电路理论基础习题答案

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1、电路理论基础习题答案第一章1-1.(a)、(b)吸收10W；(c)、(d)发出10W.1-2.–1A;–10V;–1A;–4mW.1-3.–0.5A;–6V;–15e–tV;1.75cos2tA;3Ω;1.8cos22tW.1-4.u=104i;u=-104i;u=2000i;u=-104i;0tmspmW41002250tmsimA4101-5.i,A0.4.75t0.25　1.25ms-0.4(d)1-6.0.1A.1-7.u,V1000　10t,ms(e)u,V800　10-20　t,ms(f)iA0ms11231-e-

2、tt1-8.2mF;4mC;0;4mJ.1-9.9.6V,0.192W,1.152mJ;16V,0,3.2mJ.p(W)100　12t(s)-101-10.1–e-106tA,t>0取s.1-11.3H,6(1–t)2J;3mH,6(1–1000t)2mJ;1-12.0.4F,0.1-13.供12W;吸40W;吸2W;(2V)供26W,(5A)吸10W.1-14.–40V,–1mA;–50V,–1mA;50V,1mA.1-15.0.5A,1W;2A,4W;–1A,–2W;1A,2W.1-16.10V,50W;50V,250W;

3、–3V,–15W;2V,10W.1-17.(a)2V;R耗4/3W;US:–2/3W,IS:2W;(b)–3V;R耗3W;US:–2W,IS:5W;(c)2V,–3V;R耗4W;3W;US:2W,IS:5W;1-18.24V,发72W;3A,吸15W;24V6电压源;3A↓电流源或5/3Ω电阻.1-19.0,US/RL,US;US/R1,US/R1,–USRf/R1.1-20.6A,4A,2A,1A,4A;8V,–10V,18V.1-21.K打开：(a)0,0,0;(b)10V,0,10V;(c)10V,10V,0;K闭合:(

4、a)10V,4V,6V;(b)4V,4V,0;(c)4V,0,4V;1-22.2V;7V;3.25V;2V.1-23.10Ω.1-24.14V.1-25.–2.333V,1.333A;0.4V,0.8A.1-26.80V.※第二章2-1.2.5Ω;1.6R;8/3Ω;0.5R;4Ω;1.448Ω;.R/8;1.5Ω;1.269Ω;40Ω;14Ω.2-2.,.2-3.1.618Ω.2-4.400V;363.6V;IA=.5A,电流表及滑线电阻损坏.2-6.5kΩ.2-7.0.75Ω.2-8.R1=38KΩ,R2=10/3KΩ2-

5、9.10/3A,1.2Ω;–5V,3Ω;8V,4Ω;0.5A,30/11Ω.2-10.1A,2Ω;5V,2Ω;2A;2A;2A,6Ω.2-11.–75mA;–0.5A.2-12.6Ω;7.5Ω;0;2.1Ω.2-13.4Ω;1.5Ω;2kΩ.2-14.1.6V,-24V2-15.(a)(b)–2A↓,吸20W.2-16.2-17.3A.2-18.7.33V.2-19.86.76W.2-20.1V,4W.2-21.64W.2-22.电压源发50W，电流源发1050W2-232-24.7V,3A;8V,1A.2-25.4V,2.5

6、V,2V.2-26.2Ω2-27.60V.2-28.4.5V.2-29.–18V.2-30.原构成无解的矛盾方程组;(改后)4V,10V.2-312-32.2.5A2-33.3.33kV,50kV.2-34加法运算2-35.R3(R1+R2)iS/R1.2-36.可证明IL=－uS/R3.2-37.–2;4.※第三章3-1.44V;–1+9=8V;6+9=15V;sint+0.2e–tV.3-2.155V.3-3.190mA.3-4.1.8倍.3-5.左供52W,右供78W.3-6.1V;1A;0.75A.3-7.3A;1.3

7、3mA;1.5mA;2/3A;2A.3-8.20V,–75.38V.3-9.–1A;2A;1A.3-10.5V,20V;–2V,4V.3-12.4.6V.3-13.2V;0.5A.3-14.10V,5kV.3-15.3-16.22.5V3-17.4/3V,75W;4/3V,4.69W.3-18.3V,529/12W.;1V,2.25W.3-193-20.50V.3-21.0.2A.3-22.1A.3-23.1.6V.3-24.4A;3-25.23.6V;5A,10V.3-26.3-274V3-28.※第四章4-1.141.1V

8、,100V,50Hz,0.02s,0o,–120o;120o.4-2.7.07/0oA,1/–45oA,18.75/–40.9oA.4-3.,7.75mA.4-4.10/53.13oA,10/126.87oA,10/–126.87oA,10/–53.13oA；各瞬时表达式略。