Đáp án:
a.
+ K mở:
\(R = \dfrac{{153}}{7}\Omega \)
+ K đóng:
\(\begin{array}{l}
R = 4\Omega \\
b.{I_2} = 1A
\end{array}\)
Giải thích các bước giải:
a.
+ K mở:
\(\begin{array}{l}
{R_3}nt(({R_1}nt{R_2})//{R_4})\\
{R_{12}} = {R_1} + {R_2} = 6 + 20 = 26\Omega \\
{R_{124}} = \dfrac{{{R_{12}}{R_4}}}{{{R_{12}} + {R_4}}} = \dfrac{{2.26}}{{2 + 26}} = \dfrac{{13}}{7}\Omega \\
R = {R_3} + {R_{124}} = 20 + \dfrac{{13}}{7} = \dfrac{{153}}{7}\Omega
\end{array}\)
+ K đóng:
\(\begin{array}{l}
(({R_2}//{R_3})nt{R_4})//{R_1}\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{20.20}}{{20 + 20}} = 10\Omega \\
{R_{234}} = {R_{23}} + {R_4} = 10 + 2 = 12\Omega \\
R = \dfrac{{{R_1}{R_{234}}}}{{{R_1} + {R_{234}}}} = \dfrac{{6.12}}{{6 + 12}} = 4\Omega
\end{array}\)
b.
\(\begin{array}{l}
I = \dfrac{U}{R} = \dfrac{{24}}{4} = 6A\\
U = {U_1} = {U_{234}} = 24V\\
{I_{234}} = \dfrac{{{U_{234}}}}{{{R_{234}}}} = \dfrac{{24}}{{12}} = 2A\\
{U_4} = {I_{234}}{R_4} = 2.2 = 4V\\
{U_{23}} = {U_2} = {U_{234}} - {U_4} = 24 - 4 = 20V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{20}}{{20}} = 1A
\end{array}\)
