Cho mạch điện xoay chiều gồm điện trở thuần R khôn...

C

- Hướng dẫn giải

Khi ω thay đổi để U_L1 = max thì: \({Z_{C1}} = \sqrt {\frac{{{L_1}}}{C} - \frac{{{R^2}}}{2}}  \Leftrightarrow 2Z_{C1}^2 = 2{Z_{L1}}{Z_{C1}} - {R^2}\)    (1)

+ Ta có:  

\(\begin{array}{l}
{U_C} = \frac{{U{Z_{C1}}}}{{\sqrt {{R^2} + {{\left( {{Z_{L1}} - {Z_{C1}}} \right)}^2}} }}\\
 \Leftrightarrow 40\sqrt 3  = \frac{{120{Z_{C1}}}}{{\sqrt {{R^2} + {{\left( {{Z_{L1}} - {Z_{C1}}} \right)}^2}} }}\\
 \Leftrightarrow {R^2} + {\left( {{Z_{L1}} - {Z_{C1}}} \right)^2} = 3Z_{C1}^2 \Rightarrow {R^2} = 2Z_{C1}^2 - Z_{L1}^2 + 2{Z_{L1}}{Z_{C1}}
\end{array}\)   (2)

+ Thay (2) vào (1), ta có:  

\(\begin{array}{l}
2Z_{C1}^2 = 2{Z_{L1}}{Z_{C1}} - \left( {2Z_{C1}^2 - Z_{L1}^2 + 2{Z_{L1}}{Z_{C1}}} \right)\\
 \Leftrightarrow 4Z_{C1}^2 = Z_{L1}^2 \Leftrightarrow \frac{2}{{{\omega _1}C}} = {\omega _1}{L_1} \Rightarrow {L_1}C = \frac{2}{{\omega _1^2}}
\end{array}\)    (3)

+ Khi ω thay đổi để U_L2 = max thì:  

\(\begin{array}{l}
{Z_{C2}} = \sqrt {\frac{{{L_2}}}{C} - \frac{{{R^2}}}{2}}  \Leftrightarrow {\left( {\frac{1}{{{\omega _2}C}}} \right)^2} = \frac{{{L_2}}}{C} - \frac{{{R^2}}}{2} \Leftrightarrow {\left( {\frac{1}{{{\omega _2}C}}} \right)^2} = \frac{{2{L_1}}}{C} - \frac{{{R^2}}}{2}\\
 \Rightarrow {\left( {\frac{1}{{{\omega _2}C}}} \right)^2} = \frac{{{L_1}}}{C} + \left( {\frac{{{L_1}}}{C} - \frac{{{R^2}}}{2}} \right) \Leftrightarrow {\left( {\frac{1}{{{\omega _2}C}}} \right)^2} = \frac{{{L_1}}}{C} + Z_{C1}^2 \Leftrightarrow {\left( {\frac{1}{{{\omega _2}C}}} \right)^2} = \frac{{{L_1}}}{C} + {\left( {\frac{1}{{{\omega _1}C}}} \right)^2}\\
 \Rightarrow {\left( {\frac{1}{{{\omega _2}}}} \right)^2} = C{L_1} + {\left( {\frac{1}{{{\omega _1}}}} \right)^2} \Rightarrow {\omega _2} = \sqrt {\frac{1}{{C{L_1} + {{\left( {\frac{1}{{{\omega _1}}}} \right)}^2}}}} 
\end{array}\)    (4)

+ Thay (3) vào (4), ta có: \({\omega _2} = \sqrt {\frac{1}{{\frac{2}{{\omega _1^2}} + {{\left( {\frac{1}{{{\omega _1}}}} \right)}^2}}}}  = 40\pi \sqrt 3 \) (rad/s)