30)/2(2+10)¼(cid:2)24/12¼(cid:2)2 A i10Ω¼ð3+3Þ2+210¼1A
p0325 V2Ω¼((cid:2)2)(2)¼(cid:2)4 V V10Ω¼(1)⋅10¼10 V
2Ω 10Ω
+ +
6V 30V 3A 2Ω 3A
10Ω
o0305 1.37
+ +
a a
1 1V 1 1V 1
+
1 1 1A 1 1V
b b
a a
1 1/2 R = 1.5
th
+ +
1V V = 1V
th
b b
(cid:3) (cid:4)(cid:3) (cid:4)
o0310 1.38 Itshouldhavethe value ofR ¼1:5Ω:P ¼IV ¼ 1 1 1:5 ¼1=6W.
th max 1:5+1:5 1:5+1:5
o0315 1.39 Usingsuperposition,theopen-circuitvoltageatx-yisV ¼V ¼12⋅ 40k50 +1⋅30k
oc th 10+ð40k50Þ
ð20+10k40Þ⋅ 40k10 ¼8:276+4:138¼12:414V, where40k50¼40⋅50/(40+50)¼
20+40k10
22.22Ω, 10k40¼8Ω
p0345 R ¼10k40k50¼8k50¼8⋅50/(8+50)¼6.896Ω
th
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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B978-0-12-817008-3.09989-7,09989
Instructor’s solutionsmanual: Solutions to allexercises to accompany 5
x
10Ω 20Ω
+
12V 40Ω 30Ω 1A
y
p0350 R obtained byshort-circuiting the 12V sourceand open-circuiting the 1 A
th
source.
x
6.9Ω
+
12.4V
y
o0325o0320 1.40 (a) R ¼25k(10+25)¼12.5Ω, therefore R ¼12.5Ω
th L
o0330 (b) V ¼ 6 ⋅25+12(cid:2) 12 ⋅25¼3+12(cid:2)6¼9V,therefore
th 15+10 10+15+25
P ¼I⋅V ¼ 9 ⋅9⋅ 12:5 ¼1:62 W
max 12:5+12:5 12:5+12:5
o0335 (c) PRL¼10Ω¼I⋅V ¼10+912:5⋅9⋅10+1012:5¼1:60 W
10Ω a
+ +
12V
6V
R
L
15Ω 25Ω
b
o0340 1.41 Formaximumpowertransfertoaload,theequivalentsourceandloadresistances
must be matched, that is, equalto each other.
o0345 1.42 i ¼V/R
R 1
1
o0350 1.43 i2¼(cid:7)(R2ii(cid:2)V2)/(R(cid:7)2+R3)¼[2⋅((cid:2)0.33)(cid:2)2]/(2+3)¼(cid:2)2.66/5¼(cid:2)0.532 A
(cid:7) (cid:7)
(cid:7) 8 (cid:2)2 1(cid:7)
o0355 1.44 i3¼(cid:7)(cid:7)(cid:7)(cid:2)2 5 (cid:2)2(cid:7)(cid:7)(cid:7)(cid:5)199¼(cid:2)5ð4(cid:2)51Þ9(cid:2)93ð40(cid:2)4Þ¼(cid:2)110939¼(cid:2)0:517A
(cid:2)5 0 (cid:2)3
o0360 1.45 i ¼(cid:2)0.33 A
1
o0365 1.46 i ¼i (cid:2)i ¼(cid:2)0.33(cid:2)((cid:2)0.52)¼0.19 A
R5 1 3
o0370 1.47 Yes,it resultsin a matrix with positive diagonalterms and negative off-diagonal
terms.This helpswhen checking the equations forerrors.
o0380o0375 1.48 (a) q ¼CV¼2μF⋅12V¼24⋅10(cid:2)6 coulombs (c).
0
o0385 (b) i ¼v /R¼12V/100Ω¼0.12 A.
o o
o0390 (c) τ¼RC¼100Ω⋅2μF¼200μs.
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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typesetter SPi. It is not allowed to publish this proof online or in print. This proof copy is the copyright property of the publisher and is confidential until formal publication.
