Mạch nối tiếp theo thứ tự gồm điện trở thuần R, cuộn dây có độ tự cảm L

* Hướng dẫn giải

Xét đồ thị tại \(t=0\)

\({u_{RL}}\left\{ {\begin{array}{*{20}{c}}
{{U_{0RL}} = \sqrt 6  \cdot x}\\
{{\varphi _{uRL}} = 0}
\end{array}} \right.{u_{RC}}\left\{ {\begin{array}{*{20}{c}}
{{U_{0RC}} = 2x}\\
{{\varphi _{uRC}} =  - \frac{{5\pi }}{{12}}}
\end{array}} \right.\)

Dựa trên giản đồ vector Fresnel :  

+ Định lý hàm cos: \({{\left( {{U}_{L}}+{{U}_{C}} \right)}^{2}}=U_{RL}^{2}+U_{RC}^{2}-2{{U}_{RL}}{{U}_{RC}}\cos \left( \frac{5\pi }{12} \right)\)

\(\Rightarrow {{\text{U}}_{\text{L}}}+{{\text{U}}_{\text{C}}}=\sqrt{4+2\sqrt{3}}\)

\(+\text{S}=\frac{1}{2}*\sqrt{6}*2*\sin \left( \frac{5\pi }{12} \right)=\frac{1}{2}*{{\text{U}}_{\text{R}}}\left( {{\text{U}}_{\text{L}}}+{{\text{U}}_{\text{C}}} \right)\)

\(\Rightarrow {{\text{U}}_{\text{R}}}=\sqrt{3},{{\text{U}}_{\text{L}}}=\sqrt{3},{{\text{U}}_{\text{C}}}=1\)

\(\Rightarrow {{Z}_{L}}=\text{R},{{Z}_{C}}=\frac{\text{R}}{\sqrt{3}}\Rightarrow {{Z}_{L}}{{Z}_{C}}=\frac{{{\text{R}}^{2}}}{\sqrt{3}}\Rightarrow \frac{{{\text{R}}^{2}}\text{C}}{2~\text{L}}=\frac{\sqrt{3}}{2}\)

+ Khi \({{\omega }_{2}}\) thì \({{\text{U}}_{\text{cmax}}}\Rightarrow {{\left( \frac{\text{U}}{{{\text{U}}_{\text{Cmax}}}} \right)}^{2}}=1-{{\left( 1-\frac{{{\text{R}}^{2}}\text{C}}{2~\text{L}} \right)}^{2}}\Rightarrow {{\text{U}}_{\text{cmax}}}=\frac{\text{U}}{\sqrt{1-{{\left( 1-\frac{{{\text{R}}^{2}}\text{C}}{2~\text{L}} \right)}^{2}}}}=50,45(~\text{V})\)