电工学第三版习题答案

第二章2-1解：i=10sin(314t+60°)mAt=1ms,)31001.0100sin(10si2-3解：(1)452003010021UU(2)u1滞后u275°.2-4解：)1.53sin(25)1.53sin(2521titi2-5解：)30sin30(cos50)150sin150(cos27.7021jUjU2-6解：)60sin60(cos210)30sin30(cos21021jIjI2-8解：电流与电压表的读数为有效值。2-9解：XL=ωLIjXUL2-15解：Z=R1+R2+j(XL-XC)CjIULjIURIUCLR12-28解:220605.0220401cos1cos22111UPIUPIUIPSPPPPUISIIIIIIIcos0)1sin1(cos212122112-30解:22)sin(cos31cos)1sin1(cos01211111IUZIIIIIUPIIIUU2-32:未接电容时接电容后：有功功率不变，电压不变，功率因数提高到0.9。ui11RLRuLuCuiRLRuLu1cos1UIP2122112sin1sinQQQUIQUIQC2-34:(1)求L=2.44MhRLICIILULCf21053.106.310106.303.1523.15IURUUUUIXUXXLXCCCCCLL3-1:)180314sin(22203)60314sin(22202tete3-4:电阻负载上的电流和相位UVWNRXCXLi1iNi3i2N01102210220ARUIp电感负载上的电流和相位电容负载上的电流和相位3-8:WUVUUU1I2I3I1INI2I022902210220AXUILp150222102212022IUUP033902210220AXUICp15022210120333IIUUP1201038120380150103812038090103803801203801203800380313131232323121212312312ZUIjZUIjZUIUUU233131223231211IIIIIIIII4-2解:U1=N1UN1=U1/U=220/0.2=1100U2=N2UN2=180K=N1/N2=1100/180=6.1S=U1I1I1=S/U1=1000/220I2=S/U24-3解:U1/U2=N1/N27-2解:a)ui＜6V,二级管导通，uo=6V。ui＞6V,二级管截止，uo=ui。b)ui＜6V,二级管截止，uo=ui。ui＞6V,二级管导通，uo=6V。7-3解:ui＞6V,稳压管击穿，uo=6V。0V＜ui＜6V,稳压管反偏，uo=ui。ui＜0V，稳压管正偏导通，uo=0V。b)ui＜-6V,稳压管击穿，uo=-6V。-6V＜ui＜0V,稳压管反偏，uo=ui。ui＞0V，稳压管正偏导通，uo=0V。7-5解:（1）UO=1.2U2(2)UO=0.9U2(3)UO=1.4U2(4)UO=U2(5)UO=0.45U2(6)UO=1.414U27-7解:(1)U1是B,U2是E,U3是C,NPN,硅管。(2)U1是B,U2是E,U3是C,NPN,锗管。(3)U1是C,U2是B,U3是E,PNP,硅管。(4)U1是C,U2是B,U3是E,PNP,锗管。8-1解:（1）（2）BCECCBCBIUURII（3）BBECCBCCECCCIUURRUUI（4）取UCE=0.3VBBECCBCCECCCIUURRUUIBBEBRUICCUBCIICCCERIUCCU8-2解：CBCCCCECBBECCBBCBEBBCBCCCRIIUURRUUIIIURIRIIU)()1()(8-3解：VRRRUBBBB2126211可以知道电路饱和。VURUUICECECCCC3.08-5解：直流通路+UCCRC1在交流时被短路，交流电压为零。8-7解：（1）RBRC2RLuiuo交流通路1212EBEBEBEBECCCBBBBIIRURUUIIURRRUCECCCEECCCCCEIRRURIRIUU)(（2）（3）（4）UiAuUoRsririUsUi（5）当Rs=0时，Ui=UsAu不变。Uo=Au·Ui电压放大倍数取决于电路，与信号源无关。rbeRCRLoUiUiIbIcIbIR'B微变等效电路)mA()mV(26)1()(300EbeIrbeLurR'AbebeBirrRr||'CoRr8-8解：直流通路+UCCRB1RB2RCrbebibiRB1RB2RCRLui（3）CBoBbebeLCBRRrRrrirRRRAu21210-1解：S断开：U+=U-=0V,Auf=-10R1/2R1S闭合:假设S点的电位为u111011111111RuoRuRuRuRuui10-2解：第一个运放的输出为u1，第二个运放的输出为u2.122111211uuuuiRRfuRRuuiRRfu10-3解：设Rf2和Rf3之间的电位为u1则：1102131111RfuRfuuoRfuuiRRfu10-5解：111UoRRfUoUiUo10-6解：URRfUoUURRRUiU)11(32310-7解：2110202)10101(1UoUoUoUiKKUoUiKKUo10-9解：2)331(1331)3/31(1uiRkRuoRkRuouiRkRuo10-10解：2211)31(1212)31(uiRRRRRfuiRRRRRfuo10-14解：（a）ui〈2V,uo=10Vui〉2V,uo=-10V(b)ui〈2V,uo=-10Vui〉2V,uo=10V