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B978-0-12-817008-3.09989-7,09989
6 Instructor’s solutions manual:Solutions to allexercises to accompany
o0395 1.49 q(t)¼q e(cid:2)t/τ¼24μCe(cid:2)1ms/200μs¼24e(cid:2)1/.2¼24e(cid:2)5¼24⋅0.0067¼0.162μC
0
p0435 i(t)¼i e(cid:2)t/τ¼(0.12A)e(cid:2)5¼0.809 mA.
o
o0400 1.50 1 ¼ 1 + 1 + 1 ¼106+107+108, C ¼9:009⋅10(cid:2)9C
Ceq C1 C2 C3 eq
Q¼VC ¼100⋅9:009⋅10(cid:2)9¼9:009⋅10(cid:2)7, V ¼Q¼9:009⋅10(cid:2)7¼:9009 V,
eq 1 C1 10(cid:2)6
p0450 V ¼9.009⋅10(cid:2)7/10(cid:2)7¼9.009V, V ¼9.009⋅10(cid:2)7/10(cid:2)8¼90.09 V.
2 3
o0410o0405 1.51 (a) Using t!∞ in (1.51), V(t!∞)¼V ¼12V, Q¼CV¼2⋅10(cid:2)6⋅12¼24μC
bat.
o0415 (b) Using (1.51), V(t¼τ)¼V (1(cid:2)e(cid:2)1)¼12(1(cid:2)0.368)¼120⋅632¼7.59 V;
bat.
Q¼CV¼15.2 μC.
o0425o0420 1.52 (a) W ¼1CV2¼1⋅2⋅10(cid:2)6⋅ð12Þ2¼144 microjoules (μJ)after using (1.18)
2 2
o0430 (b) Voltageon adischarging capacitor is given by (1.52).
V ¼V e(cid:2)2¼12⋅0.135¼1.62 V. Energy remaining in discharging capacitor
C batt. (cid:3)
W ¼12CV2¼12C Vbatt:e(cid:2)t=τÞ2¼12CV2be(cid:2)2t=τ Therefore,after 2time constants,
W ¼1CV2e(cid:2)4¼1⋅2⋅10(cid:2)6⋅122⋅0:018¼2:64μJ.
2 b 2
o0435 1.53 Using (1.57) i ¼i(t¼∞)¼V/R¼12/1000¼12mA.
o
o0440 (a) i(cid:2)¼i+¼12 mA.
o o
o0445 (b) i(t)¼i e(cid:2)t/τ¼i e(cid:2)t/(L/R)¼12⋅e(cid:2)2¼12⋅0.135¼1.62 mA.
o o
o0450 (c) The rate ofheat production is P¼i2R.SinceP¼dW/dt,the total energy
dissipated asheat in the resistors is
Ð Ð (cid:3) (cid:4) Ð (cid:3) (cid:4)
W ¼ ∞i2Rdt¼R ∞ i e(cid:2)t=τ 2dt¼i2R ∞e(cid:2)2t=τdt¼i2R (cid:2)τe(cid:2)2t=τ j∞¼i2Rτ
0 0 o (cid:3) (cid:4)o 0 o 2 0 o 2
¼1i2L¼1i2L¼1i L¼1 12⋅10(cid:2)3 21⋅10(cid:2)3¼72 nanojoules (nJ).
2 o 2 0 2 o 2
o0455 1.54 τ¼L/R¼10⋅10(cid:2)3/100¼0.1 ms¼100 μs.
o0465o0460 1.55 (a) v30Ω¼Rioe(cid:2)t/τ¼(30Ω)(0.15A)e(cid:2)t/(4/70)¼4.5e(cid:2)t/0.057 V
o0470 (b) τ¼L/R¼4/70¼0.057 seconds (s).
ϑ
4.5
t
o0475 1.56 Inordertofindthecurrentthroughtheswitch,letusfirstfindthevoltageacrossthe
(cid:3) (cid:4)
40Ωresistor.v40Ω¼Ri+Ldi=dt¼3(cid:3)0i+4d(cid:4)i=dt¼30 0:15+0:05e(cid:2)t=0:09 +4ddtð0:15+0:05
e(cid:2)t=0:09Þ¼4:5+1:5e(cid:2)t=0:09+4⋅0:05 (cid:2) 1 e(cid:2)t=0:09¼4:5+ð1:5(cid:2)2:22Þe(cid:2)t=0:09¼
0:09
4:5(cid:2)0:722e(cid:2)t=0:09, wherethecurrentthroughtheinductori¼0.15+0.05e(cid:2)t/0.09is
giveninExample1.10.Thecurrentthroughtheswitchisvoltagev40Ωdividedby
the40Ωresistor:iswitch¼v40Ω/40¼0.11(cid:2)0.018e(cid:2)t/0.09A.
