电路原理作业1、2及答案.ppt

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1、作业1P1.1求下面两电路中的电流I，电压U。±I+–U4V3Ω1A±I+–U4V3Ω1AEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P1.2求上题中图（a）电阻上消耗功率，电压源和电流源发出或吸收功率，验证功率平衡。P1.3将题1图（b）中3Ω短路，求电压源和电流源的功率，验证功率平衡。P1.4下图是晶体管电路模型，求电压放大倍数Uo/Ui。

2、RUiIi+–+–UoRLkIIiEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P1.5求图中电流源5A的功率和VCCS的功率，是吸收还是发出，验证功率平衡。+–U5A2Ω2ΩEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-20

3、11AsposePtyLtd.P1.6用KCL、KVL求图中电流I。2Ω3Ω6Ω±1V±1V1Ω±I2VEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.作业1解答P1.1（a）U=4V（b）I=－1AＵ＝4-3I=4-3×（-1）=7VP1.2（a）电阻上消耗功率电流源功率，发出电压源功率，发出，功率平衡。Evaluationonly.Create

4、dwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P1.33Ω短路后，电压源功率P4V=4×（-1）=-4V，吸收电流源功率P1A=4×1=4W，发出P4V+P1A=4–4=0，功率平衡。P1.4电阻R上电流受控电流源电流，电阻RL上电压，电压放大倍数，负号表示RL上实际电压极性相反。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5Clien

5、tProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P1.5两电阻2Ω并联，故电流相等，设为I，根据KCL，，得U=2V,I=1A。6Ω±1V±1V2Ω3Ω1Ω±I2VI1I2I3PIs=52=10(W)PR=I2R=1(2+2)=4(W)功率守恒。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.6Ω±

6、1V±1V2Ω3Ω1Ω±I2VI1I2I3P1.6取回路如图，由KVL，，，，最后得。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.作业2P2.1图示电路中，E=40V，R1=R3=10，R2=R4=5。用支路电流法求I1、U3。R1±E+U3¯I10.1U3R2R3R4–+10I1abcdEvaluationonly.Createdwith

7、Aspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P2.2利用节点法求上述电路中的I1、U3。P2.3利用回路电流法求图示电路中的Ua、Ub。2±12V4A2124abEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.作业2解答P2.1参考方

8、向规定如图，由KCLI1-I2-0.1U3=0I2-I3-I5=00.1U3+I5-I4=0由KVLR1I1+R2I2+R3I3=ER4I4-R3I3=10I1U3=R3I3R1±E+U3¯I1I2I30.1U3R2R3R4–+10I1abcdI4I5Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解上述方程，得：I