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Câu hỏi: Hai cuộn dây $\left( {{r}_{1}},{{L}_{1}} \right)$ và $\left( {{r}_{2}},{{L}_{2}} \right)$ mắc nối tiếp rồi mắc vào nguồn điện xoay chiều hđt U. Gọi ${{U}_{1}};{{U}_{2}}$ là hđt ở 2 đầu mỗi cuộn. Điều kiện để $U={{U}_{1}}+{{U}_{2}}$ là:
A. ${{L}_{1}}{{L}_{2}}={{r}_{1}}{{r}_{2}}.$
B. $\dfrac{{{L}_{1}}}{{{r}_{1}}}=\dfrac{{{L}_{2}}}{{{r}_{2}}}.$
C. $\dfrac{{{L}_{1}}}{{{r}_{2}}}=\dfrac{{{L}_{2}}}{{{r}_{1}}}.$
D. ${{L}_{1}}+{{L}_{2}}={{r}_{1}}+{{r}_{2}}.$
A. ${{L}_{1}}{{L}_{2}}={{r}_{1}}{{r}_{2}}.$
B. $\dfrac{{{L}_{1}}}{{{r}_{1}}}=\dfrac{{{L}_{2}}}{{{r}_{2}}}.$
C. $\dfrac{{{L}_{1}}}{{{r}_{2}}}=\dfrac{{{L}_{2}}}{{{r}_{1}}}.$
D. ${{L}_{1}}+{{L}_{2}}={{r}_{1}}+{{r}_{2}}.$
Ta có: $\left\{ \begin{aligned}
& U=I\sqrt{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}+\sqrt{{{\left( {{Z}_{L1}}+{{Z}_{L2}} \right)}^{2}}}} \\
& {{U}_{1}}=I\sqrt{\left( r_{1}^{2}+Z_{L1}^{2} \right)} \\
& {{U}_{2}}=I\sqrt{r_{2}^{2}+Z_{L2}^{2}} \\
\end{aligned} \right.$
$\Rightarrow U={{U}_{1}}+{{U}_{2}}$ $\Leftrightarrow \sqrt{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}+{{\left( {{Z}_{L1}}+{{Z}_{L2}} \right)}^{2}}}=\sqrt{r_{1}^{2}+Z_{L1}^{2}}+\sqrt{r_{2}^{2}+Z_{L2}^{2}}$
$\begin{aligned}
& \Rightarrow r_{1}^{2}+2{{r}_{1}}{{r}_{2}}+r_{2}^{2}+Z_{L1}^{2}+Z_{L2}^{2}+2{{Z}_{L1}}{{Z}_{L2}} \\
& \Rightarrow r_{1}^{2}+r_{2}^{2}+Z_{L1}^{2}+Z_{L2}^{2}+2\sqrt{\left( r_{1}^{2}+Z_{L1}^{2} \right)\left( r_{2}^{2}+Z_{L2}^{2} \right)} \\
& \Rightarrow 2{{r}_{1}}{{r}_{2}}+2{{Z}_{L1}}{{Z}_{L2}}=2\sqrt{r_{1}^{2}r_{2}^{2}+Z_{L1}^{2}Z_{L2}^{2}+r_{1}^{2}Z_{L2}^{2}+r_{2}^{2}Z_{L1}^{2}} \\
& \Rightarrow r_{1}^{2}r_{2}^{2}+Z_{L1}^{2}Z_{L2}^{2}+2{{r}_{1}}{{Z}_{L1}}{{r}_{2}}{{Z}_{L2}} =r_{1}^{2}r_{2}^{2}+Z_{L1}^{2}Z_{L2}^{2}+r_{1}^{2}Z_{L2}^{2}+r_{2}^{2}Z_{L1}^{2} \\
& \Rightarrow 2{{r}_{1}}{{r}_{2}}{{Z}_{L1}}{{Z}_{L2}}=r_{1}^{2}Z_{L2}^{2}+r_{2}^{2}Z_{L1}^{2}\Rightarrow {{\left( {{r}_{1}}{{Z}_{L2}}-{{r}_{2}}{{Z}_{L1}} \right)}^{2}}=0\Rightarrow {{r}_{1}}{{Z}_{L2}}={{r}_{2}}{{Z}_{L1}}\Rightarrow \dfrac{{{L}_{1}}}{{{r}_{1}}}=\dfrac{{{L}_{2}}}{{{r}_{2}}} \\
\end{aligned}$
& U=I\sqrt{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}+\sqrt{{{\left( {{Z}_{L1}}+{{Z}_{L2}} \right)}^{2}}}} \\
& {{U}_{1}}=I\sqrt{\left( r_{1}^{2}+Z_{L1}^{2} \right)} \\
& {{U}_{2}}=I\sqrt{r_{2}^{2}+Z_{L2}^{2}} \\
\end{aligned} \right.$
$\Rightarrow U={{U}_{1}}+{{U}_{2}}$ $\Leftrightarrow \sqrt{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}+{{\left( {{Z}_{L1}}+{{Z}_{L2}} \right)}^{2}}}=\sqrt{r_{1}^{2}+Z_{L1}^{2}}+\sqrt{r_{2}^{2}+Z_{L2}^{2}}$
$\begin{aligned}
& \Rightarrow r_{1}^{2}+2{{r}_{1}}{{r}_{2}}+r_{2}^{2}+Z_{L1}^{2}+Z_{L2}^{2}+2{{Z}_{L1}}{{Z}_{L2}} \\
& \Rightarrow r_{1}^{2}+r_{2}^{2}+Z_{L1}^{2}+Z_{L2}^{2}+2\sqrt{\left( r_{1}^{2}+Z_{L1}^{2} \right)\left( r_{2}^{2}+Z_{L2}^{2} \right)} \\
& \Rightarrow 2{{r}_{1}}{{r}_{2}}+2{{Z}_{L1}}{{Z}_{L2}}=2\sqrt{r_{1}^{2}r_{2}^{2}+Z_{L1}^{2}Z_{L2}^{2}+r_{1}^{2}Z_{L2}^{2}+r_{2}^{2}Z_{L1}^{2}} \\
& \Rightarrow r_{1}^{2}r_{2}^{2}+Z_{L1}^{2}Z_{L2}^{2}+2{{r}_{1}}{{Z}_{L1}}{{r}_{2}}{{Z}_{L2}} =r_{1}^{2}r_{2}^{2}+Z_{L1}^{2}Z_{L2}^{2}+r_{1}^{2}Z_{L2}^{2}+r_{2}^{2}Z_{L1}^{2} \\
& \Rightarrow 2{{r}_{1}}{{r}_{2}}{{Z}_{L1}}{{Z}_{L2}}=r_{1}^{2}Z_{L2}^{2}+r_{2}^{2}Z_{L1}^{2}\Rightarrow {{\left( {{r}_{1}}{{Z}_{L2}}-{{r}_{2}}{{Z}_{L1}} \right)}^{2}}=0\Rightarrow {{r}_{1}}{{Z}_{L2}}={{r}_{2}}{{Z}_{L1}}\Rightarrow \dfrac{{{L}_{1}}}{{{r}_{1}}}=\dfrac{{{L}_{2}}}{{{r}_{2}}} \\
\end{aligned}$
Đáp án B.