Đặt điện áp xoay chiều có giá trị hiệu dụng không đổi nhưng có tần số

* Hướng dẫn giải

Từ đồ thị:

\(\frac{U_{L}^{\max }}{{{U}_{L1}}}=\frac{U_{L}^{\max }}{{{U}_{L2}}}=\frac{7}{4},$ và $U_{L}^{\max }=\frac{U}{\sqrt{1-{{n}^{-2}}}}\xrightarrow{\frac{U_{L}^{\max }}{U}=\frac{14}{6}=\frac{7}{3}}\Rightarrow n=\frac{7\sqrt{10}}{20}\Rightarrow {{\cos }^{2}}{{\varphi }_{L}}=\frac{2}{1+n}=0,95.\)

\({{U}_{L}}=L\omega .\frac{U}{R}\cos \varphi \Rightarrow \frac{1}{{{\omega }^{2}}}={{\left( \frac{U}{{{U}_{L}}}.\frac{L}{R}\cos \varphi \right)}^{2}}\Rightarrow \left\{ \begin{matrix} \frac{1}{\omega _{1}^{2}}={{\left( \frac{U}{{{U}_{L1}}}.\frac{L}{R}\cos {{\varphi }_{1}} \right)}^{2}} \\ \frac{1}{\omega _{2}^{2}}={{\left( \frac{U}{{{U}_{L2}}}.\frac{L}{R}\cos {{\varphi }_{2}} \right)}^{2}} \\ \frac{1}{\omega _{L}^{2}}={{\left( \frac{U}{U_{L}^{\max }}.\frac{L}{R}\cos {{\varphi }_{L}} \right)}^{2}} \\ \end{matrix} \right.. (1).\)

\(\begin{align} & \frac{2}{\omega _{L}^{2}}=\frac{1}{\omega _{1}^{2}}+\frac{1}{\omega _{2}^{2}}\xrightarrow{\left( 1 \right)} \\ & =2.{{\left( \frac{4}{7} \right)}^{2}}.0,95=0,62. \\ \end{align}\) (2)\)

Ta có:   \({{P}_{1}}=U{{I}_{1}}\cos {{\varphi }_{1}}=U\frac{U}{{{Z}_{1}}}\cos {{\varphi }_{1}}=\frac{{{U}^{2}}}{R}{{\cos }^{2}}{{\varphi }_{1}}.\);   \({{P}_{2}}=U{{I}_{2}}\cos {{\varphi }_{2}}=U\frac{U}{{{Z}_{2}}}\cos {{\varphi }_{2}}=\frac{{{U}^{2}}}{R}{{\cos }^{2}}{{\varphi }_{2}}.\)     

       \(=>{{P}_{1}}+{{P}_{2}}=\frac{{{U}^{2}}}{R}({{\cos }^{2}}{{\varphi }_{1}}+{{\cos }^{2}}{{\varphi }_{2}})\xrightarrow{(2)}.\)       

Thế số:

\({{P}_{1}}+{{P}_{2}}={{P}_{CH}}.2.{{\left( \frac{{{U}_{L}}}{U_{L}^{\max }} \right)}^{2}}{{\cos }^{2}}{{\varphi }_{L}}=287.2.{{\left( \frac{4}{7} \right)}^{2}}0,95=178,1\ W\)