$\dfrac{\varphi _{1}}{\varphi _{2}}$ gần giá trị nào nhất?

$w_{1}>w_{2}\Rightarrow Z_{L_1}-Z_{C_1}>0$
Từ $U_{1}=U_{2}=U\dfrac{\sqrt{21}}{7}\Rightarrow U\dfrac{R}{\sqrt{R^{2}+\left(Z_{L_1}-Z_{C_1}\right)^{2}}}=\dfrac{\sqrt{21}}{7}U\Rightarrow \dfrac{Z_{L_1}-Z_{C_1}}{R}=\dfrac{2}{\sqrt{3}}$
Thay đổi $w$ để $Z_{1}=Z_{2}\Rightarrow w_{1}.w_{2}=w_{0}^{2}$ (với $w_{0}$ là tần số góc để mạch cộng hưởng) hay $w_{1}.w_{2}=\dfrac{1}{LC}\Rightarrow Z_{C_1}=Z_{L_2}$
ta có $\tan \left(\varphi _{1}-\varphi _{2}\right)=\dfrac{\tan \varphi _{1}-\tan \varphi _{2}}{1+\tan \varphi _{1}.\tan \varphi _{2}}$
mặt khác $\tan \varphi _{1}-\tan \varphi _{2}=\dfrac{Z_{L_1}-Z_{L_2}}{R}=\dfrac{Z_{L_1}-Z_{C_1}}{R}=\dfrac{2}{\sqrt{3}}\Leftrightarrow \tan \varphi_{1}.\tan \varphi _{2}=1\Rightarrow \tan \varphi _{1}=\sqrt{3},\tan \varphi _{2}=\dfrac{1}{\sqrt{3}}\Rightarrow \dfrac{\varphi _{1}}{\varphi _{2}}=2$
Đáp án A.