Câu hỏi 7

R = 10Ω; \({Z_L} = \omega L = 100\pi .{{0,1} \over \pi } = 10\Omega ;{Z_C} = {1 \over {\omega C}} = {1 \over {100\pi {{.5.10}^{ - 4}}{1 \over \pi }}} = 20\Omega \)

\(I = {{{U_L}} \over {{Z_L}}} = {{20} \over {10}} = 2A\)

U = I.Z = \(2.\sqrt {{R^2} + {{({Z_L} - {Z_C})}^2}}  = 2.\sqrt {{{10}^2} + {{(10 - 20)}^2}}  = 20\sqrt 2 V\)

\(\tan \varphi  = {{{Z_L} - {Z_C}} \over R} =  - 1 \Rightarrow \varphi  = {{ - \pi } \over 4}\)

\( \Rightarrow u = 20\sqrt 2 .\sqrt 2 .cos(100\pi t - {\pi  \over 4})V = 40.cos(100\pi t - {\pi  \over 4})V\)