Gửi 5 năm trước
Hướng dẫn giải
\(L=L_0\) để \(U_{Lmax}\)
+) Khi \(Z_L=\dfrac{R^2+Z_C^2}{Z_C} \Leftrightarrow \dfrac{Z_L-Z_C}{R}.\dfrac{Z_C}{R}=1\) thì:
\(U_{Lmax}=\dfrac{U\sqrt{R^2+Z_C^2}}{R}=\dfrac{U}{\cos \varphi_{RC}}\) khi đó: \(\vec{U} \bot \vec U_{RC}\)
Hệ quả của \(\vec{U} \bot \vec {U_{RC}}\)
+) \(U_{Lmax}^2=U^2+U_{RC}^2\)
+) \(U_{Lmax}(U_{Lmax}-U_C)=U^2\)
+) \(U_C(U_{Lmax}-U_C)=U_R^2\)
+) \(\dfrac{1}{U_R^2}=\dfrac{1}{U^2}+\dfrac{1}{U_{RC}^2}\)
+) \(\left (\dfrac{u}{U} \right )^2+\left (\dfrac{u_{RC}}{U_{RC}} \right )^2=2\)
+) \(\tan \varphi.\tan \varphi_{RC}=1\)
Hướng dẫn giải
\(L=L_0\) để \(U_{Lmax}\)
+) Khi \(Z_L=\dfrac{R^2+Z_C^2}{Z_C} \Leftrightarrow \dfrac{Z_L-Z_C}{R}.\dfrac{Z_C}{R}=1\) thì:
\(U_{Lmax}=\dfrac{U\sqrt{R^2+Z_C^2}}{R}=\dfrac{U}{\cos \varphi_{RC}}\) khi đó: \(\vec{U} \bot \vec U_{RC}\)
Hệ quả của \(\vec{U} \bot \vec {U_{RC}}\)
+) \(U_{Lmax}^2=U^2+U_{RC}^2\)
+) \(U_{Lmax}(U_{Lmax}-U_C)=U^2\)
+) \(U_C(U_{Lmax}-U_C)=U_R^2\)
+) \(\dfrac{1}{U_R^2}=\dfrac{1}{U^2}+\dfrac{1}{U_{RC}^2}\)
+) \(\left (\dfrac{u}{U} \right )^2+\left (\dfrac{u_{RC}}{U_{RC}} \right )^2=2\)
+) \(\tan \varphi.\tan \varphi_{RC}=1\)
Gửi 5 năm trước