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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typesetter SPi. It is not allowed to publish this proof online or in print. This proof copy is the copyright property of the publisher and is confidential until formal publication.
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B978-0-12-817008-3.09989-7,09989
Instructor’s solutionsmanual: Solutions to allexercises to accompany 7
o0485o0480 1.57 (a) v (0) ¼(120/20)10 ¼60 V. Bysuperposition: v (∞) (120/15)5– (240/15)5¼
C C
40(cid:2)80¼(cid:2)40.
o0490 (b)
V(t)
c
60
t
0
–40
o0495 (c) v (0)¼60¼A+B;v (∞)¼-40¼A;v (t)¼-40+100e(cid:2)30000t;RC¼(10ll10ll10)
C C C
10(cid:2)5 ¼ (10/3) 10(cid:2)5.
o0500 1.58 (Fig.1.44):α¼(1/2RC)¼2.5;ω 2¼1/LC¼6;¼(cid:2)2.5(cid:6)0.5¼-2,-3;v ¼v ¼Ae(cid:2)2t+
0 L C
Be(cid:2)3t;v (0)¼5,i (0)¼0whichgivesA+B¼5;i +iL+i ¼0whichleadstoi (0)¼C
C L R C C
dv (0)/dt¼1/25(-2A-3B)¼-v (0)/R–i (0)¼-1.Solving,weget–A–1.5B¼-12.5,
C C L
therefore A¼ -10 and B ¼15. Hencev (t) ¼-10e(cid:2)2t +15e(cid:2)3tV for t>0.
L pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1.59 (Fig. 1.28): α ¼1/2RC ¼103, ω 2 ¼ 1/LC¼64 (cid:3) 104;s ¼(cid:2)α(cid:6) α2(cid:2)ω 2 ¼ (cid:2)400,
0 1 0
o0505 2
(cid:2)1600s(cid:2)1; i ¼ Ae(cid:2)400t+Be(cid:2)1600t; i (0)¼(cid:2)2¼A+B; di (0)/dt¼v(0)/L¼(cid:2)400A
L L L
– 1600B ¼32 (cid:3)40 ¼1280;solving, we obtain A¼(cid:2)1.6, B¼(cid:2)0.4, hencei ¼
L(t)
(cid:2)1.6e(cid:2)400t(cid:2)0.4e(cid:2)1600t.
o0510 1.60 α2 ¼(1/2RC)2 ¼ω 2 ¼1/LC.
0
o0515 1.61 (Fig.1.45):v(0)¼0,i(0)¼450/250p0ffi¼ffiffiffiffiffi0ffiffi.ffi1ffiffiffi8ffiffiffiAffiffi,αp¼ffiffi1ffiffiffi/ffiffi2ffiffiRffiffiffiCffiffiffiffi¼106/(2(cid:3)625(cid:3)2)¼400, ω02¼
1/LC¼106/2(cid:3)2¼250000, ω ¼ α2(cid:2)ω 2¼ 25(cid:2)16(cid:3)100¼300,i(t)¼e(cid:2)400t(Acos
d 0
300t+Bsin300t),whichgivesA¼0.18;v (0)¼v(0)¼2di(0+)/dt¼0¼300B–400(cid:3)
L
0.18,whichgivesB¼ 0.2;i(t)¼(0.18cos300t+0.24sin300t)Amp.fort>0.
