Đặt điện áp \(u={{U}_{0}}cos100\pi t\text{ }\left( V \right)\) (t tính bằng s)

Ta có:

\(\begin{array}{l} \left\{ {\begin{array}{*{20}{c}} {{Z_L} = 150}\\ {{Z_C} = 100} \end{array}\left\{ {\begin{array}{*{20}{c}} {\tan \varphi = 1/\sqrt 3 \Rightarrow \varphi = \pi /6}\\ {Z = \sqrt {{R^2} + Z_{LC}^2} = 100\left( \Omega \right)} \end{array}} \right.} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}} {{Z_{RL}} = 100\sqrt 3 \left( \Omega \right)}\\ {{\varphi _{RL}} = \pi /3{\rm{ }}} \end{array}} \right..\\ i = {I_0}\cos \left( {100\pi t - \frac{\pi }{6}} \right)\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}} {{u_{RL}} = {U_{0RL}}\cos \left( {100\pi t - \frac{\pi }{6} + \frac{\pi }{3}} \right)}\\ {{u_C} = {U_{0C}} = \cos \left( {100\pi t - \frac{\pi }{6} - \frac{\pi }{2}} \right)} \end{array}} \right.\\ \left\{ {\begin{array}{*{20}{c}} {{u_{RL\left( {{t_1}} \right)}} = 100\sqrt 3 {I_0}\cos \left( {100\pi {t_1} + \frac{\pi }{6}} \right) = 150V}\\ {{u_C}\left( {{t_2}} \right) = 100{I_0}\cos \left( {100\pi {t_1} + \frac{{2\pi }}{3}} \right) = 150V} \end{array}} \right.\\ \Rightarrow {\left( {\frac{{150}}{{100\sqrt 3 {I_0}}}} \right)^2} + {\left( {\frac{{150}}{{100{I_0}}}} \right)^2} = 1 = > {I_0} = \sqrt 3 \;A \end{array}\)

Từ đó suy ra:

\(=>{{U}_{0}}=Z{{I}_{0}}=100.\sqrt{3}=100\sqrt{3}\left( \text{V} \right)\)