电路理论_期末考试试题及答案,推荐文档
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一二三四五总分标准分60 10 10 10 10得分2013一、求解下列各题(共60 分,没有过程不得分)1.求电流I 。
(7 分)解:列回路电流方程:I1= 3A-3I1+(3 + 2 + 3)I2- 3I3= 0-6I1-3I2+(6 + 3 + 4)I3= 0补充方程:I =-I2解方程得:I =-I2=-1.8A2.图示正弦电路中两盏白炽灯都不亮,电源电压有效值为220V,角频率为103rad/s,试通过电路分析说明灯不亮的原因。
(7 分)解:XL=wL = 103 ΩX =1= 103 Ωc wc1电容电感串联:jwL -j = 0Ωwc所以,并联的白炽灯被短路串联的白炽灯上的电压为220v,超过其额定电压,被烧坏综上:两灯均不亮2Ω4Ω 3A3Ωa6Ω3ΩIb4Ω15W/110V+u s15W/110V1H1F-13. 图示电路中, R = L = 1C= 10Ω , u S = 10 + 10 2 cos tV ,试求电流 i 的有效值 I 及电压源 u s 发出的功率。
(8 分)当U s 1 = 10V 单独作用时,电路图如下:I = U s 1 1R + R / / R = 2A 3 1 A a3当U s 2 = 10∠00V 单独作用时:X L = jwL = j 10ΩX c = - j wc= - j 10Ω即当电容与电感发生并联谐振,电路图如下: 此时,I 1 = I b =U s 2 R + R = 1 ∠00 A2∴电流 i 的有效值:1I = (1)2 + ( 3 2 )2 = 0.6 Ai = 2 + 12 cos w tA3 2 ∴ P = 2 ⨯10 + 1⨯10 = 11.67w3 2∴ I = 0.6 A P = 11.67wI 1I aRR + R-u s1I 1I bRR + Ru s2-i 1iRR + u s -RCLI =10 2 1 1 14. 图示正弦交流电路中, u (t ) = 20 cos 5tV ,求(1) 该电路的(复)阻抗 Z ,将该电路简化为两个元件串联并求元件参数。
(2) 该单口网络的平均功率和功率因数。
(8 分)解 : Z =10+j 10 Ω最简模型为 10Ω 电阻与 2H 电感串联。
平均功率: P = U 2 Re[Y ] = 2000 ⨯10 = 100W102 +102功率因数:= c os450 =≈ 0.707 2,电感性5.正弦电流电路中, u s (t ) = 16 2 cos(10t ) V ,用网孔分析法求电流 i 1(t ) 和i 2 (t ) 。
(7 分)解0.3H3Ωi 10.1F 0.1HU网孔电流方程( j 3 + 3 + j 2)I - j 2I 2 = 16∠0︒ (- j + j + j 2 + 2)I2 - j 2I = 0= (2 - j 2) A = 2 2∠ - 45 Ai 1 (t ) = 4 cos(10t - 45 ) AI2 = 2 A = 2∠0 A i 2 (t ) = 2 cos(10t ) A6. 对称三相电路,三相负载作星形连接,各相负载阻抗 Z =3+j4Ω,设对称三相电源的线电压 uAB = 380 式。
(8 分)cos(314t + 60︒) V ,试求各相负载电流的瞬时表达解: Z = 3 + j 4 = 5∠53︒ Ω2 2 + S0.2HU S2Ωi 2+ i 10.1H**0.3Hi 23Ω0.4H0.1F2Ω2H+u5Ω0.02F1HI1 3B A ⎣ ⎦线电压U AB = 380∠60︒V相电压U = U ∠ - 30︒ = 220∠30︒VA ABU 220∠30︒ 220∠30︒ ︒∴ I = A = = = 44∠ - 23 A AZ 3 + j 4 5∠53︒I = I ∠ -120︒ = 44∠ -143︒ A I = I ∠120︒ = 44∠97︒ ACA故有: i A = 44i B = 44 cos(314t - 23︒ ) Acos(314t - 143︒ ) Ai C = 44 2 cos(314t + 97︒ ) A7. 图示双口网络的传输参数 T 为T = ⎡2.550Ω⎤ 。
若 U =9V 、R =16,求⎢0.05s 0.7 ⎥ s U s 发出的功率。
(8 分)解:U = 2.5U 2 - 50I 2 I 1 = 0.05U 2 - 0.7I 2U s = U =1 16∠00VU 2 = -16I 2 ∴ I 2 = -0.1∠00 AI 1= 0.15 ∠00 A则U s 发出功率发出功率为 P = 9 ⨯ 0.15 = 1.35w8. 如图所示电路中直流电压源的电压为 24V ,且电路原已达稳态,t=0 时合上开关 S ,求t ≥ 0 时的电感电流i L 。
