Đặt điện áp 

A

- Hướng dẫn giải

Ta có:

Khi L = L1 thì UAM1 = UR1 = U

Khi L = L2 thì  \({U_{AM2}} = {U_{R2}} = \sqrt 8 U\)

 \(\begin{array}{l} {\varphi _1} + {\varphi _2} = \frac{\pi }{2}\\ \to \tan {\varphi _1}\tan {\varphi _2} = - 1\\ \to \frac{{{Z_{L1}} - {Z_{C1}}}}{R}.\frac{{{Z_{L1}} - {Z_{C1}}}}{R} = - 1\left( 1 \right) \end{array}\)

Mặt khác: ta có:

\(\begin{array}{l} \frac{{{U_{R1}}}}{{{U_{R2}}}} = \frac{1}{{\sqrt 8 }}\\ \to {I_2} = \sqrt 8 {I_1} \to {Z_1} = \sqrt 8 {Z_2}\\ \leftrightarrow \sqrt {{R^2} + {{\left( {{Z_{L1}} - {Z_C}} \right)}^2}} = \sqrt 8 \sqrt {{R^2} + {{\left( {{Z_{L2}} - {Z_C}} \right)}^2}} \\ \leftrightarrow {\left( {{Z_{L1}} - {Z_C}} \right)^2} - 7{R^2} - 8{\left( {{Z_{L2}} - {Z_C}} \right)^2} = 0\left( 2 \right) \end{array}\)

Chia cả hai vế của (2) cho (ZL2 - ZC) kết hợp với (1), Ta được:

\(\begin{array}{l}\frac{{{{\left( {{Z_{L1}} - {Z_C}} \right)}^2}}}{{{{\left( {{Z_{L2}} - {Z_C}} \right)}^2}}} - 7\frac{{\left( {{Z_{L1}} - {Z_C}} \right)}}{{\left( {{Z_{L2}} - {Z_C}} \right)}} - 8 = 0\\ \to \left[ \begin{array}{l} \frac{{\left( {{Z_{L1}} - {Z_C}} \right)}}{{\left( {{Z_{L2}} - {Z_C}} \right)}} = 1\left( {Loai} \right)\\ \frac{{\left( {{Z_{L1}} - {Z_C}} \right)}}{{\left( {{Z_{L2}} - {Z_C}} \right)}} = - 8 \end{array} \right.\\ \frac{{\left( {{Z_{L1}} - {Z_C}} \right)}}{{\left( {{Z_{L2}} - {Z_C}} \right)}} = - 8\\ \to - \frac{{\left( {{Z_{L1}} - {Z_C}} \right)}}{8} = \left( {{Z_{L2}} - {Z_C}} \right)\\ (1) \Rightarrow {\left( {{Z_{L1}} - {Z_C}} \right)^2} = 8{R^2} \end{array}\)

Hệ số công suất của mạch khi L=L:

\(\begin{array}{l} \cos {\varphi _1} = \frac{R}{{{Z_1}}} = \frac{R}{{\sqrt {{R^2} + {{\left( {{Z_{L1}} - {Z_C}} \right)}^2}} }}\\ = \frac{R}{{3R}} = \frac{1}{3} \end{array}\)

 Chọn A