Phương pháp:
*Khi f = f0 thì UC = U nên $Z_C^2 = {R^2} + {\left( {{Z_L} - {Z_C}} \right)^2} \Rightarrow \left\{ \begin{array}{l} Z_L^2 = 2\frac{L}{C} - {R^2}\left( 1 \right)\\ {Z_C} = \frac{{{R^2} + Z_L^2}}{{2{Z_L}}} = \frac{{{x^2} + 1}}{2}{Z_L} \end{array} \right.$
(Đã đặt R = xZ$_L$).
$\Rightarrow \tan \varphi =\frac{{{Z}_{L}}-{{Z}_{C}}}{R}\Leftrightarrow -0,75=\frac{{{Z}_{L}}-\frac{{{x}^{2}}+1}{2}{{Z}_{L}}}{x{{Z}_{L}}}$
$\Rightarrow x=2\Rightarrow R=2{{\Zeta }_{L}};{{\Zeta }_{C}}=\frac{{{2}^{2}}+1}{2}{{\Zeta }_{L}}=2,5{{\Zeta }_{L}}$(2)
*Khi f = f0 + 45 thì UL = U nên$Z'_{L}^{2}={{R}^{2}}+{{\left( Z{{'}_{L}}-Z{{'}_{C}} \right)}^{2}}\Rightarrow Z'_{C}^{2}=2\frac{L}{C}-{{R}^{2}}$ (3).
Từ (1) và (3)$\Rightarrow {{Z}_{L}}=Z{{'}_{C}}$ (4) .Thay (4) vào (2):
${{Z}_{C}}=2,5Z{{'}_{C}}\Leftrightarrow \frac{1}{2\pi {{f}_{0}}}=2,5.\frac{1}{2\pi \left( {{f}_{0}}+45 \right)}\Rightarrow {{f}_{0}}=30\left( Hz \right).$
Thay f0 = 30 Hz vào (2), ta được $\frac{1}{60\pi C}=2,5.100\pi L\Rightarrow \frac{1}{LC}=2,5.{{\left( 60\pi \right)}^{2}}$(5)
*${{U}_{AM}}=I{{Z}_{RC}}=U\sqrt{\frac{{{R}^{2}}+Z_{C}^{2}}{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}\notin R\Leftrightarrow {{Z}_{L}}=2{{Z}_{C}}\Leftrightarrow \frac{1}{LC}=0,5{{\left( 2\pi f \right)}^{2}}$(6)
Thay (5) vào (6):$0,5{{\left( 2\pi f \right)}^{2}}=2,5{{\left( 60\pi \right)}^{2}}\Rightarrow f=30\sqrt{5}\left( Hz \right)\Rightarrow $ Chọn B
