Lời giải:– Khi nối tắt tụ
\(\begin{array}{l}
{Z_C} = \frac{1}{{\omega C}} = 20\Omega .\\
U_{AB}^2 = {\left( {{U_{R1}} + {U_{R2}}} \right)^2} + U_L^2 = {60^2}\\
U_{MB}^2 = U_{R2}^2 + U_L^2 = {\left( {20\sqrt 5 } \right)^2} = 2000V.\\
{U_{R2}} = 10\sqrt 2 V;{U_L} = 30\sqrt 2 V
\end{array}\)
- Khi chưa nối tắt có
\(\begin{array}{l}
{U_{AM}} = \frac{{U\sqrt {R_1^2 + Z_C^2} }}{{\sqrt {{{\left( {{R_1} + {R_2}} \right)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = \frac{{60\sqrt {{{\left( {2x} \right)}^2} + {{20}^2}} }}{{\sqrt {{{\left( {2x + x} \right)}^2} + {{\left( {3x - 20} \right)}^2}} }} = 24\sqrt 5 \\
\to {\rm{ }}x{\rm{ }} = {\rm{ }}10.
\end{array}\)
Hệ số công suất mạch:
\(\cos \varphi = \frac{{{R_1} + {R_2}}}{{\sqrt {{{\left( {{R_1} + {R_2}} \right)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = \frac{{20 + 10}}{{\sqrt {{{\left( {20 + 10} \right)}^2} + {{\left( {30 - 20} \right)}^2}} }} = 0,95.\)
