a)
Ta có: \(\left\{ \begin{array}{l}R = 30\Omega \\{Z_L} = \omega L = 100\pi \dfrac{{{{2.10}^{ - 1}}}}{\pi } = 20\Omega \\{Z_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{100\pi \dfrac{{{{10}^{ - 3}}}}{{5\pi }}}} = 50\Omega \end{array} \right.\)
Tổng trở:
\(\begin{array}{l}Z = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} \\ = \sqrt {{{30}^2} + {{\left( {20 - 50} \right)}^2}} = 30\sqrt 2 \end{array}\)
b)
\(i = \dfrac{u}{{\overline Z }} = \dfrac{{140\sqrt 2 \angle 0}}{{30 + \left( {20 - 50} \right)i}} = \dfrac{{14}}{3}\angle \dfrac{\pi }{4}\)
\( \Rightarrow i = \dfrac{{14}}{3}cos\left( {100\pi t + \dfrac{\pi }{4}} \right)A\)
c) Để \({u_{AB}}\) cùng pha với \(i \Rightarrow \) mạch cộng hưởng, khi đó \({Z_L} = {Z_{{C_1}}}\)
\( \Rightarrow {C_1} = \dfrac{1}{{{Z_{{C_1}}}\omega }} = \dfrac{1}{{20.100\pi }} = \dfrac{{{{5.10}^{ - 4}}}}{\pi }F\)