(7 分) 解:2 2 +U S -RNi (0 ) =48= 6 AL - 4 + 4i L (0+ ) = i L (0- ) = 6 Ai (∞) = 48⨯ 0.5 = 4 A L 4 + 2R = 4 ⨯ 1+ 4 = 6Ω2 = L = 12 = 2s R 6 - t - t - t i L = 4(1- e 2) + 6e 2= 4 + 2e 2A4Ω+48Vi S L12H 4Ω4ΩU0 + 604I 1j 30050Ω +50Ωj 300Ω Z L60∠00 V -6ΩI 13 1 2 3 3I SC 二、 图示电路中开关无论打开还是闭合都不影响电路的工作状态,求 R L 。
(10分)解:回路电流方程 :I 1 = 1A(3 + 6)I 2 - 6I 3 - 3I 1 = 4.5V (6 +1+ R L )I 3 - I 1 - 6I 2 = 4V(6 +1)I ' - I - 6I = 0 I ' = I ∴ R L I 3 = 4∴ I 3 = 2 AR L = 2Ω三、电路如图所示,Z L 为何值时可获得最大功率?并求此最大功率。
(10 分)解:U = -200I -100I + 60 = -300I + 60 = -300 o11160U = = 30 2∠450o 1- j = 60 100 = 0.6∠00 Z = U 0 = = 50 2∠45 0 eqI SC0.6Z L = 50 2∠ - 450 U 2 1800 P = 0 = = 9w4R 200四、已知电源电压 U =150V ,角频率 ω=314rad/s ,功率表的读数为 75W ,且知 R 1= R 2= R 3=R ,I 1=I 2=I 3=I , 求 R 、L 、C 的数值。
(10 分)解:P = I 2R + I 2R + I 2R1 23I 1 = I 2 + I 3 I = I = I123则 = 600 1 * +UV*R 1R 2R 3 P = U I cos= 75wI = 0.5AR = 100ΩLC2 3IwLI 2 = U L = tanwc 3 = U c = tanRI 2 U RRI 3 U RW30 2∠450 III 1L = 0.55HC = 1.84 ⨯10-5 FR = 100Ωc R 1+K 1(t =0) i (t)-K 1F u c (t )+2 -10Vt =t 1R 2五、如图所示电路,已知u C (0- ) = 0 , t =0 时 K 1 闭合,t = ln2s 时 K 2 又闭合, u C (t ) 波形图如图所示,求两个电阻的阻值。
(10 分)解:t < 0 时U c (0+ ) = U c (0- ) = 0 0 < t < ln 2 5 时 U c 1 (∞) = 10V 1 = R 1C -tU (t ) =10(1- e R 1c) 又 t = ln 2 时U c (t ) = 5V ,得 R 1 = 1Ωt > ln 2 52 = C (R 1 / /R 2 ) U c 2(∞) = 10 ⨯ R 2R + R1 2由图可知t > ln 2后无变化 ∴U (∞) = 5V 即 10R 2 = 5R = R = 1Ωc 2R + R 21综上: R 2 = R 1 = 1Ω10A ( ln 2 , 5)5ln 2t/su c (t ) / V1 2At the end, Xiao Bian gives you a passage. Minand once said, "people who learn to learn are very happy people.". In every wonderful life, learning is an eternal theme. As a professional clerical and teaching position, I understand the importance ofcontinuous learning, "life is diligent, nothing can be gained", only continuous learning can achieve better self. Only by constantly learning and mastering the latest relevant knowledge, can employees from all walks of life keep up with the pace of enterprise development and innovate to meet the needs of the market. This document is also edited by my studio professionals, there may be errors in the document, if there are errors, please correct, thank you!“”“”。