Câu hỏi 8

+ Khi tốc độ quay của roto là n (vòng/phút):

\(\eqalign{
& P = {{{\omega ^2}{\Phi ^2}.R} \over {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}} = {{{\omega ^2}{\Phi ^2}.R} \over {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}}} = {{{\omega ^2}{\Phi ^2}.R} \over {{R^2} + {\omega ^2}{L^2} - {{2L} \over C} + {1 \over {{\omega ^2}{C^2}}}}} = {{{\Phi ^2}.R} \over {{{{R^2}} \over {{\omega ^2}}} + {L^2} - {{2L} \over {{\omega ^2}C}} + {1 \over {{\omega ^4}{C^2}}}}} \cr
& \Rightarrow P = {{{\Phi ^2}.R} \over {{1 \over {{\omega ^4}{C^2}}} + \left( {{R^2} - {{2L} \over C}} \right){1 \over {{\omega ^2}}} + {L^2}}} \cr} \)

\(\eqalign{
& {P_{\max }} \Leftrightarrow {\left[ {{1 \over {{\omega ^4}{C^2}}} + \left( {{R^2} - {{2L} \over C}} \right){1 \over {{\omega ^2}}} + {L^2}} \right]_{\min }} \Leftrightarrow {1 \over {{\omega ^2}}} = {{{{2L} \over C} - {R^2}} \over {{2 \over {{C^2}}}}} = {{2{5 \over {3\pi }}.{{6\pi } \over {{{5.10}^{ - 4}}}} - {{\left( {100\sqrt 2 } \right)}^2}} \over {{2 \over {{{\left( {{{6\pi } \over {{{5.10}^{ - 4}}}}} \right)}^2}}}}} = {1 \over {14400{\pi ^2}}} \Rightarrow \omega = 120\pi \cr
& \Rightarrow \left\{ \matrix{
{Z_L} = \omega L = 120\pi .{5 \over {3\pi }} = 200\Omega \hfill \cr
{Z_C} = {1 \over {\omega C}} = {1 \over {120\pi .{{{{5.10}^{ - 4}}} \over {6\pi }}}} = 100\Omega \hfill \cr} \right. \Rightarrow {P_{m{\rm{ax}}}} = {{{\omega ^2}{\Phi ^2}R} \over {{R^2} + {{\left( {200 - 100} \right)}^2}}} = 161,5(*) \cr} \)

+ Khi tốc độ quay của roto là 2n (vòng/phút)

\( \Rightarrow \left\{ \matrix{
{Z_L}' = 2{Z_L} = 400\Omega \hfill \cr
{Z_C}' = {{{Z_C}} \over 2} = 50\Omega \hfill \cr} \right. \Rightarrow P' = {{\omega {'^2}{\Phi ^2}R} \over {{R^2} + {{\left( {{Z_L}' - {Z_C}'} \right)}^2}}} = {{4{\omega ^2}R} \over {{R^2} + {{\left( {400 - 50} \right)}^2}}}(**)\)

 Từ (*) và (**) \( \Rightarrow {{P'} \over {{P_{m{\rm{ax}}}}}} = {{\omega {'^2}} \over {{\omega ^2}}}.{{{R^2} + {{\left( {200 - 100} \right)}^2}} \over {{R^2} + {{\left( {400 - 50} \right)}^2}}} \Leftrightarrow {{P'} \over {161,5}} = 4.{{{{\left( {100\sqrt 2 } \right)}^2} + {{100}^2}} \over {{{\left( {100\sqrt 2 } \right)}^2} + {{350}^2}}} = {{16} \over {19}} \Rightarrow P' = 136W\)