一、分别用回路电流法和结点电压法求图示电路中的电压U2。+-10V2Ω4Ω++---1U4Ω+15.0U2Ω2Ω2U解:回路电流法Il1Il2Il31ll214IIU1024242l3l2l1III1l2l15.0424UIIl324IU1l3l15.0422UIIV42U解得:A1l3IA2l1IA1l2I+-10V2Ω4Ω++---1U4Ω+15.0U2Ω2Ω2U结点电压法①②③④322nnuuU2102121214121n3n2n1uuu1n25.0Uun11uU2102141214121n3n2n1uuuV42U解得:二、试求:RL为何值时可获最大功率?最大功率为多少?-+1AI24ΩI3Ω6ΩLR解:Iu6oc06142III解得:V3ocuIiIiu423用加压求流法得Iiu63解得:6eqiuR当RL=Req=6Ω时,有eq2ocmax4RuPW375.06432三、图示电路中N仅由电阻组成,已知图(a)中U1=1V,电流I2=0.5A,求图(b)中。1ˆIN(b)+-3V0Ω10.3A1ˆI2I(a)1UN+-Ω24A解:22112211ˆˆˆˆiuiuiuiu103.0343.02ˆ2211IIIU11U5.02I解得:A8.10ˆ1Ituc四、图示电路原已稳定,t=0时将开关S闭合,试求开关闭合后的电容电压。-+36V5FS3Ω6ΩCu0t+--+10V解:14V10-36366CuV1000CCuus1056363RCVe2414e141014e01.010tttCCCCuuutu