2021電路原理模擬試卷
電路原理模擬試卷電路原理課程模擬試卷 一、 (14%)電路如圖1所示,已知12133,30,10,S S I A U V R R =220,R = 求各支路電流。二、 (14%)電路如圖2所示,已知30,3,5,S S U V I A R =求:a b -左側(cè)電路的戴維南等效電路。 1解: 12322332 S S I I I I R I R U-+=?+?=,232330201030I I I I -+=?+?= 2303I I A= 2解:開路電壓,用迭加定理011115330153333S S U R I U V=?+=?+?= 入端電阻02210233R R R R R R ?=+ 圖 1圖 2U s 2I s三、(12%)電路如圖3所示,已知123412,3,3,6,6,S S U V I A R R R R = (1)試用迭加定理求電阻2R 上的電壓2U ;(2)若保持3S I A =不變,為使得20U =,電流源S U 應(yīng)為多少?四、(12%)電路如圖4所示,正弦交流電壓源0200sin(100060),()S t V t u =+ 2410,10,10,R L H C F -=求:(1)電流()t i , (2)電壓源發(fā)出的有功功率P 和無功功率Q 。 3解:(1)電流源作用 222126S R R U I V R R ?=?=+電壓源作用"22128SU U R V R R =?=+"22214U U U V=+=(2)122212122603S S S U R R U I R U R R R R ?=?+?=+=+9S U V=-圖 3 圖 4S u 4解:(1)00020060200601051101055101010S U I R j L j j j j C R j L j =?-+-+ 0()40sin(1000105)i t t A =+(2)0cos 200cos(45)4000P U I W =?=?-=, 0sin 200sin(45)4000var Q U I =?=?-=-五、(12%)電路如圖5所示,已知200sin(1000),()S t V t u =0.03,M H =1280,0.08R L L H =,5510C F -=?,求電壓0()t u 。六、(12%)三相對(duì)稱電路如圖6所示,電源相電壓120sin au t = V ,負(fù)載阻抗 1240,90Z j Z =。求:(1)畫出單相圖; (2)計(jì)算電流A I ,1A I ,2A I 和AB I 。 5解:00112000236.91808020SU I j j R j L jC=-+-+- 圖 5 圖 6 i S u ab0()t000130236.96053.1U j M I j =?=?-= 00()53.1)u t t V =+ 6解:(1)單相圖(2)0011120039040a A U I Z j =- 002212004030a A Y U I Z = 01243536.9A A A I I I j =+=-=- 2AB A I I A = 七、 (12%)電路如圖7所示,已知100S U V =,12110,100R R L F =,開關(guān)打開已久。當(dāng)0t =時(shí)開關(guān)閉合,求電感電流()L i t 。八、(12%)電路如圖8所示,已知9S U V =, 12330,60,R R R =,1800C F =,(0)0C u -=。開關(guān)打開已久。當(dāng)0t =時(shí)開關(guān)閉合,(1)求()C u t ;(2)欲使開關(guān)閉合后()0.1t A i = (t 0),則電容電壓的初始值(0)C u -應(yīng)為多少? 圖 8 ()C t i LU IA .Z 1IA1. 7解:由三要素法,直接求電感電流211000L R = 12100(0)520S L U i A R R -=+,210S LPU i A R =1000()(0)105tt L LP L LP i t i i i ee -=+-=- 8解: (1)()()(0)(0)tC CP C CP u t u t u u e-=+-221212()0t S SC U U u t R R e R R R R -=?+-?+ 1231211()8080010R R R C R R ?=+?=?=+10()66tC u t e -=-(2) 由11111()(0)0.1(0)0.1ttP P i t i i i ei e-+=+-=+-得 1(0)0.1i +=由迭加定理21232121312123(0)(0)S CU u R i R R R R R R R R R R R R +=-?+ (0)920.160803C u +=-?, (0)(0)6C C u u V +-=