Câu 1: 

Ta có: 

\(\begin{array}{l}
\dfrac{\lambda }{2} = 2 \Rightarrow \lambda  = 4\left( {cm} \right)\\
f = \dfrac{v}{\lambda } = \dfrac{{40}}{4} = 10\left( {Hz} \right)
\end{array}\)

Câu 2: 

Ta có: 

\(\begin{array}{l}
{Z_L} = \omega L = 100\pi .\dfrac{{0,8}}{\pi } = 80\Omega \\
{Z_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{100\pi .\dfrac{{{{2.10}^{ - 4}}}}{\pi }}}50\Omega \\
Z = \sqrt {{R^2} + {{\left( {{Z_C} - {Z_L}} \right)}^2}}  = \sqrt {{{40}^2} + {{30}^2}}  = 50\Omega \\
{U_0} = {I_0}Z = 3.50 = 150V\\
\tan \varphi  = \dfrac{{{Z_L} - {Z_C}}}{R} = \dfrac{{80 - 50}}{{40}} = \dfrac{3}{4}\\
 \Rightarrow \varphi  = 0,64\left( {rad} \right)\\
 \Rightarrow u = 150\cos \left( {100\pi t + 0,64} \right)
\end{array}\)

Câu 3: 

Ta có: 

\(\begin{array}{l}
{Z_L} = \omega L = 100\pi .\dfrac{1}{\pi } = 100\Omega \\
{Z_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{100\pi .\dfrac{{{{200.10}^{ - 6}}}}{\pi }}}50\Omega \\
Z = \sqrt {{R^2} + {{\left( {{Z_C} - {Z_L}} \right)}^2}}  = \sqrt {{{50}^2} + {{50}^2}}  = 50\sqrt 2 \Omega \\
{I_0} = \dfrac{{{U_0}}}{Z} = \dfrac{{220\sqrt 2 }}{{50\sqrt 2 }} = 4,4A\\
\tan \varphi  = \dfrac{{{Z_L} - {Z_C}}}{R} = \dfrac{{100 - 50}}{{50}} = 1\\
 \Rightarrow \varphi  = \dfrac{\pi }{4}\left( {rad} \right)\\
 \Rightarrow i = 4,4\cos \left( {100\pi t - \dfrac{\pi }{4}} \right)
\end{array}\)

Câu 4: 

Ta có: 

\(\begin{array}{l}
{Z_L} = \omega L = 100\pi .\dfrac{{0,8}}{\pi } = 80\Omega \\
{Z_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{100\pi .\dfrac{{{{10}^{ - 3}}}}{{6\pi }}}} = 60\Omega \\
Z = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}  = \sqrt {{{20}^2} + {{20}^2}}  = 20\sqrt 2 \Omega \\
{I_0} = \dfrac{{{U_0}}}{Z} = \dfrac{{220\sqrt 2 }}{{20\sqrt 2 }} = 11A\\
{U_{0R}} = {I_0}R = 11.20 = 220V\\
{U_{0L}} = {I_0}{Z_L} = 11.80 = 880V\\
{\left( {\dfrac{{{u_R}}}{{{U_{0R}}}}} \right)^2} + {\left( {\dfrac{{{u_L}}}{{{U_{0L}}}}} \right)^2} = 1\\
 \Rightarrow {\left( {\dfrac{{110\sqrt 3 }}{{220}}} \right)^2} + {\left( {\dfrac{{{u_L}}}{{880}}} \right)^2} = 1\\
 \Rightarrow {u_L} = 440V
\end{array}\)