电路试卷答案2007年攻读入学考试试题

2007年攻读入学考试试考试科目 电 ：（18%）图一所示电路R12R26，US9V1，非线性电阻伏安（1）（2）I。I

UR此外，利用节点法、回路法对于分析复杂电路十分快捷。1 1

I2R2U KVL92I126∴I1 UOC62I1B1-2：6I2I22I1IKVL2I122I16I2∴RU2I

I3I1

U6I2(11)U1 2n 1U(11 2 nI1 2则有I12n

12

∴R2Ω1

I3R2I3R2UIb233U

I6I6 Ub11 1KVL92I12I1

∴I3

2（方法同思路一）R2Ω U UU U则62IU2II2

I1 I23A（18%）图二所示电路，已知无源二端口网络P在电源频率为时的开路参数（Z参数为16

10j，若U120VR12R3，试求：（1）1110

4j ZU1（2）22端电压U1I21I1 12UPS本题双口网络的参数方程的应用以及等效阻抗的概念及求法U1U116jI110UU210jI14U2RLI2∴

U1有开路参数Z2-1TUS 6jUS 6j6j 则从111ZU136j10j6j1 36j10U116U116jI110U210jI14I1 则解得：U2

129U1U1USI1RS120U2

V 则 I1

R

U1U1I1Z160

U116jI110

I2

435

A

P P U2

129jV 5现求从22看去的戴维南等效电路 1则U116jI110U2U210jI14U1U2U22U116U116jI110U210jI141PUUIR120 1 I1I1

2418jV5

∴

UOC129j

（18%）US12V，R10Ω，L0.1H，C0.01F，20Ω，Kt=0时KKiL(t)和uC(t)。RR Ruc（1）uC(0)iL(0)iL

20iL

uCR则i(0)Us120.6A

uC(0)20iL(0)ILIL(s) ① S10S20IL SUC则有12100.1S)I

L(s)

(s)

(s)

120.06SS(300.1SU(s)20IL(S)12S100 10100 1.2S2372SS(300.1S)(S利用拉斯反变换求uC(t)，iL

(s)

0.4

0.4

SL∴i(t)(0.40.2e300t)(t)L1.2S2372S 120 4Uc(s)

S(300.1)(S

S S∴u(t)(8120e10t4e300t)(t)V （15%）图四所示电路，P

L0.1H，当uS(t)

2100sin(1000t450)(tV，且iL(00A时，电感电流iL(tsin1000t(tA当uS(t)100(t)V，且iL(00.5A时，电感电流iL(t)uSPPa.by()y(0，c（当uS(t

2100sin(1000t450)

且iL(00AUUSZLjwL100

IL

RuSRuSL

则US

1002 20（2）当uS(t)100(t)V且iL(0)0.5AiL(0iL(00.5Ai()1001A 时间常数L0.1

则有i(ti([i(0)i()]e10.5e1000tA(t 输入输出方程u(t)Ri(tLdiL diL(t)1000i10u 通解i(tAe1000tA特解iL(t)iL(0)

则i(t)Ae1000t1L由换路定则知iL(0iL(0LiL(0)A1Li(t)10.5e1000tL

AiL(0)iL(0)0Ai()1001A 时间常数L0.1 则i(t1e1000tAdiLRi L

(t)

SILiL(0)SILL则i(t0.5e1000tALi(t)i(t)

(t)10.5e1000tA(t 则Is100S0.051 SLitt0.5te1000t10.5e1000tLaiaiLbuSPPabuabKustR当uS(tUUS

100A则有I 2A R R

（2）当uS(t)100(t)V且iL(00.5A开路电压

(s)s

则Is100S0.051 Sitt0.5te1000t10.5e1000tLSIL（18%）R1R210L1HC1F（1）H(sIL(s)；（2）列出电路的状态方程；（3）i(0)0A

(0)0V，且US

L uS(t)(tV时，试说明uR2(t)

