解答:解答:开关K闭合时:开关K打开时:Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewords
解答:由已知的电路图可知:选B解答:由已知的电路图可知:(1)相量分析:(2)由相量图分析: 电阻上的电流大小:电容上的电流大小:Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.asposewo
解答:由: 或: 由以上数据得出:的虚部为零即为实数由解答:先求由下图可以求出:画出时刻的等效电路如下图:由图可知:Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: :products.
解答分析:去耦等效电路如下图所示: 相量等效电路如下图所示:(1)只有电感与电容串联的支路发生串联谐振时(短路时)电流i的值最小且等于0即:(2)只有电感与电容串联的支路发生开路时电流i的值最大且等于即:Created with an evaluation copy of Aspose.Words. To discover the full versions of our APIs
2008一填空题(共12小题每题5分):1[ 1 ][1W]2. [8W ]3. U[8V ].4 [-36W ]516V0A32V4A4A6789380100111250二解:根据叠加定理有 u=K1 usK2 Is代如已知数据得 4K1K2=0 K1 = 12
2009填空题14 W 212.5 V 3W45 6 7 8 9 10750W 110A 120.50.1931sin(100πt-72.35O)0.02239sin(3×100πt-83.94 O) A解:用叠加原理求解当12V电压源单独作用时电压所以当电压源为0时电
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2010一.填空题1)NA发出功率NB吸收功率NA吸收功率NB发出功率2) 2A15V3) 1A1S 4)R5)6)7)8)27 W9)10)11)10V0.2A12)二.取网孔为回路且顺时针绕行列出结点电压方程:解得 三. 回路方程解得受控源中电流故受控源的功率为零四. 当IS1单独作用时有 当IS2单独作用时有 根据叠加原理两个电源共同作用时―uocReq5ΩI 若
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