Phương pháp giải:

Dùng phương pháp đại số: Xác định công thức tính \(\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{0}}}\),\(\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{1}}}\)và \(\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{2}}\)từ mối quan hệ giữa U_L và U_Lmax. Sau đó sử dụng dữ kiện còn lại tính \(\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{0}}}\).

Lời giải chi tiết:

Khi L = L₀ thì U_Lmax nên ta có:

\({{\text{Z}}_{\text{L0}}}\text{ = }\frac{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}{{{\text{Z}}_{\text{C}}}}\) và \(\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{0}}}\text{ = }\frac{\text{R}}{{{\text{Z}}_{\text{0}}}}\text{ = }\frac{\text{R}}{\sqrt{{{\text{R}}^{\text{2}}}\text{ + }\frac{{{\text{R}}^{\text{4}}}}{\text{Z}_{\text{C}}^{\text{2}}}}}=\frac{{{\text{Z}}_{\text{C}}}}{\sqrt{\text{Z}_{\text{C}}^{\text{2}}\text{ + }{{\text{R}}^{\text{2}}}}}=\frac{\sqrt{\text{Z}_{\text{C}}^{\text{2}}\text{ + }{{\text{R}}^{\text{2}}}}}{{{\text{Z}}_{\text{L0}}}}\)

Khi L = L₁ hoặc L = L₂ thì điện áp hiệu dụng giữa hai đầu cuộn cảm có giá trị như nhau và bằng U_L:

\(\frac{\text{U}{{\text{Z}}_{\text{L1}}}}{{{\text{Z}}_{\text{1}}}}\text{ = }\frac{\text{U}{{\text{Z}}_{\text{L2}}}}{{{\text{Z}}_{\text{2}}}}\Leftrightarrow \frac{{{\text{Z}}_{\text{L1}}}}{\sqrt{{{\text{R}}^{\text{2}}}\text{+}{{\left( {{\text{Z}}_{\text{L1}}}-{{\text{Z}}_{\text{C}}} \right)}^{\text{2}}}}}\text{=}\frac{{{\text{Z}}_{\text{L2}}}}{\sqrt{{{\text{R}}^{\text{2}}}\text{+}{{\left( {{\text{Z}}_{\text{L2}}}-{{\text{Z}}_{\text{C}}} \right)}^{\text{2}}}}}\Leftrightarrow \frac{{{\text{R}}^{\text{2}}}\text{+}{{\left( {{\text{Z}}_{\text{L1}}}-{{\text{Z}}_{\text{C}}} \right)}^{\text{2}}}}{\text{Z}_{\text{L1}}^{\text{2}}}=\frac{{{\text{R}}^{\text{2}}}\text{+}{{\left( {{\text{Z}}_{\text{L2}}}-{{\text{Z}}_{\text{C}}} \right)}^{\text{2}}}}{\text{Z}_{\text{L2}}^{\text{2}}}\)

\(\Leftrightarrow \frac{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}{\text{Z}_{\text{L1}}^{\text{2}}}-\frac{\text{2}{{\text{Z}}_{\text{C}}}}{{{\text{Z}}_{\text{L1}}}}=\frac{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}{\text{Z}_{\text{L2}}^{\text{2}}}-\frac{\text{2}{{\text{Z}}_{\text{C}}}}{{{\text{Z}}_{\text{L2}}}}\Leftrightarrow \left( {{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}} \right)\left( \frac{1}{\text{Z}_{\text{L1}}^{\text{2}}}-\frac{1}{\text{Z}_{\text{L2}}^{\text{2}}} \right)=2{{\text{Z}}_{\text{C}}}\left( \frac{1}{{{\text{Z}}_{\text{L1}}}}-\frac{1}{{{\text{Z}}_{\text{L2}}}} \right)\)

\(\Leftrightarrow \frac{1}{{{\text{Z}}_{\text{L1}}}}+\frac{1}{{{\text{Z}}_{\text{L2}}}}=2\frac{{{\text{Z}}_{\text{C}}}}{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}\Leftrightarrow \frac{1}{{{\text{Z}}_{\text{L1}}}}+\frac{1}{{{\text{Z}}_{\text{L2}}}}=\frac{2}{{{\text{Z}}_{\text{L0}}}}\)

Theo đề bài ta có:

\(\frac{{{\text{U}}_{\text{L}}}}{{{\text{U}}_{\text{Lmax}}}}\text{ = }\frac{\text{U}\text{.}{{\text{Z}}_{\text{L1}}}}{\sqrt{{{\text{R}}^{\text{2}}}\text{+}{{\left( {{\text{Z}}_{\text{L1}}}-{{\text{Z}}_{\text{C}}} \right)}^{\text{2}}}}}.\frac{\text{R}}{\text{U}\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}=\frac{\text{R}}{{{\text{Z}}_{\text{1}}}}\text{.}\frac{{{\text{Z}}_{\text{L1}}}}{\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}=\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{1}}}.\frac{{{\text{Z}}_{\text{L1}}}}{\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}\)

\(\Rightarrow \text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{1}}}\text{.}\frac{{{\text{Z}}_{\text{L1}}}}{\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}\text{ = k}\Rightarrow \text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{1}}}=\frac{\text{k}\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}{{{\text{Z}}_{\text{L1}}}}\)

Tương tự ta có:      \(\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{2}}=\frac{\text{k}\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}{{{\text{Z}}_{\text{L2}}}}\)

 Mặt khác: \(\text{cos}{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{1}}}\text{ + cos}{{\text{ }\!\!\varphi\!\!\text{ }}_{2}}\text{ = nk}\Leftrightarrow \frac{\text{k}\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}{{{\text{Z}}_{\text{L1}}}}+\frac{\text{k}\sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}}{{{\text{Z}}_{\text{L2}}}}=\text{nk}\Leftrightarrow \sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}\left( \frac{1}{{{\text{Z}}_{\text{L1}}}}+\frac{1}{{{\text{Z}}_{\text{L2}}}} \right)=\text{n}\)

\(\Leftrightarrow \sqrt{{{\text{R}}^{\text{2}}}\text{ + Z}_{\text{C}}^{\text{2}}}.\frac{2}{{{\text{Z}}_{\text{L0}}}}=\text{n}\Leftrightarrow 2\text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{0}}}=\text{n}\Leftrightarrow \text{cos }{{\text{ }\!\!\varphi\!\!\text{ }}_{\text{0}}}=\frac{\text{n}}{\text{2}}\)

Chọn C