Đáp án:
\(\begin{array}{l}
a.{R_2} = 12\Omega \\
b.\\
{I_A} = \dfrac{{60}}{{119}}A\\
{I_k} = \dfrac{9}{{119}}A
\end{array}\)
Dòng điện đi từ B đến A
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{34}} = {R_3} + {R_4} = 4 + 8 = 12\Omega \\
R = \dfrac{U}{I} = \dfrac{{12}}{{0,5}} = 24\Omega \\
{R_{1234}} = R - {R_5} - {R_A} = 24 - 15 - 1 = 8\Omega \\
\dfrac{1}{{{R_{1234}}}} = \dfrac{1}{{{R_{12}}}} + \dfrac{1}{{{R_{34}}}}\\
\Rightarrow \dfrac{1}{8} = \dfrac{1}{{{R_{12}}}} = \dfrac{1}{{12}}\\
\Rightarrow {R_{12}} = 24\Omega \\
{R_2} = {R_{12}} - {R_1} = 24 - 12 = 12\Omega \\
b.\\
{R_{13}} = \dfrac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \dfrac{{12.4}}{{12 + 4}} = 3\Omega \\
{R_{24}} = \dfrac{{{R_2}{R_4}}}{{{R_2} + {R_4}}} = \dfrac{{12.8}}{{12 + 8}} = 4,8\Omega \\
R = {R_A} + {R_{13}} + {R_{24}} + {R_5} = 1 + 3 + 4,8 + 15 = 23,8\Omega \\
I = {I_A} = {I_{13}} = {I_{24}} = \dfrac{U}{R} = \dfrac{{12}}{{23,8}} = \dfrac{{60}}{{119}}A\\
{U_1} = {U_{13}} = {I_{13}}{R_{13}} = \dfrac{{60}}{{119}}.3 = \dfrac{{180}}{{119}}V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{\dfrac{{180}}{{119}}}}{{12}} = \dfrac{{15}}{{119}}A\\
{U_2} = {U_{24}} = {I_{24}}{R_{24}} = \dfrac{{60}}{{119}}.4,8 = \dfrac{{288}}{{119}}V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{\frac{{288}}{{119}}}}{{12}} = \dfrac{{24}}{{119}}A\\
{I_2} > {I_1}
\end{array}\)
\(\begin{array}{l}
{I_2} = {I_1} + {I_k}\\
\Rightarrow {I_k} = {I_2} - {I_1} = \dfrac{{24}}{{119}} - \dfrac{{15}}{{119}} = \dfrac{9}{{119}}A
\end{array}\)
Dòng điện qua khóa k đi từ B đến A
