Question

Đã trả lời bởi chuyên gia

Answer 1

Chọn đáp án C

Với $L = L_{0} \Rightarrow U_{L \max} \Rightarrow Z_{L 0} = \frac{R^{2} + Z_{C}^{2}}{Z_{C}};$ $\cos φ_{0} = \frac{R}{Z_{0}} = \frac{R}{\sqrt{R^{2} + \frac{R^{4}}{Z_{C}^{2}}}} = \frac{Z_{C}}{\sqrt{R^{2} + Z_{C}^{2}}}$

Với $L = L_{1}$ và $L = L_{2}$ thì $U_{L}$ bằng nhau $\frac{Z_{L 1}}{\sqrt{R^{2} + {(Z_{L 1}^{2} - Z_{C})}^{2}}} = \frac{Z_{L 2}}{\sqrt{R^{2} + {(Z_{L 2}^{2} - Z_{C})}^{2}}}$

$\frac{R^{2} + (Z_{L 1}^{2} - Z_{C})}{Z_{L 1}^{2}} = \frac{R^{2} + (Z_{L 2}^{2} - Z_{C})}{Z_{L 2}^{2}} \Leftrightarrow \frac{R^{2} + Z_{C}^{2}}{Z_{L 1}^{2}} - \frac{2 Z_{C}}{Z_{L I}} = \frac{R^{2} + Z_{C}^{2}}{Z_{L 2}^{2}} - \frac{2 Z_{C}}{Z_{L 2}}$

$\Leftrightarrow (R^{2} + Z_{C}^{2}) (\frac{1}{Z_{L 1}^{2}} - \frac{1}{Z_{L 2}^{2}}) = 2 Z_{C} (\frac{1}{Z_{L 1}^{2} - Z_{L 2}^{2}}) \Leftrightarrow \frac{1}{Z_{L 1}} + \frac{1}{Z_{L 2}} = \frac{2}{Z_{L 0}}$

Theo đề bài ta có:

$\frac{U_{L}}{U_{L \max}} = \frac{U . Z_{L 1}}{\sqrt{R^{2} + {(Z_{L 2}^{2} - Z_{C})}^{2}}} . \frac{R}{U \sqrt{R^{2} + Z_{C}^{2}}} = \frac{R}{Z_{2}} \frac{Z_{L 2}}{\sqrt{R^{2} + Z_{C}^{2}}} = \cos φ_{2} . \frac{Z_{L 2}}{\sqrt{R^{2} + Z_{C}^{2}}} \Rightarrow \cos φ_{2} = \frac{k \sqrt{R^{2} + Z_{C}^{2}}}{Z_{L 2}}$

Từ đây suy ra: $\cos φ_{1} + \cos φ_{2} = n k \Leftrightarrow \sqrt{R^{2} + Z_{C}^{2}} (\frac{1}{Z_{L 1}^{2} + Z_{L 2}^{2}}) = n$

$\Leftrightarrow \sqrt{R^{2} + Z_{C}^{2}} . \frac{2}{Z_{L 0}} = n \Leftrightarrow \cos φ_{0} = \frac{n}{2}$

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Answer 2