英文版电路原理课后题答案
电路分析基础(英文版)课后答案第二章
2Some Circuit Simpli¯cationTechniquesDrill ExercisesDE2.116k64=12:8−;12:8+7:2=20−;20k30=12−[a]v=5(12)=60V[b]p5A(del)=(5)(60)=300W[c]i A=60=30=2A i C=3(64)=(80)=2:4Ai B=5¡2=3A p10−=(2:4)210=57:6WDE2.2[a]v o(no load)=200(75)=100=150V[b]75k150=50k−,therefore v o=200(50)=75=133:3V[c]i=200=25;000=8mA,p25k=(8£10¡3)2(25;000)=1:6W[d]Maximum dissipation at no load since v o is maximump=v2o75;000=0:3W2728CHAPTER2.Some Circuit Simpli¯cation TechniquesDE2.3v30=6+4(0:825)=9:3V;i30=v3030=0:31Ai6=i30+0:825=1:135A;i10=0:825+0:31=1:135A¡v30¡6i b+v20¡10i10=0¢::v20=9:3+16(1:135)=27:46Vi20=27:4620=1:373A;i5=i6+i20=2:508Ai30=0:31A;i6=1:135A;i10=1:135A;i20=1:373A;and i5=2:508ADE2.4Problems29i =7212=6A [a]v =7212(8)=48V ;i 120V =120¡57:620=3:12A [b]v a =6(9:6)=57:6V ;p 120V (del)=120i a =374:40WDE 2.5[a]110V source actingalone:R e =10(14)24=356−i 0=1105+35=6=13213Av 0o=µ356¶µ13213¶=77013V 4A source actingalone:5−k 10−=50=15=10=3−10=3+2=16=3−30CHAPTER2.Some Circuit Simpli¯cation Techniques 16=3k12=48=13−Hence our circuit reduces to:It follows thatv00a=4(48=13)=(192=13)Vandv00o=¡v00a(16=3)(10=3)=¡58v00a=¡(120=13)V¢::v o=v0o+v00o=77013¡12013=50V[b]p=v2o10=250WDE2.670-V source acting alone:v0=70¡4i0bi0s=v0b2+v010=i0a+i0b70=20i0a+v0bi0a=70¡v0b 20Problems31¢::i 0b=v 0b 2+v 010¡70¡v 0b 20=1120v 0b +v 010¡3:5v 0=v 0b+2i 0b ¢::v 0b =v 0¡2i 0b¢::i 0b =1120(v 0¡2i 0b )+v 010¡3:5or i 0b =1342v 0¡7042¢::v 0=70¡4µ1342v 0¡7042¶orv 0=322094=161047V 50-V source actingalone:v 00=¡4i 00bv 00=v 00b +2i 00bv 00=¡50+10i 00d ¢::i 00d=v 00+5010i 00s =v 00b 2+v 00+5010i 00b=v 00b 20+i 00s =v 00b 20+v 00b 2+v 00+5010=1120v 00b +v 00+5010v 00b =v 00¡2i 00b¢::i 00b =1120(v 00¡2i 00b )+v 00+5010or i 00b =1342v 00+10042Thus,v 00=¡4µ1342v 00+10042¶orv 00=¡20047V Hence,v =v 0+v 00=161047¡20047=141047=30V32CHAPTER2.Some Circuit Simpli¯cation Techniques ProblemsP2.1[a]p4−=i2s4=(12)24=576W p18−=(4)218=288W p3−=(8)23=192W p6−=(8)26=384W[b]p120V(delivered)=120i s=120(12)=1440W[c]p diss=576+288+192+384=1440WP2.2[a]From Ex.3-1:i1=4A,i2=8A,i s=12Aat node x:¡12+4+8=0,at node y:12¡4¡8=0[b]v1=4i s=48V v3=3i2=24Vv2=18i1=72V v4=6i2=48Vloop abda:¡120+48+72=0;loop bcdb:¡72+24+48=0;loop abcda:¡120+48+24+48=0P2.31R eq=16+110+115=1030=13;R eq=3−v(2+8+5)−=(20)(3)=60V;i(2+8+5)−=60=15=4A P5−=(4)2(5)=80WP2.4[a]R eq=2+2+(1=4+1=5+1=20)¡1=6−i g=120=6=20Av4−=120¡(2+2)20=40Vi o=40=4=10AProblems33 