Từ đồ thị ta thấy uMB khi k đóng sớm pha hơn so với khi k mở góc π/3; và ${U_{MB\left( {mo} \right)}} = {U_{MB\left( {dong} \right)}} = 100\left( V \right)$
$\begin{gathered}
\frac{{U\sqrt {{r^2} + {{({Z_L} - {Z_C})}^2}} }}{{\sqrt {9{r^2} + {{({Z_L} - {Z_C})}^2}} }} = \frac{{U\sqrt {{r^2} + Z_L^2} }}{{\sqrt {9{r^2} + Z_L^2} }} \Rightarrow {Z_C} = 2{Z_L} \hfill \\
\tan {\varphi _{MBm}} = \frac{{{Z_L} - {Z_C}}}{r} = \frac{{ - {Z_L}}}{r} \hfill \\
\tan {\varphi _d} = \frac{{{Z_L}}}{{9r}} \hfill \\
\tan {\varphi _{MBd}} = \frac{{{Z_L}}}{r} = \frac{{{Z_L}}}{r} \to {\tan ^{ - 1}}\left( {\frac{{ - {Z_L}}}{{9r}}} \right) + {\tan ^{ - 1}}\left( {\frac{{ - {Z_L}}}{r}} \right) + {\tan ^{ - 1}}\left( {\frac{{{Z_L}}}{{9r}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{Z_L}}}{r}} \right) = \frac{\pi }{3} \hfill \\
{\tan ^{ - 1}}\left( {\frac{{ - x}}{9}} \right) + {\tan ^{ - 1}}\left( { - x} \right) + {\tan ^{ - 1}}\left( {\frac{x}{9}} \right) + {\tan ^{ - 1}}\left( x \right) = 60 \hfill \\
\to x = \sqrt 3 \to {Z_L} = r\sqrt 3 \hfill \\
\end{gathered} $
Ta có $\frac{{100}}{{\sqrt 2 }} = \frac{{U\sqrt {{r^2} + 3{r^2}} }}{{\sqrt {9{r^2} + 3{r^2}} }}$=>U=122,4744871V