o0520 1.62 Inductor(L¼8H)currentis3A,whichcorrespondstostoredenergy½Li2¼36J.Heat
dissipated in R is
Ð∞ Ð∞
p0575 v2dt ¼ 2667.7 t2e(cid:2)5.29t dt¼ 2667.7 2! ¼ 36J
R 5:293
0 0
o0525 1.63 Inductor(L¼8H)currentis3A,whichcorrespondstostoredenergy½Li2¼36J.Heat
dissipated in Ris
p0585 Ð∞v2dt ¼ Ð∞ð64:6e(cid:2)0:5tsin2:6tÞ2dt ¼74.52Ð∞e(cid:2)tsin22.6t dt ¼35.9Ð3J
R R¼56
o0530 1.64 v(t)Ð0¼est(A10+tA2)givenbyEq.(1.70).v(00)¼V0¼A1.iL(t)¼1/L vdt¼1/L[A1est/s+
A testdt];i (0) ¼ 1/L[V /s -A /s2 ];
2 L 0 2
p0595 i (t) ¼est/L [V /s –(V s – Ls2 I )(st(cid:2)1)] ¼ est [I + (V /L – I s)t]
L 0 0 0 s2 0 0 0
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
Date:19/11/19 Time:19:14:07 PageNumber:7
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B978-0-12-817008-3.09989-7,09989
8 Instructor’s solutions manual:Solutions to allexercises to accompany
o0535 1.65 Repeatthe calculation outlined in Problem1.64,exceptnowapply itto the
underdampÐed case. Ð
o0540 1.66 vC(t)¼1/C idt¼(2(cid:3)108)(0.21(cid:3)10(cid:2)3) e(cid:2)300(cid:8)tsin14140(cid:9)tdt.Afterintegrating,setting
t¼0, we obtain v (0) ¼ (2 (cid:3)108)(0.21(cid:3)10(cid:2)3) (cid:2)14140 ¼ (cid:2)0:6⋅109 ¼ -3 V,where
Ð C 3002+141402 3002+141402
eatsin pt dt] ¼eat (asin pt(cid:2)pcospt)/(a2 + p2)was used.
o0545 1.67 Using the derived current iÐ(t), we obtain that i(0)м 10(cid:2)2 mA. Toverify initial
voltage,weusev (t)¼1/C idt¼(2(cid:3)108)(10(cid:2)2) e(cid:2)300tcos14140tdt¼(2(cid:3)106)
C
((cid:2)300cos14140t(cid:2)14140sin14140t)/(3002+141402))whichfor v (0)¼ (2(cid:3)106)
C
(-300)/(2(cid:3)108)¼ -3V.
Chapter 2: Solutions to problems
s0015
o0555o0550 2.1 (a) i lags vby 10o.
o0560 (b) i leadsv by5o. ilags v by 355°becausecos(x(cid:2)5°)¼ cos(x+355°).
o0565 (c) i leads v by 160(cid:7)because i can also be expressedas (cid:2)5 cos(vt(cid:2)30°)¼ 5 cos
(vt(cid:2)210(cid:7)),orilagsvby200(cid:7)becauseicanalsobeexpressedas(cid:2)5cos(vt(cid:2)30(cid:7))
¼5 cos(vt + 150(cid:7)).
o0570 2.2 Att¼0, the voltage equation becomes40¼50cosθ, cosθ¼0.8 and θ¼ (cid:6)0.8.
Since voltage is decreasing at t¼0, v¼50cos(wt+37°).
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o0575 2.3 22+32ejtan(cid:2)1ð3=2Þ¼3:6ej56:5°, 32+52ejtan(cid:2)1ð(cid:2)5=3Þ¼5:8e(cid:2)j59°,
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
72+92ejtan(cid:2)1ð9=(cid:2)7Þ¼11:4ej127:9°:
o0580 2.4 v(t)¼5coswt¼ Re[5]ejwt, phasor is 5; v(t)¼ Re[5e(cid:2)j90°]ejwt, phasor is 5e(cid:2)j90°;
v(t)¼ Re[V ejθ]ejwt, phasor is V ejθ;v(t)¼ Re[120ej30°]ejwt,phasor is 120ej30°.
p p
o0585 2.5 Re[j10]ejwt¼ Re10ej(wt+π/2)¼10cos(wt+π/2)¼(cid:2)10sinwt, Re[10+j10]ejwt¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Re 102+102ejπ=4+jwt¼14:1cosðwt+π=4Þ,Re½10(cid:2)j10(cid:8)ejwt¼14:1 cosðwt(cid:2)π=4Þ:
(cid:7) (cid:3) (cid:4)
2.6 Re½j10(cid:8)ejwt(cid:7) w¼377 ¼(cid:2)10sin 377⋅10(cid:2)3 ¼(cid:2)10sin :377¼(cid:2)10ð:368Þ¼(cid:2)3:68,Re½10+j10]
o0590 t¼10(cid:2)3s
ej.377¼14.1cos(.377+π/4)¼14.1⋅0.4¼5.69, Re[10(cid:2)j10]ej0.377¼14.1cos
(.377(cid:2)π/4)¼14.1⋅0.92¼12.94.
