Đặt điện áp xoay chiều u = U

C

- Hướng dẫn giải

Chọn C

U_C = U_Cmax khi ω =$\frac{1}{L}\sqrt{\frac{L}{C}-\frac{R^2}{2}}$  (1) và U_Cmax= $\frac{2UL}{R\sqrt{4LC-R^2{C}^2}}$ (*)

Khi đó Z_L = ωL = $\sqrt{\frac{L}{C}-\frac{R^2}{2}}$   ;  Z_{C ­}= $\frac{1}{C\omega }=\frac{L}{C}\frac{1}{\sqrt{{\displaystyle \frac{L}{C}-\frac{R^2}{2}}}}$

U_RL = $\frac{U\sqrt{R^2+Z_L^2}}{\sqrt{R^2+{(Z_L-Z_C)}^2}}$  ;  U_Cmax=$\frac{U{Z}_C}{\sqrt{R^2+(Z_L-Z_C{)}^2}}$

 =>$\frac{U_{RL}}{U_{Cmax}}=\frac{\sqrt{R^2+Z_L^2}}{Z_C}=\frac{\sqrt{5}}{3}$

=> 9$\left(R^2+Z_L^2\right)$=5$Z_C^2$  => 9(R² +$\frac{L}{C}-\frac{R^2}{2}$ ) = 5$Z_C^2$  

=> 9( $\frac{L}{C}+\frac{R^2}{2}$ )C²($\frac{L}{C}-\frac{R^2}{2}$) =5L² => 9C²( $\frac{L^2}{C^2}-\frac{R^4}{4}$) = 5L²

=> 4L² =$\frac{9R^4{C}^2}{4}$  => 4L= 3R²C (**)

U_Cmax = $\frac{2UL}{R\sqrt{4LC-R^2{C}^2}}=\frac{2UL}{R\sqrt{C(4L-R^{2}C)}}$  =$\frac{2UL}{R\sqrt{2C{R}^{2}C}}=\frac{2U}{\sqrt{2}}\frac{L}{R^{2}C}$

 = $\frac{2U}{\sqrt{2}}.\frac{3}{4}$= 90V

=> U = 60$\sqrt{2}$ (V)