Câu hỏi 5

Ta có:

$\eqalign{
& {{U.n} \over {\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = {{U.3n} \over {\sqrt {{R^2} + {{\left( {3{Z_L} - {{{Z_C}} \over 3}} \right)}^2}} }} \leftrightarrow {R^2} + {\left( {3{Z_L} - {{{Z_C}} \over 3}} \right)^2} = 9\left( {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} \right) \cr
& \leftrightarrow 8{R^2} + {{80} \over 9}Z_C^2 - 16{Z_L}{Z_C} = 0 \leftrightarrow 8{R^2} + {{80} \over 9}Z_C^2 - 16\omega L{1 \over {\omega C}} = 0 \to {Z_C} = 600\Omega = {1 \over {\omega C}} \to \omega = 50\pi \cr} $

mặt khác: f = p.n = 5.n => n = 10p = 5 vòng/s

=> Chọn A