Đáp án:
\(\begin{array}{l}
a.\\
{I_A} = 0,4A\\
b.\\
{I_A} = 1,2A
\end{array}\)
Giải thích các bước giải:
Hiệu điện thế nào vậy bạn.
\(\begin{array}{l}
a.\\
(({R_2}nt{R_4})//{R_1})nt{R_3}\\
{R_{24}} = {R_2} + {R_4} = 6 + 18 = 24\Omega \\
{R_{124}} = \dfrac{{{R_1}{R_{24}}}}{{{R_1} + {R_{24}}}} = \dfrac{{6.24}}{{6 + 24}} = 4,8\Omega \\
R = {R_{124}} + {R_3} = 6 + 4,8 = 10,8\Omega \\
{I_{124}} = I = \dfrac{U}{R} = \dfrac{{21,6}}{{10,8}} = 2A\\
{U_{24}} = {U_{124}} = {\rm{I}}{{\rm{R}}_{124}} = 2.4,8 = 9,6V\\
{I_A} = {I_{24}} = \dfrac{{{U_{24}}}}{{{R_{24}}}} = \dfrac{{9,6}}{{24}} = 0,4A\\
b.\\
(({R_2}//{R_3})nt{R_1})//{R_4}\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{6.6}}{{6 + 6}} = 3\Omega \\
{R_{123}} = 3 + 6 = 9\Omega \\
R = \dfrac{{{R_{123}}{R_4}}}{{{R_{123}} + {R_4}}} = \dfrac{{9.18}}{{9 + 18}} = 6\Omega \\
I = \dfrac{U}{R} = \dfrac{{21,6}}{6} = 3,6A\\
{U_4} = {U_{123}} = U = 21,6V\\
{I_{23}} = {I_{123}} = \dfrac{{{U_{123}}}}{{{R_{123}}}} = \dfrac{{21,6}}{9} = 2,4A\\
{U_2} = {U_{23}} = {I_{23}}{R_{23}} = 2,4.3 = 7,2V\\
{I_A} = {I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{7,2}}{6} = 1,2A
\end{array}\)
