Đáp án:
\(\begin{array}{l}
C3.{R_1} = 3{R_2}\\
C4.{R_2} = 1,1\Omega \\
C5.{R_2} = 50\Omega \\
C6.{S_2} = 0,1333m{m^2}
\end{array}\)
Giải thích các bước giải:
C3.
Ta có:
\(\begin{array}{l}
\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho \dfrac{l}{{{S_1}}}}}{{\rho \dfrac{l}{{{S_2}}}}} = \dfrac{{{S_2}}}{{{S_1}}} = \dfrac{6}{2} = 3\\
\Rightarrow {R_1} = 3{R_2}
\end{array}\)
C4.
Ta có:
\(\begin{array}{l}
\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho \dfrac{l}{{{S_1}}}}}{{\rho \dfrac{l}{{{S_2}}}}} \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{S_2}}}{{{S_1}}}\\
\Rightarrow \dfrac{{5,5}}{{{R_2}}} = \dfrac{{2,5}}{{0,5}}\\
\Rightarrow {R_2} = 1,1\Omega
\end{array}\)
C5.
Ta có:
\(\begin{array}{l}
\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho \dfrac{{{l_1}}}{{{S_1}}}}}{{\rho \dfrac{{{l_2}}}{{{S_2}}}}} \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{l_1}{S_2}}}{{{l_2}{S_1}}}\\
\Rightarrow {R_2} = \dfrac{{{R_1}{l_2}{S_1}}}{{{l_1}{S_2}}} = \dfrac{{500.50.0,1}}{{100.0,5}} = 50\Omega
\end{array}\)
C6.
Ta có:
\(\begin{array}{l}
\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho \dfrac{{{l_1}}}{{{S_1}}}}}{{\rho \dfrac{{{l_2}}}{{{S_2}}}}} \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{l_1}{S_2}}}{{{l_2}{S_1}}}\\
\Rightarrow {S_2} = \dfrac{{{R_1}{l_2}{S_1}}}{{{R_2}{l_1}}} = \dfrac{{120.50.0,2}}{{45.200}} = 0,1333m{m^2}
\end{array}\)