i(15+5)−=40=(15+5)=2Av o=(5)(2)=10V[b]i15−=2A;P15−=(2)2(15)=60W[c]P120V=(120)(20)=2:4kWP2.5[a]R eq=R k R=R22R=R2[b]R eq=R k R k R k¢¢¢k R(n R's)=R kR n¡1=R2=(n¡1)R+R=(n¡1)=R2nR=Rn[c]One solution:700−=200−+500−=1000=5+1000=2=1k−k1k−k1k−k1k−k1k−+1k−k1k−[d]One solution:5:5k−=5k−+0:5k−=2k−+2k−+1k−+0:5k−=2k−+2k−+2k−2+2k−4=2k−+2k−+2k−k2k−+2k−k2k−k2k−k2k−34CHAPTER2.Some Circuit Simpli¯cation TechniquesP2.6[a]12−k24−=8−Therefore,R ab=8+2+6=16−[b]1R eq=124k−+130k−+120k−=15120k−=18k−R eq=8k−;R eq+7=15k−1R ab=115k−+130k−+115k−=530k−=16k−R ab=6k−P2.7[a]For circuit(a)R ab=15k(18+48k16)=10−For circuit(b)1 R e =120+115+120+14+112=3060=12R e=2−R e+16=18−18k18=9−R ab=10+8+9=27−For circuit(c)48k16=12−12+8=20−20k30=12−12+18=30−30k15=10−10+10+20=40−R ab=40k60=24−[b]P a=20210=40WP b=144227=768WP c=62(24)=864WProblems35P2.8[a]5k20=100=25=4−5k20+9k18+10=20−9k18=162=27=6−20k30=600=50=12−R ab=5+12+3=20−[b]5+15=20−30k20=600=50=12−20k60=1200=80=15−3k6=18=9=2−15+10=25−3k6+30k20=2+12=14−25k75=1875=100=18:75−26k14=364=40=9:1−18:75+11:25=30−R ab=2:5+9:1+3:4=15−[c]3+5=8−60k40=2400=100=24−8k12=96=20=4:8−24+6=30−4:8+5:2=10−30k10=300=40=7:5−45+15=60−R ab=1:5+7:5+1:0=10−P2.9[a]R cond=845(0:0397)=33:5465−R total=2(1=2)R cond=33:5465−P loss=(2000)2(33:5465)=134:186MWP calif=800(2)¡134:186=1465:814MWE±ciency=(1465:814=1600)£100=91:61%[b]P calif=2000¡134:86=1865:814MWE±ciency=93:29%[c]P loss=(3000)2¢2¢(1=3)¢845¢(0:0397)=201:279MWP oregon=3000MW;P calif=3000¡201:279=2798:7MWE±ciency=(2798:70=3000)£100=93:29%P2.10i10k=(18)(15)40=6:75mAv15k=¡(6:75)(15)=¡101:25V i3k=18¡6:75=11:25mAv12k=¡(12)(11:25)=¡135Vv o=¡101:25¡(¡135)=33:75V36CHAPTER 2.Some Circuit Simpli¯cation TechniquesP 2.11[a]v 1k =11+5(30)=5V v 15k =1515+60(30)=6Vv x =v 15k ¡v 1k =6¡5=1V [b]v 1k =v s6(1)=v s =6v 15k=v s75(15)=v s =5v x =(v s =5)¡(v s =6)=v s =30P 2.1260k 30=20−i 30−=(25)(75)125=15A v o =(15)(20)=300V v o +30i 30=750V v g ¡12(25)=750v g =1050VP 2.135−k 20−=4−;4−+6−=10−;10k 40=8−;Therefore,i g =1258+2=12:5A i 6−=(40)(12:5)50=10A;i o =(5)(10)25=2A P 2.14[a]40k 10=8−i 75V =7510=7:5A 8+7=15−i 4+3−=7:5µ3045¶=5A15k 30=10−i o =¡5µ1050¶=¡1A[b]i 10−=i 4+3−+i o =5¡1=4AP 10−=(4)2(10)=160WP2.15[a]v9−=(1)(9)=9Vi2−=9=(2+1)=3Ai4−=1+3=4A;v25−=(4)(4)+9=25Vi25−=25=25=1A;i3−=i25−+i9−+i2−=1+1+3=5A;v40−=v25−¡v3−=25¡(¡5)(3)=40Vi40−=40=40=1Ai5k20−=i40−+i25−+i4−=1+1+4=6Av5k20−=(4)(6)=24Vv32−=v40−+v5k20−=40+24=64Vi32−=64=32=2A;i10−=i32−+i5k20−=2+6=8Av