Mắc nối tiếp: Điện trở R, cuộn dây thuần cảm L và...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Ta có

 \(\begin{array}{l}4L = C{R^2} =  > 4{Z_L} = \frac{{{R^2}}}{{2{Z_C}}} =  > {R^2} = 4{Z_L}{Z_C}\\{f_1} = {f_0}:\cos {\varphi _1} = \frac{R}{Z} = \frac{R}{{\sqrt {{R^2} + {{\left( {{Z_{{L_1}}} - {Z_{{C_1}}}} \right)}^2}} }} = \frac{R}{{\sqrt {4{Z_{{L_1}}}{Z_{{C_1}}} + {{\left( {{Z_{{L_1}}} - {Z_{{C_1}}}} \right)}^2}} }} = \frac{R}{{\sqrt {{{\left( {{Z_{{L_1}}} + {Z_{{C_1}}}} \right)}^2}} }} = \frac{R}{{\left( {{Z_{{L_1}}} + {Z_{{C_1}}}} \right)}}\end{array}\)

Có \(\cos {\varphi _1} = 0,8 =  > R = \left( {{Z_{{L_1}}} + {Z_{{C_1}}}} \right)\)

Mặt khác \({R^2} = 4{Z_{{L_1}}}{Z_{{C_1}}} =  > 4{Z_{{L_1}}}{Z_{{C_1}}} = 0,{8^2}{\left( {{Z_{{L_1}}} + {Z_{{C_1}}}} \right)^2} =  > Z_{_{{L_1}}}^2 + Z_{_{{C_1}}}^2 - 4,25{Z_{{L_1}}}{Z_{{C_1}}} = 0 =  > \left[ \begin{array}{l}{Z_{{L_1}}} = 4{Z_{{C_1}}}\\{Z_{{L_1}}} = \frac{1}{4}{Z_{{C_1}}}\end{array} \right.\)

Nếu \({Z_{{L_1}}} = 4{Z_{{C_1}}} =  > R = 0,8\left( {{Z_{{L_1}}} + {Z_{{C_1}}}} \right) = 4{Z_{{C_1}}}\)

Khi \({f_2} = \frac{{{f_0}}}{2}\) thì \(\left\{ \begin{array}{l}{Z_{{L_2}}} = \frac{{{Z_{{L_1}}}}}{2} = 2{Z_{{C_1}}}\\{Z_{{C_2}}} = 2{Z_{{C_1}}}\end{array} \right.\) =>cộng hưởng \( =  > \cos \varphi  = 1\)

Nếu \({Z_{{L_1}}} = \frac{1}{4}{Z_{{C_1}}} =  > R = 0,8\left( {{Z_{{L_1}}} + {Z_{{C_1}}}} \right) = 4{Z_{{L_1}}}\)

Khi \({f_2} = \frac{{{f_0}}}{2}\) thì \(\left\{ \begin{array}{l}{Z_{{L_2}}} = \frac{{{Z_{{L_1}}}}}{2}\\{Z_{{C_2}}} = 2{Z_{{C_1}}} = 8{Z_{{L_1}}}\end{array} \right.\)

=>Hệ số công suất là \(\cos {\varphi _2} = \frac{R}{{\sqrt {{R^2} + {{\left( {{Z_{{L_2}}} - {Z_{{C_2}}}} \right)}^2}} }} = \frac{{4{Z_{{L_1}}}}}{{\sqrt {16Z_{{L_1}}^2 + {{\left( {\frac{1}{2}{Z_{{L_1}}} - 8{Z_{{L_1}}}} \right)}^2}} }}0,47\) gần đáp án C nhất

Chọn C