[电路分析基础4.ppt

2.定理的证明 + 替代 叠加 A中独立源置零 下 页 上 页 a b i + – u N A u a b + – A a b i + – u N0 u a b i + – A Req 返 回 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. 下 页 上 页 i + – u N a b Req Uoc + - 返 回 a b i + – u N A 等效 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. 3.定理的应用 （1）开路电压Uoc 的计算 等效电阻为将一端口网络内部独立电源全部置零(电压源短路，电流源开路)后，所得无源一端口网络的输入电阻。常用下列方法计算： （2）等效电阻的计算 戴维宁等效电路中电压源电压等于将外电路断开时的开路电压Uoc，电压源方向与所求开路电压方向有关。计算Uoc的方法视电路形式选择前面学过的任意方法，使易于计算。 下 页 上 页 返 回 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 3 方法更有一般性。 当网络内部不含有受控源时可采用电阻串并联和△－Y互换的方法计算等效电阻； 开路电压，短路电流法 外加电源法（加电压求电流或加电流求电压）； 下 页 上 页 u a b i + – N0 Req i a b Req Uoc + - u + - a b u i + – N0 Req 返 回 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. 外电路可以是任意的线性或非线性电路，外电路发生改变时，含源一端口网络的等效电路不变(伏-安特性等效)。 当一端口内部含有受控源时，控制电路与受控源必须包含在被化简的同一部分电路中。 下 页 上 页 注意 例1 计算Rx分别为1.2?、5.2?时的电流I I Rx a b + – 10V 4? 6? 6? 4? 解 断开Rx支路，将剩余一端口网络化为戴维宁等效电路： 返 回 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. 求等效电阻Req Req=4//6+6//4=4.8? Rx =1.2?时， I= Uoc /(Req + Rx) =0.333A Rx =5.2?时， I= Uoc /(Req + Rx) =0.2A 下 页 上 页 Uoc = U1 - U2