g=10(8)+v32−=80+64=144V:[b]P20−=(v5k20−)220=24220=28:8WP2.16[a]Let i s be the current oriented down through the resistors.Then,i s=V sR1+R2+¢¢¢+R k+¢¢¢+R nandv k=R k i s=R kR1+R2+¢¢¢+R k+¢¢¢+R nV s[b]i s=2005+15+30+10+40=2Av1=2(5)=10V v2=2(15)=30V v3=2(30)=60V v4=2(10)=20V v5=2(40)=80VP2.17[a]v o=2525(20)=20V[b]v o=255+R eR eR e=(20)(12)32=7:5k−v o=2512:5(7:5)=15V[c]v o25=2025=0:80[d]v o25=1525=0:60P2.18[a]No load:v o=R2R1+R2V s=¾V s¢::¾=R2R1+R2 Load:v o=R eR1+R eV s=¯V s¢::¯=R eR e+R1R e=R2R LR2+R L¢::¯=R2R LR1R2+R L(R1+R2)But R1+R2=R2¾¢::R1=R2¾¡R2¢::¯=R2R LR2³R2¾¡R2´+R L R2¾¯=R LR 2³1¾¡1´+R L ¾or ¯R 2µ1¾¡1¶+¯R L ¾=R L ¯R 2µ1¾¡1¶=R L Ã1¡¯¾!¢::R 2=(¾¡¯)¯(1¡¾)R LR 1=(1¡¾)¾R 2=þ¡¯¾¯!R L[b]R 1=(0:9¡0:7)0:63(126)k −=40k −R 2=(0:9¡0:7)(0:7)(0:1)(126)k −=360k −P 2.19[a]Let v o be the voltage across the parallel branches,positive at the upperterminal,theni g =v o G 1+v o G 2+¢¢¢+v o G N =v o (G 1+G 2+¢¢¢+G N )It follows thatv o =i g(G 1+G 2+¢¢¢+G N )The current in the k th branch is i k =v o G k ;Thus,i k =i g G k[G 1+G 2+¢+G N ][b]i 6:25=1142(0:16)[4+0:4+1+0:16+0:1+0:05]=32mAP 2.20R e =48£103=500−¢::XG =1500=2mS i 1=2i 2=2(10i 3)=20i 4i 2=10i 3=10i 4i 3=i 48=20i4+10i4+i4+i4=32i4¢::i4=832=0:25mAR4=v gi4=40:25£10¡3=16k−i3=i4=0:25mA ¢::R3=16k−i2=10i4=2:5mAR2=v gi2=42:5£10¡3=1:6k−i1=20i4=5mAR1=v gi1=45£10¡3=800−P2.21[a]i o=120=40k−=3mA[b]v a=(3)(20)=60Vi a=v a100=0:6mAi b=4¡3:6=0:4mAv b=60¡(0:4)(15)=54Vi g=0:4¡54=30=¡1:4mAp75V(developed)=(75)(1:4)=105mWCheck:p4mA(developed)=(60)(4)=240mWX P dev=105+240=345mWX P dis=(¡1:4)2(15)+(1:8)2(30)+(0:4)2(15)+(0:6)2(100)+(3)2(20)=345mWP2.22Apply source transformations to both current sources to geti o=¡66=¡1mAP2.23[a]¢::v o=1(240)=120V;i o=120=24=5A2[b]p300V=¡12:5(300)=¡3750WTherefore,the300V source is developing3.75kW.[c]¡10+i6−+7:5¡12:5=0;¢::i6−=15Av10A+4(10)+6(15)=0;¢::v10A=¡130Vp10A=10v10A=¡1300WTherefore the10A source is developing1300W.[d]X p dev=3750+1300=5050Wp4−=100(4)=400Wp40−=(7:5)2(40)=2250Wp6−=(15)2(6)=1350Wp42−=(5)2(42)=1050WX p diss=400+1350+2250+1050=5050W(CHECKS)P2.24Applying a source transformation to each current source yieldsNow combine the20V and10V sources into a single voltage source and the5−,4−and1−resistors into a single resistor to getNow use a source transformation on each voltage source,thuswhich can be reduced to¢::i o=(1:25)(8)=1A10P2.25First,¯nd the