18. Tổng hợp bài tập điện xoay chiều (Phần 3)

+ Khi tốc độ quay của roto là $n$ (vòng/phút): $\begin{array}{l}P = \frac{{{\omega ^2}{\Phi ^2}.R}}{{{R^2} + {{\left( {{Z_L} – {Z_C}} \right)}^2}}} = \frac{{{\omega ^2}{\Phi ^2}.R}}{{{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}}} = \frac{{{\omega ^2}{\Phi ^2}.R}}{{{R^2} + {\omega ^2}{L^2} – \frac{{2L}}{C} + \frac{1}{{{\omega ^2}{C^2}}}}}\\ = \frac{{{\Phi ^2}.R}}{{\frac{{{R^2}}}{{{\omega ^2}}} + {L^2} – \frac{{2L}}{{{\omega ^2}C}} + \frac{1}{{{\omega ^4}{C^2}}}}} \Rightarrow P = \frac{{{\Phi ^2}.R}}{{\frac{1}{{{\omega ^4}{C^2}}} + \left( {{R^2} – \frac{{2L}}{C}} \right)\frac{1}{{{\omega ^2}}} + {L^2}}}\\{P_{\max }} \Leftrightarrow {\left( {\frac{1}{{{\omega ^4}{C^2}}} + \left( {{R^2} – \frac{{2L}}{C}} \right)\frac{1}{{{\omega ^2}}} + {L^2}} \right)_{\min }}\\ \Leftrightarrow \frac{1}{{{\omega ^2}}} = \frac{{\frac{{2L}}{C} – {R^2}}}{{\frac{2}{{{C^2}}}}} = \frac{{2\frac{5}{{3\pi }}.\frac{{6\pi }}{{{{5.10}^{ – 4}}}} – {{\left( {100\sqrt 2 } \right)}^2}}}{{\frac{2}{{{{\left( {\frac{{6\pi }}{{{{5.10}^{ – 4}}}}} \right)}^2}}}}} = \frac{1}{{14400{\pi ^2}}}\\ \Rightarrow \omega  = 120\pi  \Rightarrow \left\{ \begin{array}{l}{Z_L} = \omega L = 120\pi .\frac{5}{{3\pi }} = 200\Omega \\{Z_C} = \frac{1}{{\omega C}} = \frac{1}{{120\pi .\frac{{{{5.10}^{ – 4}}}}{{6\pi }}}} = 100\Omega \end{array} \right.\\ \Rightarrow {P_{m{\rm{ax}}}} = \frac{{{\omega ^2}{\Phi ^2}R}}{{{R^2} + {{\left( {200 – 100} \right)}^2}}} = 161,5(*)\end{array}$

+ Khi tốc độ quay của roto là $2n$ (vòng/phút) $\Rightarrow \left\{ \begin{array}{l}{Z_L}^\prime  = 2{Z_L} = 400\Omega \\{Z_C}^\prime  = \frac{{{Z_C}}}{2} = 50\Omega \end{array} \right. \Rightarrow P’ = \frac{{\omega {‘^2}{\Phi ^2}R}}{{{R^2}}} + {\left( {{Z_L}^\prime  – {Z_C}^\prime } \right)^2} = \frac{{4{\omega ^2}R}}{{{R^2} + {{\left( {400 – 50} \right)}^2}}}(**)$

Từ (*) và (**)   

${\rm{}} \Rightarrow \frac{{P’}}{{{P_{m{\rm{ax}}}}}} = {\rm{ }}\frac{{\omega {‘^2}}}{{{\omega ^2}}}.\frac{{{R^2} + {{\left( {200 – 100} \right)}^2}}}{{{R^2} + {{\left( {400 – 50} \right)}^2}}} \Leftrightarrow \frac{{P’}}{{161,5}} = 4.\frac{{{{\left( {100\sqrt 2 } \right)}^2} + {{100}^2}}}{{{{\left( {100\sqrt 2 } \right)}^2} + {{350}^2}}} = \frac{{16}}{{19}} \Rightarrow P’ = 136W$