電工學(xué)第三版習(xí)題答案

第二章2-1解：i=10sin(314t+60°)mA10sin(1000.001s1)it=1ms,3U100302-3解：(1)1U200452(2)u滯后u75°.12i52sin(t53.1)2-4解：1i52sin(t53.1)22-5解：U70.7(cos150jsin150)12U50(cos30jsin30)22-6解：I10(cos30jsin30)12I10(cos60jsin60)222-8解：電流與電壓表的讀數(shù)為有效值。2-9解：X=ωLLUjXIL2-15解：Z=R1+R2+j(X-X)LCUIRRUIjLL1CUIjC2-28解:PUIcos11PUcos12200.540I11P60I22U220II(cos1sin1)11II022III12SUIPPP12PcosS2-30解:UU0II(cos1sin1)11PUcos1I11I3I(cossin)1III21UZ2I22-32:未接電容時(shí)iRPUI1cos1uRuLuL接電容后：iRuRuLuCL有功功率不變，電壓不變，功率因數(shù)提高到0.9。ICUILIRLQUIsin111QUIsin222QQQC122-34:1(1)f2LC0求L=2.44MhXL15.3LXXLCUIX215.330.6CCU10UCUU3.06C10UR1.53Ie22202sin(314t60)3-1:e32202sin(314t180)3-4:i1URiNNNXLXCi3Wi2V電阻負(fù)載上的電流和相位IUp22022A101R100電感負(fù)載上的電流和相位IUp22A10220UU120X22PL900I222102215022電容負(fù)載上的電流和相位U220UU120Ip22A103PX3C900II21022150333UWI2UI1UINI2I1IUV33-8:U380012U38012023U38012031U3800I12109012Zj3812U38012023I1015023Z23j38U38012031I1012031Z3138III11231III22312III331234-2解:U=NUN=U/U=220/0.2=11001111U=NUN=180222K=N1/N=1100/180=6.12S=UII=S/U=1000/2201111I=S/U224-3解:U/U=N/N21217-2解:a)u＜6V,二級(jí)管導(dǎo)通，iu=6V。ou＞6V,二級(jí)管截止，iu=u。oib)u＜6V,二級(jí)管截止，u=u。iiou＞6V,二級(jí)管導(dǎo)通，u=6V。io7-3解:u＞6V,穩(wěn)壓管擊穿，u=6V。io0V＜u＜6V,穩(wěn)壓管反偏，iu=u。oiu＜0V，穩(wěn)壓管正偏導(dǎo)通，iu=0V。ob)u＜-6V,穩(wěn)壓管擊穿，iu=-6V。o-6V＜u＜0V,穩(wěn)壓管反偏，iu=u。oiu＞0V，穩(wěn)壓管正偏導(dǎo)通，iu=0V。o7-5解:（1）U=1.2UO2(2)U=0.9UO2(3)U=1.4U2O(4)U=UO2(5)U=0.45U2O(6)U=1.414UO27-7解:(1)U是B,U是E,U是C,NPN,硅管。312(2)U是B,U是E,U是C,NPN,鍺管。312(3)U是C,U是B,U是E,PNP,硅管。312

(4)U是C,U是B,U是E,PNP,鍺管。3128-1解:IUUCC（1）BERBBIICBUUIRCCCECC（2）IICBUURCCCEBIB（3）UUICCCECRCUURCCBEBIB（4）取U=0.3VCEUUICCCECRCUURCCBEBIB8-2解：U(II)RIRUBECCCBCBBIICBUUICCBER(1)RBBCUU(II)RCECCCBC8-3解：RU6122VB1RRBB1B2可以知道電路飽和。UUCCICCERCU0.3VCE8-5解：直流通路+UCC交流通路uoRLRC2uiRBR在交流時(shí)被短路，交流電壓為零。C18-7解：（1）UURB2RRB1BCCB2UUUIIRBBEBCEREEIIE1BUUIRIREECCCCCEU(RR)ICCCEC（2）IIIbciRLR'BrbeIbUURCoi微變等效電路（3）26(mV)r300()(1)beI(mA)ER'AuLrberR'||rrbeiBberRoC（4）riUiUsriRsUoAuUi（5）當(dāng)Rs=0時(shí)，Ui=UsAu不變。Uo=Au·Ui電壓放大倍數(shù)取決于電路，與信號(hào)源無(wú)關(guān)。8-8解：直流通路+UCCRCRRB2B1iibbruiRbeRLB1RRB2C（3）RRRLAuB2CrberirRbeB1rRRCoB210-1解：S斷開(kāi)：U=U=0V,Auf=-10R1/2R1+-S閉合:假設(shè)S點(diǎn)的電位為u1uiu1u1u1R1R1R1u1uoR110R110-2解：第一個(gè)運(yùn)放的輸出為u1，第二個(gè)運(yùn)放的輸出為u2.u1RfR1uiu2u1RfRuiR1Ru21u2u110-3解：設(shè)Rf2和Rf3之間的電位為u1Rf1u1uiR1u1uou10u1Rf3Rf2Rf1則：10-5解：Uo1UiUoRfUo1R110-6解：R3R2R3UUiUUUo(1Rf)UR110-7解：10KUo1(1)Ui10KUo220K10KUiUoUo1U