Gửi 5 năm trước
Hướng dẫn giải
\(C=C_0 \) để \(U_{Cmax}\)
+) Khi \(Z_C=\dfrac{U\sqrt{R^2+Z_L^2}}{Z_L} \Leftrightarrow \dfrac{Z_C-Z_L}{R}.\dfrac{Z_L}{R}=1\)
\(U_{Cmax}=\dfrac{U\sqrt{R^2+Z_L^2}}{R}=\dfrac{U}{\cos \varphi_{RL}}\) khi đó \(\vec{U} \bot \vec{U_{RL}}\)
Hệ quả của \(\vec{U} \bot \vec{U_{RL}}\)
+) \(U_{Cmax}^2=U^2+U_{RL}^2\)
+) \(U_{Cmax}(U_{Cmax}-U_L)=U^2\)
+) \(U_L(U_{Cmax}-U_L)=U_R^2\)
+) \(\dfrac{1}{U_R^2}=\dfrac{1}{U^2}+\dfrac{1}{U_{RL}^2}\)
+) \(\left ( \dfrac{u}{U} \right )^2+\left ( \dfrac{u_{RL}}{U_{RL}} \right )^2=2\)
+) \(\tan \varphi.\tan \varphi_{RL}=1\)
Hướng dẫn giải
\(C=C_0 \) để \(U_{Cmax}\)
+) Khi \(Z_C=\dfrac{U\sqrt{R^2+Z_L^2}}{Z_L} \Leftrightarrow \dfrac{Z_C-Z_L}{R}.\dfrac{Z_L}{R}=1\)
\(U_{Cmax}=\dfrac{U\sqrt{R^2+Z_L^2}}{R}=\dfrac{U}{\cos \varphi_{RL}}\) khi đó \(\vec{U} \bot \vec{U_{RL}}\)
Hệ quả của \(\vec{U} \bot \vec{U_{RL}}\)
+) \(U_{Cmax}^2=U^2+U_{RL}^2\)
+) \(U_{Cmax}(U_{Cmax}-U_L)=U^2\)
+) \(U_L(U_{Cmax}-U_L)=U_R^2\)
+) \(\dfrac{1}{U_R^2}=\dfrac{1}{U^2}+\dfrac{1}{U_{RL}^2}\)
+) \(\left ( \dfrac{u}{U} \right )^2+\left ( \dfrac{u_{RL}}{U_{RL}} \right )^2=2\)
+) \(\tan \varphi.\tan \varphi_{RL}=1\)
Gửi 5 năm trước