Th¶e venin equivalent with respect to R o.P2.26100−k25−=20−¢::i=400=5A60+20v0o=20i=100V100−k60−=37:5−i=50025+37:5=8Av00o=37:5i=300Vv o=v0o+v00o=100+300=400V P2.27i0o=10025=4A15−k30−=10−i00o=¡5025=¡2A¢::i o=i0o+i00o=4¡2=2A P2.2815=2i0¢+50i1+3i0¢Problems4715=2i 0¢+12i 02i 0¢=i 01+i 02;i 01=27=26A;i 0¢=51=26A¢::i 02=1213A;v 0o=9613V ¡2i 00¢=5i 001+3i 00¢¢::i 00¢=¡i 001i 002=i 00¢¡i 001=2i 00¢4i 002+(8+i 002)8=¡2i 00¢¢::i 002=¡6413A;i 001=3213A;i 00¢=¡3213A ¢::8+i 002=4013A ¢::v 00o=8µ4013¶=32013V ¢::v o =v 0o +v 00o =9613+32013=32V48CHAPTER2.Some Circuit Simpli¯cation TechniquesP2.29[a]The evolution of the circuit shown in Fig.P2.29is illustrated in the following steps:[b]Starting at the left end of the circuit and working toward the right end,aseries of source transformations yields:Problems49V R=4 4R (2R)=V R8P2.30[a]The evolution of the circuit in Fig.P2.30can be shown in two steps,thus:[b]Moving from left to right,a series of source transformations yields:50CHAPTER2.Some Circuit Simpli¯cation Techniquesv o=V R=84R(2R)=V R16Problems51 P2.31Eq.(2.34)v o=12V R(Switch1)Eq.(2.35)v o=14V R(Switch2)Eq.(2.36)v o=18V R(Switch3)Eq.(2.37)v o=116V R(Switch4)Given V R=16V:Switch Position v o12340000v o=0V000V R v o=116V R=1V00V R0v o=18V R=2V00V R V R v o=116V R+18V R=3V0V R00v o=14V R=4V0V R0V R v o=14V R+116V R=5V0V R V R0v o=14V R+18V R=6V0V R V R V R v o=14V R+18V R+116V R=7VV R000v o=12V R=8VV R00V R v o=12V R+116V R=9VV R0V R0v o=12V R+18V R=10VV R0V R V R v o=12V R+18V R+116V R=11VV R V R00v o=12V R+14V R=12VV R V R0V R v o=12V R+14V R+116V R=13VV R V R V R0v o=12V R+14V R+18V R=14VV R V R V R V R v o=12V R+14V R+18V R+116V R=15V52CHAPTER2.Some Circuit Simpli¯cation TechniquesThis page intentionally left blank。
工程电路分析 课后答案(英文原版)
∫
0.1
− 0.08 1 0
i (t )dt
−5t
∫ [− 2 + 3e ]dt
− 0.08 0 -0.08
+
∫ [− 2 + 3e ]dt
0.1 3t 0 0.1 0
= − 2 + 3e −5t
+ − 2 + 3e3t
= 0.1351 + 0.1499 = 285 mC
Engineering Circuit Analysis, 6th Edition
∞
0
x n e − ax dx
=
n! a n +1
where n is a positive integer and a > 0,
we find the energy delivered to be = 18/(200)2 =0 - 1800/(200)3
Engineering Circuit Analysis, 6th Edition
t −100 t (c) Pabs = 30∫0 idt + 20 3e
360 e −100t
[
]
2
t = 8 ms
= 72.68 W
[
]
2
t = 8 ms
= - 36.34 W
(
)
t = 8 ms
t −100 t = 3e −100t ′ dt ′ + 60e −100t 90e ∫ 0
Engineering Circuit Analysis, 6th Edition
Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.
