[电路分析第4章修分解方法.ppt

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[电路分析第4章修分解方法.ppt

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[电路分析第4章修分解方法

22? 例2: a b 60? 60? 60? 20? 20? 20? 44? 只含电阻 R 结论: 只含电阻单口网络等效为一个电阻 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2.含独立源电路例3: 5? 0.3A 5? 1.5V 0.2A 0.5A 5? 2? 3? 0.5A 1V Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 含独立源和电阻电路或 IS RS US RS 结论 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例 1：求图(a)单口网络的等效电路。将电压源与电阻的串联等效变换为电流源与电阻的并联。将电流源与电阻的并联变换为电压源与电阻的串联等效。 (二) 等效化简的方法——逐步化简 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例2：求 I I 8? 2? 1V 3? 9V 6V 6? 8? I Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. (三)含受控源电路的等效电路 1. 只含受控源和电阻单口网络解：例1、求 ab 端钮的等效电阻(也叫ab端输入电阻)。 a b 50 I I 100? 10? + Uab _ Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 例2、求 ab 端钮的等效电阻。 Rab = 600? a b 1.5k? 1.5k? 1.5k? 750 I1 I1 结论 1、含受控源和电阻的单口网络等效为电阻； 2、受控量支路和未知量支路保留不变换。 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2