Công suất tiêu thụ của mạch: \(P = \frac{{{U^2}R}}{{{Z^2}}} = \frac{{{U^2}R}}{{{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}}\)

\({P_{\max }} \Leftrightarrow {\left[ {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} \right]_{\min }}\)

\( \Leftrightarrow {Z_L} = {Z_C} \Leftrightarrow {\omega _0}L = \frac{1}{{{\omega _0}C}} \Rightarrow \omega _0^2 = \frac{1}{{LC}}\)

Với hai giá trị của tần số góc cho cùng hệ số công suất, ta có: \({\omega _1}{\omega _2} = \omega _0^2\)

Mặt khác: \(\cos {\varphi _1} = \frac{R}{{\sqrt {{R^2} + {{\left( {{\omega _1}L - \frac{1}{{{\omega _1}C}}} \right)}^2}} }}\)

\(\cos {\varphi _1} = \frac{R}{{\sqrt {{R^2} + \left( {\omega _1^2{L^2} - 2.\frac{L}{C} + \frac{1}{{\omega _1^2{C^2}}}} \right)} }}\)

\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \frac{R}{{\sqrt {{R^2} + \left( {\omega _1^2{L^2} - 2{L^2}.\frac{1}{{LC}} + \frac{{{L^2}}}{{\omega _1^2}}.\frac{1}{{{L^2}{C^2}}}} \right)} }}\)

\(\cos {\varphi _1} = \frac{R}{{\sqrt {{R^2} + \left( {\omega _1^2{L^2} - 2{L^2}.\omega _0^2 + \frac{{{L^2}}}{{\omega _1^2}}.\omega _0^4} \right)} }}\)

\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \frac{R}{{\sqrt {{R^2} + {L^2}.\left( {\omega _1^2 - 2.\omega _0^2 + \frac{{\omega _0^4}}{{\omega _1^2}}} \right)} }}\)

\({\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \frac{R}{{\sqrt {{R^2} + {L^2}.{{\left( {{\omega _1} - \frac{{\omega _0^2}}{{{\omega _1}}}} \right)}^2}} }}{\mkern 1mu} {\mkern 1mu}  = \frac{R}{{\sqrt {{R^2} + {L^2}.{{\left( {{\omega _1} - {\omega _2}} \right)}^2}} }}\)

Theo bài ra ta có: \(\left\{ {\begin{array}{*{20}{l}}
{\cos {\varphi _1} = 0,5}\\
{{\omega _1} - {\omega _2} = 200\pi {\mkern 1mu} \left( {rad/s} \right)}\\
{L = \frac{{\sqrt 3 }}{{4\pi }}H}
\end{array}} \right.\)

\( \Rightarrow \cos {\varphi _1} = \frac{R}{{\sqrt {{R^2} + {L^2}.{{\left( {{\omega _1} - {\omega _2}} \right)}^2}} }} = 0,5\)

\( \Leftrightarrow \frac{{{R^2}}}{{{R^2} + {{\left( {\frac{{\sqrt 3 }}{{4\pi }}} \right)}^2}.{{\left( {200\pi } \right)}^2}}} = \frac{1}{4} \Leftrightarrow \frac{{{R^2}}}{{{R^2} + 7500}} = \frac{1}{4} \Rightarrow R = 50\Omega \)