电工学原理与应用第七版英文版答案
电工学原理与应用第七版英文版答案1、Actually, we don't know whether this news comes from a reliable（）or not. [单选题] *A. source(正确答案)B. originC. basisD. base2、Tony is a quiet student, _______ he is active in class. [单选题] *A. soB. andC. but(正确答案)D. or3、How _______ it rained yesterday! We had to cancel(取消) our football match. [单选题] *A. heavily(正确答案)B. lightC. lightlyD. heavy4、He often comes to work early and he is _______ late for work. [单选题] *A. usuallyB. never(正确答案)C. oftenD. sometimes5、_______ is on September the tenth. [单选题] *A. Children’s DayB. Teachers’Day(正确答案)C. Women’s DayD. Mother’s Day6、Last week they _______ in climbing the Yuelu Mountain. [单选题] *A. succeeded(正确答案)B. succeedC. successD. successful7、They took _____ measures to prevent poisonous gases from escaping. [单选题] *A.efficientB.beneficialC.validD.effective(正确答案)8、_________ along the old Silk Road is an interesting and rewarding experience. [单选题]*A. TravelB. Traveling(正确答案)C. Having traveledD. Traveled9、I’d like to go with you, ______ I’m too busy. [单选题] *A. orB. andC. soD. but(正确答案)10、We had a party last month, and it was a lot of fun, so let's have _____ one this month. [单选题] *A.otherB.the otherC.moreD.another(正确答案)11、It’s reported that there are more than 300?_______ smokers in China. [单选题] *A. million(正确答案)B. millionsC. million ofD. millions of12、I _______ Zhang Hua in the bookstore last Sunday. [单选题] *A. meetB. meetingC. meetedD. met(正确答案)13、Her ideas sound right, but _____ I'm not completely sure. [单选题] *A. somehow(正确答案)B. somewhatC. somewhereD. sometime14、--What are the young people doing there？--They are discussing how to _______?thepollution in the river. [单选题] *A. come up withB. talk withC. deal with(正确答案)D. get on with15、I will _______ at the school gate. [单选题] *A. pick you up(正确答案)B. pick up youC. pick you outD. pick out you16、—______ pencils are these?—They are Tony’s.（）[单选题] *A. WhatB. WhereC WhoD. Whose(正确答案)17、We have ______ homework today. （）[单选题] *A. too manyB. too much(正确答案)C. much tooD. very much18、--Which is Tom?--He is _______ of the two boys. [单选题] *A. tallB. tallerC. the taller(正确答案)D. the tallest19、I haven’t met him _____ the last committee meeting. [单选题] *A. forB. since(正确答案)C. atD. before20、I usually read English _______ six o’clock _______ six thirty in the morning. [单选题] *A. from;?atB. from; to(正确答案)C. at; atD. at; to21、The trouble turned out to have nothing to do with them. [单选题] *A. 由…引发的B. 与…有牵连C. 给…带来麻烦D. 与…不相干(正确答案)22、The house is well decorated _____ the disarrangement of a few photos. [单选题] *A. exceptB. besidesC. except for(正确答案)D. in addition to23、20．Jerry is hard-working. It’s not ______ that he can pass the exam easily. [单选题] * A．surpriseB．surprising (正确答案)C．surprisedD．surprises24、62．--There is? ? ? ? ? sale on in the shop today. Let’s go together.--Please wait? ? ? ? ? ?minute. I’ll finish my homework first. [单选题] *A．a; theB．a; a(正确答案)C．the; aD．