Lời giải:+Giả sử U = aω trong đó a là hằng số
+ UC = I ZC = \(\frac{a\omega {{Z}_{C}}}{\sqrt{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}\) ; UC max = \(\frac{a}{RC}\) = 220 V => a = 220RC => U = 220RC ω
+Tại f1 = 15 Hz và f2 = 60 Hz thì UC1 = UC2 => ω2L - \(\frac{1}{{{\omega }_{2}}C}\) = \(\frac{1}{{{\omega }_{1}}C}\) - ω1L => ω0 = \(\sqrt{{{\omega }_{1}}{{\omega }_{2}}}\)
f0 = \(\sqrt{{{f}_{1}}{{f}_{2}}}\) = \(\sqrt{15.60}\) = 30 Hz
+ UR = I R = \(\frac{220\text{RC }\!\!~\!\!\text{ }\!\!\omega\!\!\text{ R}}{\sqrt{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}\) = \)\frac{440.\omega L~\frac{{{R}^{2}}C}{2L}}{\sqrt{\frac{1}{{{C}^{2}}{{\omega }^{2}}}+{{R}^{2}}-2\frac{L}{C}+{{L}^{2}}{{\omega }^{2}}}}\)
U= \(\frac{440\frac{{{R}^{2}}C}{2L}}{\sqrt{\frac{1}{{{L}^{2}}{{C}^{2}}{{\omega }^{4}}}-2\left( 1-\frac{{{R}^{2}}C}{2L} \right)\frac{1}{LC{{\omega }^{2}}}+1}}\) = \)\frac{440~\left( 1-{{n}^{-1}} \right)}{\sqrt{{{\left( \frac{60\pi }{\omega } \right)}^{4}}-2{{n}^{-1}}{{\left( \frac{60\pi }{\omega } \right)}^{2}}+1}}\)
\(\Rightarrow {{\left( \frac{60\pi }{60\pi } \right)}^{2}}\)+ \({{\left( \frac{60\pi }{78\pi } \right)}^{2}}\)= 2 n-1 => n-1 = \(\frac{269}{338}\) => UR max = \(\frac{U}{\sqrt{1-{{n}^{-2}}}}\) = \(\frac{440\left( 1~-{{n}^{-1}} \right)}{\sqrt{1-{{n}^{-2}}}}\) = 148,35 V
