应用叠加定理求图(a)所示电路中的电流[tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex]和电压[tex=0.786x1.286]sgM90Q/VISKeSqiI8AMXRw==[/tex]。[img=370x277]17a90ce22a582c2.png[/img]

2022-06-03

应用叠加定理求图(a)所示电路中的电流[tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex]和电压[tex=0.786x1.286]sgM90Q/VISKeSqiI8AMXRw==[/tex]。[img=370x277]17a90ce22a582c2.png[/img]

答案：

图(b)为[tex=1.786x1.286]ZUCZfagblGnyc1pUqP2exw==[/tex]电压源单独作用的分电路，图(c)为[tex=1.286x1.0]N1sDdbn1Sa5N1z7WJ61ATw==[/tex]电流源单独作用的分电路。由图(b)所示电路得[img=793x280]17a90cea29da98d.png[/img][tex=11.071x5.357]Uhu2/1bhRl3dw2iQmc0iycsI+74nr+0EuNAQeknYG3HwcDp4RrGZVMsb6RLXEWeVecJCAVKUXemVTDNGDxjpWSYLLAgMeqh3PfWSgmejTQSKHJavt4eLGaZETsC463XcsZKLtazvqNetxQ/kmvS8WcvXx/3xDM9bFmpofKibX7mHO6fSRIIuhy+X2MylpcUnkFx1uOJlwFX2B6HyT6NZLdI8jF5Snt9i4+WPY2zRwcT2lQ2/kbcshLKOmAsBbjLJ[/tex]由图(c)得[tex=11.0x5.643]Uhu2/1bhRl3dw2iQmc0iyd7f1DgxMHdAV2d7NUcTvX6Fo7YllkuIDkUn2tLqHu5Z64WVig4+sz2bGLL283WxrHAxX4/FbtNh27plo3Ez8uc3jKBGqezSwt/nn5g8UUZedFNtQK5WwxPVqdAMHmFUHr3hVhPbanJL5CUJlhOwh/qL21GhXqWvr4jbZjUipNbFkaIN09s8jDHonztNfMHNokopz22hjjyBufcJ8Z899FaGh56hvNOw+fTidXBVOYzXSbUTSWy8UFupwD6sLxWoTuihdHgFX/WY9VYuF3rxeaw=[/tex]由叠加定理得[tex=10.214x3.071]a0s3MH7cLIdmiBRR0YN06/5iQNSUaMwiEanaJ+UGwjH5Qh8OugFACszn1lgqfQi5jQa3eeOIu1qw7FNKlP3XQCTNt+Saf8D+rNFpFyBYZdCJ0mLJ//X2ST8AB8HaG/WApK3dZbdTArpXNbp6ONQxcFi4x5JGcKjRE8BwcE9CW0o=[/tex]

举一反三

内容

0
求如题1-30图所示电路中的电压 [tex=0.929x1.286]YoBoYoHhrieJAbBhElPZhQ==[/tex] 和电流 [tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex] 。[img=375x263]17984e237a6ff39.png[/img]
1
如图[tex=0.571x1.286]PEWVZEEj5wMHw+B9HlN8XQ==[/tex]所示电路，用戴维南定理求电流[tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex]．[img=221x155]1798d6f74774b83.png[/img]
2
用叠加原理求图电路中的电流[tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex]．[img=324x191]17975e233c0676a.png[/img]
3
电路如图所示, 用诺顿定理求流过[tex=1.214x1.286]o47RZRZhxPwtQ4v6GqTDNQ==[/tex]电阻的电流[tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex]。[img=332x264]17d28bfb37c69b8.png[/img]
4
电路如图所示，有一直流电动机，经两条电阻为[tex=4.0x1.286]kHsD7CacPD3iUEWq3ngTQQt9dKvRdbCcaiOqW48S1EE=[/tex]的导线接在[tex=2.214x1.286]l/itewrgP1krfAKfomqs2Q==[/tex]的电源上，已知电动机消耗的功率为[tex=2.5x1.286]srMJyd+m411UPU59Y/QiFA==[/tex]，求电动机的端电压[tex=0.786x1.286]sgM90Q/VISKeSqiI8AMXRw==[/tex]和取用的电流[tex=0.571x1.286]RM7SKoKhXo5BhokAgZJ3fQ==[/tex]．[img=316x237]179752ddd39e1f5.png[/img]