the; the25、David ______ at home when I called at seven o’clock yesterday evening. （）[单选题] *A. didn’tB. doesn’tC. wasn’t(正确答案)D. isn’t26、Ordinary books, _________ correctly, can give you much knowledge. [单选题] *A. used(正确答案)B. to useC. usingD. use27、My daughter is neither slim nor fat and she’d like a _______ skirt. [单选题] *A. largeB. medium(正确答案)C. smallD. mini28、3．—Will you buy the black car?No, I won't. I will buya(n) ________ one because I don't have enough money. [单选题] *A．cheap(正确答案)B．expensiveC．highD．low29、I have to _______ my glasses, without which I can’t read the book. [单选题] *A. put upB. put awayC. put downD. put on(正确答案)30、Tom didn’t _______ his exam again. It was a pity. [单选题] *A. failB. winC. pass(正确答案)D. beat。
电气工程及其自动化专业英语苏小林课后答案
电气工程及其自动化专业英语苏小林课后答案【篇一：电气工程及其自动化准耶英语】/p> characterize描绘…的特征，塑造人物，具有….的特征property 性质，财产equal in magnitude to 在数量（数量级）上等同于 convert 转换converter 转换器time rate 时间变化率mathematically 从数学上来讲differentiatev 区分，区别in honor of 为纪念某人 name in honor of为纪念某人而以他命名electromotive force （ e m f ）电动势voltaic battery 伏打电池，化学电池an element 一个电器元件interpret 口译，解释，说明potential difference/voltage 电势差/电压 expend 花费，消耗instantaneous 瞬时的，促发的passive sign convention 关联参考方向the law of conservation of energy 能量守恒定律 reference polarity 参考极性electron 电子 electronic 电子的 electric 电的，电动的 time-varying 时变的 constant-valued 常量的metallic 金属的be due to 是因为，由于，归功于building block 模块coulomb库伦，ampere安培，joule焦耳，volt伏特，watt瓦特，work 功变量u（t），i(t)是电路中最基本的概念。
他们描述了电路中的各种关系。
电荷量的概念是解释电现象的基本原理，电荷量也是电路中最基本的量。
电荷也是构成物质的原子的电器属性，量纲是库伦。
我们从初等物理可以得知所有物质是由基本组成部分原子组成，而原子又包括电子（electron），质子（proton）和中子（neutron）我们都知道电荷e是带负电的电子，在数量上等于1.60210*1019 c, 而质子携带同等电荷量的正电荷，相同数量的质子，电子使原子呈现电中性（neutrally charged）。
电路分析基础(英文版)课后答案第三章
0 = ¡26i1 ¡ 90i2 + 124i3
[a] Solving, i1 = 5 A; therefore the 80 V source is delivering 400 W to the circuit.
[b] Solving, i3 = 2:5 A; therefore p8− = (6:25)(8) = 50 W
v1 + v1 ¡ v2 = 4:5
1
8
53
54 CHAPTER 3. Techniques of Circuit Analysis
v2 + v2 ¡ v1 + v2 ¡ 30 = 0
12 8
4
Solving, v1 = 6 V v2 = 18 V Thus, i = (v1 ¡ v2)=8 = ¡1:5 A v = v2 + 2i = 15 V
DE 3.8 Use the lower node as the reference node. Let v1 = node voltage across the 7.5 − resistor and v2 = node voltage across the 2.5 − resistor. Place the dependent voltage source inside a supernode between the node voltages v and v2. The node voltage equations are
3
Techniques of Circuit Analysis
Drill Exercises
DE 3.1 [a] 11,8 resistors, 2 independent sources, 1 dependent source
《测控技术与仪器专业英语》张凤登UNIT-4-电路-参考译文及练习答案
(4-9)
已知电流 i=dq/dt 的，将该变量代入上式，可以消除方程中的积分。由此可得到一个二阶微 分方程
L d 2q dq q +R + =e 2 dt dt C
练习答案 1. Use the information from the reading passage to complete the answer below. (1) It has an effect on total line voltage. (2) Their individual characteristics are unchanged. (3) Current is commonly used as reference. (4) They will be in phase. (5) The difference of XL and XC must be determined. (6) They are the Pythagorean Theorem-based formula, phasoral layout and triangulation. (7) It represents total impedance. (8) It refers to the fact that XL and XC are equal, and line voltage and current are in phase. 