Nhận xét: R, ZC không đổi \( \Rightarrow {\varphi _{{u_{RC}}/i}} = const \Rightarrow \alpha  = const\)

Trường hợp 1: L = L1, ta có:

\({\varphi _{{u_{MB1}}/u}} =  - \dfrac{\pi }{2} \Rightarrow \dfrac{U}{{\sin \alpha }} = \dfrac{{{U_{MB}}}}{{\sin \beta }} \Rightarrow \dfrac{{{U_0}}}{{\sin \alpha }} = \dfrac{{{U_{01}}}}{{\sin \beta }} = \dfrac{{{U_{01}}}}{{\sin \left( {{{90}^0} - \alpha } \right)}}\)

Trường hợp 2: L = L2, ta có:

\({\varphi _{{u_{MB2}}/u}} =  - \dfrac{\pi }{3} \Rightarrow \dfrac{{{U_0}}}{{\sin \alpha }} = \dfrac{{{U_{01}}}}{{\sin \left( {{{120}^0} - \alpha } \right)}}\)

Trường hợp 3: L = L3, ta có:

\({\varphi _{{u_{MB3}}}} =  - \dfrac{{2\pi }}{3} \Rightarrow \dfrac{{{U_0}}}{{\sin \alpha }} = \dfrac{{320}}{{\sin \left( {{{60}^0} - \alpha } \right)}}\)

\(\begin{array}{l} \Rightarrow \dfrac{{{U_0}}}{{\sin \alpha }} = \dfrac{{{U_{01}}}}{{\sin \left( {{{90}^0} - \alpha } \right)}} = \dfrac{{{U_{01}}}}{{\sin \left( {{{120}^0} - \alpha } \right)}} = \dfrac{{320}}{{\sin \left( {{{60}^0} - \alpha } \right)}}\\ \Rightarrow \sin \left( {{{90}^0} - \alpha } \right) = \sin \left( {{{120}^0} - \alpha } \right)\\ \Rightarrow {90^0} - \alpha  + {120^0} - \alpha  = {180^0} \Rightarrow \alpha  = {15^0}\\ \Rightarrow \dfrac{{{U_{01}}}}{{\sin \left( {{{90}^0} - {{15}^0}} \right)}} = \dfrac{{320}}{{\sin \left( {{{60}^0} - {{15}^0}} \right)}} \Rightarrow {U_{01}} = 437,128\,\,\left( V \right)\end{array}\)

Chọn C.