Đặt điện áp xoay chiều u = U0cos2πft(V) (U0 không đổi, f thay đổi được)

* Hướng dẫn giải

Ta có:     

\(\begin{array}{l}
 + P = \frac{{{U^2}{{\cos }^2}\varphi }}{R} \Rightarrow \frac{{{P_1}}}{{{P_2}}} = \frac{{{{\cos }^2}{\varphi _1}}}{{{{\cos }^2}{\varphi _2}}} \Leftrightarrow \frac{{20}}{{32}} = \frac{5}{8} = \frac{{{{\cos }^2}{\varphi _1}}}{{{{\cos }^2}{\varphi _2}}}\\
 + \cos \varphi  = \frac{R}{Z}\frac{{Z_2^2}}{{Z_1^2}} = \frac{5}{8} \Leftrightarrow \frac{{{R^2} + Z_{C2}^2}}{{{R^2} + Z_{C1}^2}} = \frac{5}{8}\\
 + {Z_C} = \frac{1}{{\omega C}} = \frac{1}{{2\pi fC}}\\
 \Rightarrow {Z_{C2}} = \frac{{{Z_{C1}}}}{2}\frac{{{R^2} + Z_{C2}^2}}{{{R^2} + 4Z_{C2}^2}} = \frac{5}{8}\\
 \Rightarrow 8\left( {{R^2} + Z_{C2}^2} \right) = 5\left( {{R^2} + 4Z_{C2}^2} \right) \Rightarrow {R^2} = 4Z_{C2}^2 = Z_{C1}^2\\
 + \frac{{{P_3}}}{{{P_1}}} = \frac{{{{\cos }^2}{\varphi _3}}}{{{{\cos }^2}{\varphi _1}}} = \frac{{Z_1^2}}{{Z_3^2}} = \frac{{{R^2} + Z_{C1}^2}}{{{R^2} + \frac{{Z_{C1}^2}}{9}}}\frac{{{P_3}}}{{20}} = \frac{{{R^2} + {R^2}}}{{{R^2} + \frac{{{R^2}}}{9}}}\\
 \Rightarrow {P_3} = 36W
\end{array}\)