Đáp án:C

Giải thích các bước giải:

 ta có

$\begin{array}{l}
{I_{01}} = {I_{02}} \Rightarrow {Z_1} = {Z_2}\\
 \Rightarrow \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}  = \sqrt {{R^2} + Z_L^2} \\
 \Rightarrow {Z_C} - {Z_L} = {Z_L} \Rightarrow {Z_C} = 2{Z_L}\\
\tan {\varphi _1} = \frac{{{Z_L} - {Z_C}}}{R} = \frac{{ - {Z_L}}}{R}\\
\tan {\varphi _2} = \frac{{{Z_L}}}{R}\\
\left\{ \begin{array}{l}
{\varphi _2} - {\varphi _1} = \frac{\pi }{3}\\
{\varphi _1} =  - {\varphi _2}
\end{array} \right. \Rightarrow {\varphi _1} =  - \frac{\pi }{6} = {\varphi _u} - {\varphi _{{i_1}}} = {\varphi _u} - \frac{\pi }{4} \Rightarrow {\varphi _u} = \frac{\pi }{{12}}\\
U = 60 \Rightarrow {U_0} = 60\sqrt 2 \left( V \right)\\
 \Rightarrow pt:u = 60\sqrt 2 \cos \left( {\omega t + \frac{\pi }{{12}}} \right)
\end{array}$