Đặt điện áp \({{u}_{AB}}=30\cos \left( 100\pi t \right)\left( V \right)\)

+ Khi \(C={{C}_{0}}:\ {{U}_{MN}}\left( {{U}_{L}} \right)\) đạt max \(\to \) cộng hưởng.

Khi đó ta có \({{U}_{R}}=U=15\sqrt{2}\ V;{{Z}_{L}}={{Z}_{{{C}_{0}}}}\).

Có: \({{U}_{AN}}=\sqrt{U_{R}^{2}+U_{L}^{2}}\to {{U}_{L}}=\sqrt{U_{AN}^{2}-U_{R}^{2}}=15\sqrt{6}\ V\Rightarrow \frac{{{Z}_{L}}}{R}=\frac{{{U}_{L}}}{{{U}_{R}}}=\sqrt{3}\).

+ Khi \(C=0,5{{C}_{0}}:\ {{Z}_{C}}'=2{{Z}_{{{C}_{0}}}}=2{{Z}_{L}}\).

\(\begin{array}{l}
 \to Z' = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}^\prime } \right)}^2}}  = \sqrt {{{\left( {\frac{{{Z_L}}}{{\sqrt 3 }}} \right)}^2} + {{\left( {{Z_L} - 2{Z_L}} \right)}^2}}  = \frac{2}{{\sqrt 3 }}{Z_L}\\
\frac{U}{{Z'}} = \frac{{{U_L}}}{{{Z_L}}} \to {U_L} = \frac{{{Z_L}}}{{Z'}}.U = \frac{{15\sqrt 6 }}{2}V \to {U_{0L}} = 15\sqrt 3 \;V\\
\tan \varphi  = \frac{{{Z_L} - {Z_C}^\prime }}{R} = \frac{{{Z_L} - 2{Z_L}}}{{{Z_L}/\sqrt 3 }} =  - \sqrt 3  \to \varphi  = \frac{{ - \pi }}{3} \to {\varphi _i} = {\varphi _u} - \varphi  = \frac{\pi }{3} \to {\varphi _i} + \frac{\pi }{2} = \frac{{5\pi }}{6}\\
 \Rightarrow {u_{MN}} = 15\sqrt 3 \cos \left( {100\pi t + \frac{{5\pi }}{6}} \right)\left( V \right)
\end{array}\)