o0595 2.7 V ¼5e(cid:2)j20°, V ¼7ej30°, V +V ¼5(cos20° (cid:2)jsin20°)+7(cos30° +jsin30°)
1 2 1 2
¼5ð0:94(cid:2)j0:34Þ+7ð0:87+j0:5Þ¼10:76+j1:79
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 10:762+1:792ejtan(cid:2)11:79=10:76¼10:9ej9:4°, v +v ¼10:9cosðwt+9:4°Þ
1 2
o0600 2.8 Z¼R+jwL¼10+jw4Ω
o0605 2.9 V¼IZ¼4(10+j377⋅20⋅10(cid:2)3)¼40+j30.16¼50.1ej37.6°, v(t)¼50.1cos
(377t+37.6°)
o0610 2.10 Z¼R⋅jwL¼jwRLðR(cid:2)jwLÞ¼wRLðwL+jRÞ¼6:28⋅105ð628:3+j103Þ¼283.1+j450.4Ω
R+jwL (cid:8)R2+(cid:9)ðwLÞ2 R2+w2L2 106+(cid:3)0:395⋅106 (cid:4)
o0615 2.11 Z¼R+jwL 1 =ðjwL+1=jwcÞ¼R+jwL= 1(cid:2)w2Lc
jwc
o0620 2.12 I¼ V¼ 200 ¼1+j1, since Ileads V,thisis a capacitive circuit
Z 100(cid:2)j100
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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Instructor’s solutionsmanual: Solutions to allexercises to accompany 9
R L C
=
100Ω j100Ω −j200Ω 100Ω −j100Ω
I
V V + V
R L C
V = 200
o0625 2.13 Using voltage division V ¼V R ¼200⋅100¼192:31+j38:46¼196:1 expðj11:3°Þ,
R SR+1=jwc 100(cid:2)j20
v ¼196.1cos(1000t+11.3°).
R
100Ω
+
— 1H
200 coswt 50μF
o0630 2.14 200¼I (100(cid:2)j20)(cid:2)I (100(cid:2)j20)
1 2
p0705 0¼(cid:2)I (100(cid:2)j20)+I (100(cid:2)j20+j 1000)
1 2
p0710 Solving, we obtain I ¼(cid:2)j0.2, thereforei (t)¼0.2cos(1000t+270°)
2 2
100
200 +— I1 I2 j1000
—j20
o0635 2.15 1¼I (j2(cid:2)j1)(cid:2)I j2
1 2
p0720 0¼(cid:2)I j2+I (2+j2)
1 2
p0725 Solving for I we obtain I ¼0.5+j0.5A
2 2
—j1
1 +— I1 j2 I2 2
o0640 2.16 Z¼ R+ 1/jωC ¼100 (cid:2) j 100, Y¼1/Z¼ 1/(R (cid:2)jX )¼R+jX ¼0.005 + j 0.005 S,
R2+X2
G¼ 0.005 S, B¼+0.005 S
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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10 Instructor’s solutions manual: Solutions to all exercises to accompany
o0650o0645 2.17 (a) f ¼w /2π¼1/2πRC, R¼1/2πf C¼1/6.28⋅400⋅0.5⋅10(cid:2)6¼795.8Ω
o o o
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o0655 (b) V =V ¼1= ð1+ðw=w Þ2¼1= 1+ð600=400Þ2¼0:55
o i o
o0660 2.18 f ¼1/2πRC¼1/2π104⋅10(cid:2)6¼15.9 Hz
0
o0670o0665 2.19 (a) w ¼2πf ¼1/RC, f ¼1/2πRC, C¼1/2πf R¼1/2π⋅400⋅103¼0.398 μF
o 0 o o
o0675 (b) SinceV /V (cid:9)e+jθ, itis aphase-leadnetwork
o qi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffi
o0680 (c) Vo=Vi¼1= 1+ð400=200Þ2¼1= p5,ffiffiffidB power gain
¼20 log V =V ¼(cid:2)20log 5¼(cid:2)20⋅0:349¼(cid:2)6:99dB
