Lời giải:Ta có:
\(\begin{array}{l}
\tan \varphi = \frac{{{Z_L} - {Z_C}}}{R} \Rightarrow {Z_L} = R\tan \varphi + {Z_C}\\
{U_L} = I.{Z_L} = \frac{U}{Z}.{Z_L}\\
\Rightarrow {U_L} = \frac{{U\cos \varphi }}{R}\left( {R\tan \varphi + {Z_C}} \right) = \frac{U}{R}\left( {R\sin \varphi + {Z_C}\cos \varphi } \right)\\
\Rightarrow {U_L} = \frac{U}{R}\sqrt {{R^2} + Z_C^2} \left( {\frac{R}{{\sqrt {{R^2} + Z_C^2} }}\sin \varphi + \frac{{{Z_C}}}{{\sqrt {{R^2} + Z_C^2} }}\cos \varphi } \right)
\end{array}\)
+ Mặt khác:
\(\begin{array}{l}
\Rightarrow {U_L} = \frac{U}{R}\sqrt {{R^2} + Z_C^2} \left( {\sin {\varphi _0}\sin \varphi + \cos {\varphi _0}\cos \varphi } \right)\\
\Rightarrow {U_L} = {U_{L - \max }}\cos \left( {\varphi - {\varphi _0}} \right)\\
\Rightarrow {U_L} = {U_{L - \max }}\cos \left( {\frac{{{\varphi _1} - {\varphi _2}}}{2}} \right)\\
\Rightarrow {\varphi _1} - {\varphi _2} = {\varphi _{uL2}} - {\varphi _{uL1}} = 0,2\pi + 0,1\pi = 0,3\pi \\
\Rightarrow {U_L} = {U_{L - \max }}\cos \left( {\frac{{{\varphi _1} - {\varphi _2}}}{2}} \right)\\
\Rightarrow {U_{L - \max }} = \frac{{{U_{L1}}}}{{\cos \left( {\frac{{{\varphi _1} - {\varphi _2}}}{2}} \right)}} = \frac{{20\sqrt 2 }}{{\cos \left( {0,15\pi } \right)}} = 31,74\left( V \right)
\end{array}\)
