Đặt điện áp XC có U không đổi vào 2 đầu mạch R,L,C nối tiếp có L thay đổi

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:  

 \(\begin{array}{l}
 \Rightarrow {U_L} = \frac{U}{R}\sqrt {{R^2} + Z_C^2} \left( {\sin {\varphi _0}\sin \varphi  + \cos {\varphi _0}\cos \varphi } \right)\\
 \Rightarrow {U_L} = {U_{L - \max }}\cos \left( {\varphi  - {\varphi _0}} \right)\\
 \Rightarrow {U_L} = {U_{L - \max }}\cos \left( {\frac{{{\varphi _1} - {\varphi _2}}}{2}} \right)\\
 \Rightarrow {\varphi _1} - {\varphi _2} = {\varphi _{uL2}} - {\varphi _{uL1}} = 0,2\pi  + 0,1\pi  = 0,3\pi \\
 \Rightarrow {U_L} = {U_{L - \max }}\cos \left( {\frac{{{\varphi _1} - {\varphi _2}}}{2}} \right)\\
 \Rightarrow {U_{L - \max }} = \frac{{{U_{L1}}}}{{\cos \left( {\frac{{{\varphi _1} - {\varphi _2}}}{2}} \right)}} = \frac{{20\sqrt 2 }}{{\cos \left( {0,15\pi } \right)}} = 31,74\left( V \right)
\end{array}\)