10 o i
o0685 2.20 ω ¼ 1/√LC, C ¼ 1/(2π f )2 L ¼ 1/(2π (cid:3) 870 (cid:3) 103)2 (cid:3) 20 (cid:3) 10(cid:2)6 ¼ 0.00167 μF ¼
o o
1.67(cid:3)10(cid:2)p9ffiffiffi¼ffiffiffiffi1.67 nF (cid:3) (cid:4)
o0690 2.21 w ¼1= LC, L¼1=ð2πf Þ2C¼1= 6:28⋅52⋅106 210(cid:2)11¼0:937μH
o pffiffiffiffiffiffiffi opffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o0700o0695 2.22 (a) f ¼1=2π LC¼1=2π 0:1⋅10(cid:2)3⋅0:01⋅10(cid:2)6¼159:15 kHz;V ¼jw LI
o sffiffiffiffiffiffiffiffiffiffi L o
rffiffiffi
1 V L V 10(cid:2)4 1 1 1 V
¼jpffiffiffiffiffiffiffi⋅L⋅ ¼j ⋅ ¼j ⋅ ¼j1000;V ¼ ⋅I¼ pffiffiffiffiffiffiffiffiffiffiffiffi ⋅
rLCffiffiffi R sffifficffiffiffiffiffiffiRffiffi 10(cid:2)8 0:1 C jwc j 1=LC⋅C R
L V 10(cid:2)4 1 V
¼(cid:2)j ⋅ ¼(cid:2)j ⋅ ¼(cid:2)j 1000, V ¼RI¼R ¼V ¼1V:
c R 10(cid:2)8 0:1 R R
o0705 (b) VoltagesacrossLandCare1000timeslargerthantheappliedvoltage.However,
they are 180°out ofphase,such that V ¼0V.
LC
o0710 2.23 f the same.V ¼j10 V, V ¼(cid:2)j10 V, V ¼1 V. The voltage magnification is only
o L C R
tentimesinthiscaseastheincreasedRinthecircuitdissipatesmorepowerasthe
energychanges back and forth betweenL and C.
o0715 2.24 p¼vi¼v(v/R)¼v2/R¼(1coswt)2/0.1¼10cos2wtqwffiffiaffiffiffitffitffiffisffiffi.ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi
o0725o0720 2.25 (a) f ¼ 1(cid:2)R2C=L=2π LC (from Example2.5) ¼ 1(cid:2)52⋅10(cid:2)7=10(cid:2)3=6:28
o pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(cid:10) 10(cid:2)7⋅10(cid:2)3¼ð0:999Þ=62:8⋅10(cid:2)6¼15:92 kHz:Q¼w L=R¼2π⋅15:92⋅
o
103⋅10(cid:2)3=5¼20, B¼w =Q¼R=L¼5⋅103rad=sec¼795:8Hz:
o
o0730 (b) R mustbe decreased by afactor of2to 2.5Ω.
o0735 2.26 Q¼wo/Δw¼f0/Δf¼100(cid:3)⋅106/70⋅(cid:4)103(cid:7)¼1428.6 (cid:3) (cid:4) (cid:3) (cid:4)
o0740 2.27 FromExample 2.5, Z ¼ R2+w2L2 =R(cid:7) pffiffiffiffiffi¼ R2+L=C =R¼ 52+10(cid:2)3=10(cid:2)7 =
5¼2005Ω o o wo¼1= LC
o0745 2.28 Vo¼IRZi¼R+jwL=Vð1i(cid:2)w2LCÞ⋅1(cid:2)jww2LLC¼ViRð1(cid:2)wj2wLCLÞ+jwL |Vo/Vi| ¼[1+(R/wL)2
(1(cid:2)w2LC)](cid:2)1/2
I R I I
R C L
+— C L VO
V=10V
i
Plonus,978-0-12-817008-3
CTomo pp.rboyt:eNcEtE tLhAeK AriNgDhAtsN .oPf tShtaeg aeu:Rtheovirs(ess)2 anCdha pptuebrlNios.h:eSro lwutieo nisnMfoarnmua lyoTuitl ethNaatm teh:iPsl oPnDusF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and
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