Cho mạch điện RLC mắc nối tiếp theo thứ tự R, L, C...

D

- Hướng dẫn giải

Ta có:

\(\begin{array}{l} {I_1} = {I_2}\\ \Leftrightarrow \frac{U}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_{{C_1}}}} \right)}^2}} }} = \frac{U}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_{{C_2}}}} \right)}^2}} }}\\ \Leftrightarrow {\left( {{Z_L} - {Z_{{C_1}}}} \right)^2} = {\left( {{Z_L} - {Z_{{C_2}}}} \right)^2}\\ {Z_{{C_1}}} \ne {Z_{{C_2}}} \Rightarrow {Z_L} - {Z_{{C_1}}} = - \left( {{Z_L} - {Z_{{C_2}}}} \right)\\ \Rightarrow {Z_L} = \frac{{{Z_{{C_1}}} + {Z_{{C_2}}}}}{2}\,\,\,\,\,(1) \end{array}\)

Khi P = Pmax thì mạch xảy ra hiện tượng cộng hưởng điện :

\(\begin{array}{l} \Rightarrow {Z_L} = {Z_C}\,\,\,(2)\\ (1)\, + (2) \to {Z_C} = \frac{{{Z_{{C_1}}} + {Z_{{C_2}}}}}{2}\\ \Rightarrow \frac{1}{C} = \frac{1}{2}\left( {\frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}} \right)\\ \Rightarrow C = \frac{{2\left( {{C_1} + {C_2}} \right)}}{{{C_1}{C_2}}}\\ = 2\left( {\frac{{\frac{{{{2.10}^{ - 4}}}}{\pi } + \frac{{{{2.10}^{ - 4}}}}{{3\pi }}}}{{\frac{{{{2.10}^{ - 4}}}}{\pi }.\frac{{{{2.10}^{ - 4}}}}{{3\pi }}}}} \right) = \frac{{{{2.10}^{ - 4}}}}{\pi }F \end{array}\)

Chọn D