Question

Đã trả lời bởi chuyên gia

Answer 1

tham khảo Hỏi đáp VietJack

Answer 2

* Khi chưa mắc thêm R:

\({I_1} = {I_2} = \dfrac{U}{{{R_1} + {R_2}}}\)

* Sau khi mắc thêm R:

\(\begin{array}{l}
{R_{td}} = {R_1} + \dfrac{{R{R_2}}}{{R + {R_2}}} = \dfrac{{{R_1}{R_2} + R{R_1} + R{R_2}}}{{R + {R_2}}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{U\left( {R + {R_2}} \right)}}{{{R_1}{R_2} + R{R_1} + R{R_2}}}\\
{I_R} = \dfrac{{{R_2}}}{{R + {R_2}}}.I = \dfrac{{U{R_2}}}{{{R_1}{R_2} + R{R_1} + R{R_2}}} = 0,012A\\
I - {I_1} = {4.10^{ - 3}}\\
\Rightarrow \dfrac{{U\left( {R + {R_2}} \right)}}{{{R_1}{R_2} + R{R_1} + R{R_2}}} - \dfrac{U}{{{R_1} + {R_2}}} = {4.10^{ - 3}}\\
\Rightarrow \dfrac{{\dfrac{{U\left( {R + {R_2}} \right)}}{{{R_1}{R_2} + R{R_1} + R{R_2}}} - \dfrac{U}{{{R_1} + {R_2}}}}}{{\dfrac{{U{R_2}}}{{{R_1}{R_2} + R{R_1} + R{R_2}}}}} = \dfrac{{{{4.10}^{ - 3}}}}{{0,012}}\\
\Rightarrow \dfrac{{R + {R_2}}}{{{R_2}}} - \dfrac{{\dfrac{1}{{{R_1} + {R_2}}}}}{{\dfrac{{{R_2}}}{{{R_1}{R_2} + R{R_1} + R{R_2}}}}} = \dfrac{1}{3}\\
\Rightarrow \dfrac{R}{{{R_2}}} - \dfrac{{{R_1}{R_2} + R{R_1} + R{R_2}}}{{{R_2}\left( {{R_1} + {R_2}} \right)}} = - \dfrac{2}{3}\\
\Rightarrow \dfrac{{R{R_1} + R{R_2} - \left( {{R_1}{R_2} + R{R_1} + R{R_2}} \right)}}{{{R_2}\left( {{R_1} + {R_2}} \right)}} = - \dfrac{2}{3}\\
\Rightarrow \dfrac{{{R_1}{R_2}}}{{{R_2}\left( {{R_1} + {R_2}} \right)}} = \dfrac{2}{3}\\
\Rightarrow \dfrac{1}{{1 + \dfrac{{{R_2}}}{{{R_1}}}}} = \dfrac{2}{3} \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{2}\\
\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = 2
\end{array}\)

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Answer 3

* Khi chưa mắc thêm R:

\({I_1} = {I_2} = \dfrac{U}{{{R_1} + {R_2}}}\)

* Sau khi mắc thêm R:

\(\begin{array}{l}
{R_{td}} = {R_1} + \dfrac{{R{R_2}}}{{R + {R_2}}} = \dfrac{{{R_1}{R_2} + R{R_1} + R{R_2}}}{{R + {R_2}}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{U\left( {R + {R_2}} \right)}}{{{R_1}{R_2} + R{R_1} + R{R_2}}}\\
{I_R} = \dfrac{{{R_2}}}{{R + {R_2}}}.I = \dfrac{{U{R_2}}}{{{R_1}{R_2} + R{R_1} + R{R_2}}} = 0,012A\\
I - {I_1} = {4.10^{ - 3}}\\
\Rightarrow \dfrac{{U\left( {R + {R_2}} \right)}}{{{R_1}{R_2} + R{R_1} + R{R_2}}} - \dfrac{U}{{{R_1} + {R_2}}} = {4.10^{ - 3}}\\
\Rightarrow \dfrac{{\dfrac{{U\left( {R + {R_2}} \right)}}{{{R_1}{R_2} + R{R_1} + R{R_2}}} - \dfrac{U}{{{R_1} + {R_2}}}}}{{\dfrac{{U{R_2}}}{{{R_1}{R_2} + R{R_1} + R{R_2}}}}} = \dfrac{{{{4.10}^{ - 3}}}}{{0,012}}\\
\Rightarrow \dfrac{{R + {R_2}}}{{{R_2}}} - \dfrac{{\dfrac{1}{{{R_1} + {R_2}}}}}{{\dfrac{{{R_2}}}{{{R_1}{R_2} + R{R_1} + R{R_2}}}}} = \dfrac{1}{3}\\
\Rightarrow \dfrac{R}{{{R_2}}} - \dfrac{{{R_1}{R_2} + R{R_1} + R{R_2}}}{{{R_2}\left( {{R_1} + {R_2}} \right)}} = - \dfrac{2}{3}\\
\Rightarrow \dfrac{{R{R_1} + R{R_2} - \left( {{R_1}{R_2} + R{R_1} + R{R_2}} \right)}}{{{R_2}\left( {{R_1} + {R_2}} \right)}} = - \dfrac{2}{3}\\
\Rightarrow \dfrac{{{R_1}{R_2}}}{{{R_2}\left( {{R_1} + {R_2}} \right)}} = \dfrac{2}{3}\\
\Rightarrow \dfrac{1}{{1 + \dfrac{{{R_2}}}{{{R_1}}}}} = \dfrac{2}{3} \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{1}{2}\\
\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = 2
\end{array}\)

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