CLuSiCLuSiLuR2（二）（

UC(s)11SSUSILUL(s)

I(s)SIL(s) 1SS∴US

I

C即：US(s)C

(20S2101S10S

LILL

LL∴H(s)

10S20S2101S

UC(s)则

SUn

(1S n 解得U S(10S U 20S2101S L∴I(s)Un1LS

(10S20S2101S

USH(s)

10S1SSUSIL1SSUSIL1SSUS I1I2IL

(10S)I1(s)SI2(s)US

1SI1(s)(1

10S)I2(s)(S210SI1(s)20S2101S10ULS2SI2(s)20S2101S10US则I

(s)I1(s)I

(s)

10S20S2101S

US

∴H(s)

10S20S2101S选择为状态变量uC(t、iL(t5-45-5（实线为树枝）CLuSCLuSiL

uC(t) 3325 CduC(t)

(t)

LdiL(t)u(t)

R

(t)us(t)uC(t)uR2uR2(t)R2(IR1(t)ILu(t)1u(t)5i(t)1u 2 2i(t)1

(t)1i(t)1u 20

20duC(t)1

1i(t)

u 20 2 20diL(t)1

(t)5i(t)1u 2 2duc 1 1dt

2uc20

i

1L

5L

uCuC 5-6,根据叠加定理：uLa21uCa22iL1u(t5-

auL1uC uCC

2i(t5-

aiC

auLi2i i2i 3u(t5-

iC buLsuu2∴i1u(t)1i(t)1uuu2

s sss 20

20u1u(t)5i(t)1u 2 2

C

uLdiL duC(t)1

(t)1i(t)

1u 20 2 20diL(t)1

(t)5i(t)1u 2 2duc 1 1dt

2uc20

i

1

L L

iR5-10：

uR(11)uuSuC1010 ∴u1u1u

2 2 ∴ uu1

1u 2

2 d 1d 1d d则 R2 S C5 2 2 cduCuR LdiLud

R L LC d2 R

5)duR2

1dR1dRdt 2

2dtd2 1d2即 R25.05 R2 dt

R

2dt∴5.0520 ∵ （15%）图六所示电路，已知1-2段为无损耗均匀传输线，其特征阻抗ZC100,电源f108HZ,波速V3108m/s；求：（1）欲使u(t相位超前u(t45o，则无损线长度l在上面小题确定的无损线长度的中点处接入阻抗ZuR(t)与uS(t) 1l1l2'uR(t' uS ZjZtanZ

ZCCjZ2tanZCZ

cotxjZcot2x 设设USUIRZ IRZUR

R

R

cot若使URURZ

Zcot2x

v3108 IxarccotR3m I ZC uR(t)相位超前uS电路中电流超前电压，为容性6-2：RR

cot2

x3131Z24'uR' uS ZZZjZtanx cZjZtan

1l1 Z121ZjZtanx0Zj100tan ZjZcotxj100tan 又Z2Z

Z100

50jtancot 2Z2jZctanxk(ZcjZ2tank11k21Z2Z2又Z ZZ 850252252j(Ω)1jcot8

Z50252252(ΩZ（18%）图七所示电路，N为线性电阻网络，已知电压源US18V，ab端口开路电压（1ab（2）ab（3）I1的变化量I1是多少？IIIUSNU（1） I则I121.2A

I则I

121.5(8

abRL2Ω时，UabI2RL2.4V。当短路时，有Uab0V1 2 1 2则U1Uˆ1

U2

I22 ˆ2 I1IˆI 7-I IRRRUSRUUab开路电压为

US

2VReqReq

R1R3

8Ω

82 R 3 abRL2Ω时I21I1.21.2(R2RL)1

6 2abI21I1.51.5R261A1 则II

（15%）8US2000VR120ΩR250ΩN1N212N3N451I1以及电压源US URS VARKCLKVL解：8-R R2 U12I2

25I1I1 2 25I1I1I1