Đặt điện áp u = U0cos(ωt)( U0

B

- Hướng dẫn giải

Theo giả thuyết bài toán:

\(\begin{array}{l}
{Z_L} = {Z_{L0}};{Z_L} = 3{Z_{L0}}({U_C}) \Rightarrow {Z_{L0}} + 3{Z_{L0}} = 2{Z_C}\\
Cho\,\,\,{Z_{L0}} = 1 \Rightarrow {Z_C} = 2\\
{Z_L} = 2{Z_{L0}};{Z_L} = 6{Z_{L0}}({U_L})\\
 \Rightarrow \frac{1}{{2{Z_{L0}}}} + \frac{1}{{6{Z_{L0}}}} = \frac{{2{Z_C}}}{{{R^2} + Z_C^2}} \Rightarrow \frac{1}{{\left( 2 \right)}} + \frac{1}{{\left( 6 \right)}} = \frac{{2.\left( 2 \right)}}{{{R^2} + \left( {{2^2}} \right)}} \Rightarrow {R^2} = 2\\
 \Rightarrow \frac{{{U_L}}}{{{U_C}}} = \frac{{\frac{{2{Z_{L0}}}}{{\sqrt {{R^2} + {{\left( {2{Z_{L0}} - {Z_C}} \right)}^2}} }}}}{{\frac{{{Z_C}}}{{\sqrt {{R^2} + {{\left( {{Z_{L0}} - {Z_C}} \right)}^2}} }}}} = \frac{{\frac{2}{{\sqrt {2 + {{\left( {2 - 2} \right)}^2}} }}}}{{\frac{2}{{\sqrt {2 + {{\left( {1 - 2} \right)}^2}} }}}} = \sqrt {\frac{3}{2}} 
\end{array}\)