Chọn D.

Khi C = Co

\({U_{AN}} = \frac{{U\sqrt {{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }}{{\sqrt {{{\left( {R + r} \right)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = min\)

Ta có:

\({U_{AN}} = \frac{U}{{\sqrt {1 + \frac{{{R^2} + 2Rr}}{{{r^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}} }}\;\left( 1 \right)\)

\(\left( 1 \right) \Rightarrow {U_{AN}} = min\) khi 

\({Z_C} = {Z_L}\)

\(\Rightarrow {Z_L} = {Z_{C0}}\;\left( 2 \right)\)

Khi \(C = \frac{{{C_0}}}{4}\) thì \({Z_C} = 4{Z_{C0}}\)

\({U_{AM}} = {U_C} = max\)

\(\Rightarrow \;{Z_C} = \frac{{{{\left( {R + r} \right)}^2} + Z_L^2}}{{{Z_L}}}\)

\(\left( {4{Z_{C0}}} \right) = \frac{{{{\left( {2r + r} \right)}^2} + {{\left( {{Z_{C0}}} \right)}^2}}}{{\left( {{Z_{C0}}} \right)}}\)

\(\Rightarrow \left\{ {\begin{array}{*{20}{c}} {r = \frac{{{Z_{C0}}}}{{\sqrt 3 }}}\\ {R = \frac{{2{Z_{C0}}}}{{\sqrt 3 }}} \end{array}} \right.\)

Phương trình điện áp hai đầu đoạn mạch MN (phức hóa)

\(\overline {{u_{MN}}} = \frac{{\overline {u\;} }}{{\overline {Z\;} }}.\overline {{Z_{MN}}} \)

\(\overline {{u_{MN}}} = \frac{{\left( {120\angle - \frac{\pi }{6}} \right)}}{{\left[ {\left( {\sqrt 3 } \right) + \left( {1 - 4} \right)i} \right]}}.\left[ {\left( {\frac{1}{{\sqrt 3 }}} \right) + \left( 1 \right)i} \right] = 40\angle \frac{\pi }{2}\)

Vậy \({u_{MN}} = 40{\rm{cos}}\left( {100\pi t + \frac{\pi }{2}} \right)V\)