2. Fill in the following blanks with the words or expressions given below. Change the forms where necessary. (1) triangulation (2) Impedance (3) reactive (4) capacitors (5) inductive (6) Resistive (7) magnetic (8) electromotive 3. Select the best choices to complete the article. (1) transformer (2) down (3) generated (4) little (5) current (6) resistance (7) less (8) low (9) Direct (10) alternating current
电路分析基础(英文版)课后答案第五章
i1 + i2 = i DE 5.5 v1 = 0:5 £ 10
6 t 0+
240 £ 10¡6 e¡10x dx ¡ 10 = ¡12e¡10t + 2 V
t 0+
v2 = 0:125 £ 106 v1 (1) = 2 V; W =
·
Z
240 £ 10¡6 e¡10x dx ¡ 5 = ¡3e¡10t ¡ 2 V
Z
v (t) v (0+ )
155
Z dy R t dx =¡ y L 0+
¯v(t) µ ¶ ¯ R ¯ ln y ¯ =¡ t +
v (0 )
L
v (t) R =¡ t ln + v (0 ) L v (t) = v (0 )e
+ ¡(R=L)t
"
#
µ
¶
;
Vs v (0 ) = ¡ Io R = Vs ¡ Io R R
60(240) = 48 mH 300 [b] i(0+ ) = 3 + ¡5 = ¡2 A 125 Z t (¡0:03e¡5x ) dx ¡ 2 = 0:125e¡5t ¡ 2:125 A [c] i = 6 0+ 50 Z t [d] i1 = (¡0:03e¡5x ) dx + 3 = 0:1e¡5t + 2:9 A 3 0+ i2 = 25 Z t (¡0:03e¡5x ) dx ¡ 5 = 0:025e¡5t ¡ 5:025 A 6 0+
v (0+ ) = ¡9:6 + 38:4 = 28:8 V [b] v = 0 when 38:4e¡1200t = 9:6e¡300t
电工学原理及应用英文精编版第四版课后练习题含答案-(2)
电工学原理及应用英文精编版第四版课后练习题含答案电工学原理是电气工程中的核心内容之一，掌握电工学原理是成为一名合格电气工程师的必备条件。
而对于英语为第二语言的学生来说，阅读英文电工学原理教材是一项挑战。
因此，本文介绍了电工学原理及应用英文精编版第四版的课后练习题含答案，以便学生们更好地掌握电工学原理。
电工学原理及应用英文精编版第四版简介电工学原理及应用英文精编版第四版是一本介绍电学基础和电力系统基本原理的英文教材。
本书从电学基础开始，逐步介绍了电磁现象、电路、交流电路、电力系统等内容。
此外，本书还涉及最新的用于电力系统控制和保护的数字技术。
本书不仅涵盖了电气工程专业的基础知识，而且也可以作为电子工程、石油化工、化学工程等其他工程学科的基础教材。
课后练习题本书第四版增加了大量习题，这些习题涵盖了电学基础、电磁场、电路、交流电路、电力系统、数字技术等方面的内容。
这些习题分为以下几类：多项选择题•谐振电路是一种能将_转化为_的电路。
(能量；电压)•___代表电量的多少，单位是库仑。
(Q)•___是范德瓦尔斯力引起的电势差。
(塞贝克效应)填空题•一个含有一台发电机的电力系统被称为___。
•巴尔定理被用来计算电路中的___流。
计算题•计算一个电压为220V的交流电路的功率因数，电感L=50mH，电阻R=25Ω，电容C=25μF。
（答案：0.8）问答题•什么是戴维南-楚克定理？•为什么在电力系统中使用三相电？以上仅是本书习题中的一部分，但它们代表了本书中的各个章节的重要知识点。
通过做这些习题，学生们可以更好地理解电工学的基础知识，并逐步掌握电力系统的原理和应用。
习题答案为了方便学生自学，本文还提供了一些习题的答案。
需要注意的是，本文提供的答案仅供参考。
多项选择题答案•谐振电路是一种能将能量转化为电压的电路。
•Q代表电量的多少，单位是库仑。
•塞贝克效应是范德瓦尔斯力引起的电势差。
填空题答案•一个含有一台发电机的电力系统被称为发电系统。
电路 电路原理 尼尔森Riedel 第九版 课后习题答案第 章
V (ω) = Vm
πδ(ω)
+
1 jω
ejωτ /2 − e−jωτ /2
= j2Vmπδ(ω) sin
ωτ 2
+
2Vm ω
sin
ωτ 2
=
(Vmτ ) sin(ωτ /2) ωτ /2
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
+ ej2t
−
e−j2t
+
4ej3t
− 4ej2t}
=
1 πt
3e−j2t − j2
3ej2t
+
4ej3t
− 4e−j3t j2
电路英文版第三章答案
4 × 4 + 9 = 25 V.
The current in the 25Ω resistor is 1 A. The current in the 3Ω resistor is 1 + 3 + 1 = 5 A. Therefore,
the voltage across the 40Ω resistor is 25 + 3(5) = 40 V.
of i4:
i1 = 4i2 = 4(8i3) = 5(32i4)
i2 = 8i3 = 5(8i4)
i3 = 5i4
3
NOW use KCL at the top node to relate the branch currents to the current supplied by the source.
6 A. Therefore the voltage across the 32Ω resistor is 40 + 6(4) = 64 V.
The current in the 32Ω resistor is 64/32 = 2 A. The current in the 10Ω resistor is 6 + 2 = 8 A.
Therefore,
vg = (8)(10) + 64 = 144V
[b]
The current in the Thus, the power
20Ω resistor dissipated by
is i20Ω = the 20Ω
(5)(20 resisto6r
+ is
5)
=
1.2
A
P20Ω = (1.2)2(20) = 288W