Đặt điện áp u = U√2cos100πt (V) vào đoạn mạch R, L, C nối tiếp

+ Ta có:  

\(\begin{array}{l}
{Z_C} = \frac{1}{{\omega C}} = 15\Omega \\
S = {U_L} + 2{U_C} = I\left( {{Z_{L1}} + 2{Z_C}} \right)\\
 \Rightarrow S = \frac{{U\left( {{Z_{L1}} + 2{Z_C}} \right)}}{{\sqrt {{R^2} + {{\left( {{Z_{L1}} - {Z_C}} \right)}^2}} }} = U\sqrt {\frac{{{{\left( {{Z_{L1}} + 2{Z_C}} \right)}^2}}}{{{R^2} + {{\left( {{Z_{L1}} - {Z_C}} \right)}^2}}}} \\
y = \frac{{{{\left( {{Z_{L1}} + 2{Z_C}} \right)}^2}}}{{{R^2} + {{\left( {{Z_{L1}} - {Z_C}} \right)}^2}}}
\end{array}\)

+ Tính đạo hàm hàm số của y theo x:  

 \(\begin{array}{l}
{y^/} = \frac{{2\left( {x + 2{Z_C}} \right)\left[ {{R^2} + {{\left( {x - {Z_C}} \right)}^2}} \right] - 2\left( {x - {Z_C}} \right)\left[ {{{\left( {x + 2{Z_C}} \right)}^2}} \right]}}{{M{S^2}}}\\
 \Leftrightarrow \left[ {{R^2} + {{\left( {x - {Z_C}} \right)}^2}} \right] - \left( {x - {Z_C}} \right)\left( {x + 2{Z_C}} \right) = 0\\
 \Rightarrow {R^2} + {x^2} - 2{Z_C}x + Z_C^2 - {x^2} - 2{Z_C}x + x{Z_C} + 2Z_C^2 = 0\\
 \Rightarrow {R^2} - 3{Z_C}x + 3Z_C^2 = 0\\
 \Rightarrow x = \frac{{{R^2} + 3Z_C^2}}{{3{Z_C}}} = \frac{{{{60}^2} + {{3.15}^2}}}{{3.15}} = 95\Omega \\
 \Rightarrow {L_1} = \frac{{0,95}}{\pi }
\end{array}\)