Gửi 5 năm trước
Hướng dẫn giải
+) \(\tan \varphi_{RC}.\tan \varphi=-\dfrac{1}{2} \) khi \(U_{Cmax}\) với \(\omega\) thay đổi
+) \(\tan \varphi_{RL}.\tan \varphi=-\dfrac{1}{2}\) khi \(U_{Lmax}\) với \(\omega\) thay đổi
+) Khi \(R^2=\dfrac{L}{C} \Leftrightarrow \left\{\begin{matrix}\tan \varphi_{RL}.\tan \varphi_{RC}=1\\ \vec{U_{RL}} \bot \vec{U_{RC}}\end{matrix}\right.\)
+) \(\left\{\begin{matrix}\varphi_{AM}-\varphi_{MB}=\pm \dfrac{\pi}{2} \rightarrow \tan \varphi_{AM}.\tan \varphi_{MB}=-1\\ \varphi_{AM}+\varphi_{MB}=\pm \dfrac{\pi}{2} \rightarrow \tan \varphi_{AM}.\tan \varphi_{MB}=1\end{matrix}\right.\)
+) \(\tan(\varphi _{AM}-\varphi_{MB})=\dfrac{\tan \varphi_{AM}-\tan \varphi_{MB}}{1+\tan \varphi_{AM}.\tan\varphi_{MB}}\)
+) Cho mạch \(RL\) có \(u=A\cos^2(\omega t+\varphi)\) khi đó:
\(I=\sqrt{I_1^2+I_2^2}\) với \(\left\{\begin{matrix}I_1=\dfrac{A}{2R}\\ I_2=\dfrac{A}{\sqrt{8(R^2+Z_L^2)}}\end{matrix}\right.\)
Hướng dẫn giải
+) \(\tan \varphi_{RC}.\tan \varphi=-\dfrac{1}{2} \) khi \(U_{Cmax}\) với \(\omega\) thay đổi
+) \(\tan \varphi_{RL}.\tan \varphi=-\dfrac{1}{2}\) khi \(U_{Lmax}\) với \(\omega\) thay đổi
+) Khi \(R^2=\dfrac{L}{C} \Leftrightarrow \left\{\begin{matrix}\tan \varphi_{RL}.\tan \varphi_{RC}=1\\ \vec{U_{RL}} \bot \vec{U_{RC}}\end{matrix}\right.\)
+) \(\left\{\begin{matrix}\varphi_{AM}-\varphi_{MB}=\pm \dfrac{\pi}{2} \rightarrow \tan \varphi_{AM}.\tan \varphi_{MB}=-1\\ \varphi_{AM}+\varphi_{MB}=\pm \dfrac{\pi}{2} \rightarrow \tan \varphi_{AM}.\tan \varphi_{MB}=1\end{matrix}\right.\)
+) \(\tan(\varphi _{AM}-\varphi_{MB})=\dfrac{\tan \varphi_{AM}-\tan \varphi_{MB}}{1+\tan \varphi_{AM}.\tan\varphi_{MB}}\)
+) Cho mạch \(RL\) có \(u=A\cos^2(\omega t+\varphi)\) khi đó:
\(I=\sqrt{I_1^2+I_2^2}\) với \(\left\{\begin{matrix}I_1=\dfrac{A}{2R}\\ I_2=\dfrac{A}{\sqrt{8(R^2+Z_L^2)}}\end{matrix}\right.\)
Gửi 5 năm trước