Đặt diện áp u = U√2cosωt (f thay đổi, U tỉ lệ thuậ...

C

- Hướng dẫn giải

Ta có:

\(\begin{array}{l}
I = \frac{U}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = \frac{{k\omega }}{{\sqrt {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} }}\\
 \Rightarrow I = \frac{k}{{\sqrt {\frac{{{R^2}}}{{{\omega ^2}}} + {L^2} - \frac{{2L}}{C}\frac{1}{{{\omega ^2}}} + \frac{1}{{{C^2}}}\frac{1}{{{\omega ^4}}}} }}
\end{array}\) 

+ Ta có:            

\(\left\{ \begin{array}{l}
{Z_{L2}} = 1,5{Z_{L1}}\\
{Z_{C2}} = \frac{{{Z_{C1}}}}{{1,5}}
\end{array} \right.{R^2} + \frac{{13}}{9}Z_{C1}^2 - 2{Z_{C1}}{Z_{L1}} = 0\)

Và:

\(\left\{ \begin{array}{l}
{Z_{L3}} = 0,5{Z_{L1}};{Z_{L4}} = 2{Z_{L1}}\\
{Z_{C3}} = 2{Z_{C1}};{Z_{C4}} = 0,5{Z_{C1}}
\end{array} \right.{Z_{L1}} = {Z_{C1}}\)

+ Khi f5 thì:  

\(\tan \left( { - \frac{\pi }{4}} \right) = \frac{{ - {Z_{C5}}}}{R} \Rightarrow {Z_{C5}} = R\frac{1}{{2\pi {f_5}}} = \frac{{\sqrt 5 }}{3}\frac{1}{{2\pi {f_1}}} \Rightarrow {f_5} = \frac{{3{f_1}}}{{\sqrt 5 }} = \frac{{3.60}}{{\sqrt 5 }} = 80,5Hz\)