Từ đồ thị, ta có: \(\frac{T}{2} = 10\;{\rm{ms}} \Rightarrow {\rm{T}} = 20\;{\rm{ms}} \Rightarrow {\rm{\omega }} = \frac{{2{\rm{\pi }}}}{{\rm{T}}} = 100{\rm{\pi }}({\rm{rad}}/{\rm{s}})\)

\({{\rm{U}}_{{\rm{OAN}}}} = 100\;{\rm{V}} = \sqrt {{\rm{U}}_{{\rm{OR}}}^2 + {\rm{U}}_{{\rm{OL}}}^2} {\rm{ v\`a  }}{{\rm{U}}_{{\rm{OMB}}}} = 75\;{\rm{V}} = \sqrt {{\rm{U}}_{{\rm{OR}}}^2 + {\rm{U}}_{{\rm{OC}}}^2} \)

Tại t = 0: \[\left\{ \begin{array}{l}{u_{AN}} =  - 75V =  - \frac{3}{4}{U_{0AN}}\\{u_{MB}} = 50V = \frac{2}{3}{U_{0MB}}\end{array} \right. \Rightarrow {u_{AN}} \bot {u_{MB}}\]

\( \Rightarrow \frac{1}{{{\rm{U}}_{{\rm{OR}}}^2}} = \frac{1}{{{\rm{U}}_{{\rm{OAN}}}^2}} + \frac{1}{{{\rm{U}}_{{\rm{OMB}}}^2}} \Rightarrow {{\rm{U}}_{{\rm{OR}}}} = 60\;{\rm{V}} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{{\rm{U}}_{{\rm{OL}}}} = 80\;{\rm{V}}}\\{{{\rm{U}}_{{\rm{OC}}}} = 45\;{\rm{V}}}\end{array}} \right.\)

Cảm kháng: \({{\rm{Z}}_{\rm{L}}} = \frac{{{{\rm{U}}_{0\;{\rm{L}}}}}}{{{{\rm{I}}_0}}} = 40\Omega  \Rightarrow {\rm{L}} = \frac{{{{\rm{Z}}_{\rm{L}}}}}{\omega } = \frac{{40}}{{100\pi }} = 127{\rm{mH}}\)

Dung kháng: \({{\rm{Z}}_{\rm{C}}} = \frac{{{{\rm{U}}_{0{\rm{C}}}}}}{{\rm{I}}} = 22,5\Omega  \Rightarrow {\rm{C}} = \frac{1}{{{{\rm{Z}}_{\rm{C}}}.{\rm{\omega }}}} = 141,{5.10^{ - 6}}\;{\rm{F}} = 141,5{\rm{\mu F}}\). Chọn D.