Đặt điện áp xoay chiều có dạng \(u = U\sqrt 2 cos\left( {2\pi f} \right)V\) vào hai đầu đoạn mạch

* Hướng dẫn giải

Khi \(f = {f_1}\) và \(f = {f_2}\) thì mạch có cùng công suất P₀ , ta có:

\(\begin{array}{l} {P_1} = {P_2} = {P_0}\\ \Leftrightarrow cos{\varphi _1} = cos{\varphi _2}\\ \Leftrightarrow \frac{R}{{\sqrt {{R^2} + {{\left( {{Z_{L1}} - {Z_{C1}}} \right)}^2}} }} = \frac{R}{{\sqrt {{R^2} + {{\left( {{Z_{L2}} - {Z_{C2}}} \right)}^2}} }}\\ \Leftrightarrow {Z_{L1}} + {Z_{L2}} = {Z_{{C_1}}} + {Z_{C2}}\\ \Leftrightarrow L\left( {{\omega _1} + {\omega _2}} \right) = \frac{1}{C}\left( {\frac{1}{{{\omega _1}}} + \frac{1}{{{\omega _2}}}} \right)\,\,\,(1)\\ \Rightarrow \frac{1}{{LC}} = {\omega _1}{\omega _2} \end{array}\)

Để  \({U_{{C_{max}}}}\) khi đó \({\omega _3} = \frac{1}{{LC}} - \frac{{{R^2}}}{{2{L^2}}}\)

Theo đề bài ta có: 

\(\begin{array}{l} R = \sqrt {\frac{L}{C}} \Rightarrow {R^2} = \frac{L}{C}\\ \Rightarrow {R^2} = {Z_{L1}}{Z_{C1}}\\ \Rightarrow \omega _3^2 = \frac{1}{{LC}} - \frac{{\frac{L}{C}}}{{2{L^2}}} = \frac{1}{{2LC}}\,\,(2) \end{array}\)

Lại có  \(\frac{{{f_1} + {f_2}}}{{{f_3}}} = \frac{{{\omega _1} + {\omega _2}}}{{{\omega _3}}} = \frac{9}{2}\)  (3)

Từ (1), (2) ta suy ra:  \({\omega _1}{\omega _2} = 2\omega _3^2\)

Kết hợp với (3) ta suy ra:

\(\begin{array}{l} \left\{ {\begin{array}{*{20}{l}} {{\omega _1} = 8{\omega _2} = 4{\omega _3}}\\ {{\omega _2} = \frac{{{\omega _3}}}{2}} \end{array}} \right.\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{Z_{{L_1}}} = 8{Z_{L2}} = 4{Z_{L3}}}\\ {{Z_{C1}} = \frac{{{Z_{C2}}}}{8} = \frac{{{Z_{C3}}}}{4}} \end{array}} \right. \end{array}\)

Ta có: 

\(\begin{array}{l} {Z_{L1}} + {Z_{L2}} = {Z_{C1}} + {Z_{{C_2}}}\\ \Rightarrow {Z_{L1}} + \frac{{{Z_{L1}}}}{8} = {Z_{C1}} + 8{Z_{C1}}\\ \Rightarrow {Z_{L1}} = 8{Z_{C1}}\\ P = \frac{{{U^2}R}}{{{Z^2}}} = \frac{{{U^2}R}}{{Z_{C3}^2 - Z_{L3}^2}}\\ {P_0} = \frac{{{U^2}R}}{{{R^2} + {{\left( {{Z_{{L_1}}} - {Z_{{C_1}}}} \right)}^2}}} = \frac{{{U^2}R}}{{Z_{L1}^2 - {Z_{L1}}{Z_C} + Z_{C1}^2}}\\ \Rightarrow \frac{{{P_0}}}{P} = \frac{{Z_{C3}^2 - Z_{L3}^2}}{{Z_{L1}^2 - {Z_{L1}}{Z_{C1}} + Z_{C1}^2}}\\ \Leftrightarrow \frac{{{P_0}}}{P} = \frac{{16Z_{C1}^2 - 4Z_{C1}^2}}{{64Z_{C1}^2 - 8Z_{C1}^2 + Z_{C1}^2}} = \frac{{12}}{{57}} = \frac{4}{{19}} \end{array}\)