Đáp án:
\(\begin{array}{l}
a.R = 3,6\Omega \\
b.\\
{U_1} = {U_2} = 6V\\
{U_4} = {U_3} = 12V\\
{I_1} = 3A\\
{I_2} = 2A\\
{I_3} = 3A\\
{I_4} = 2A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{2.3}}{{2 + 3}} = 1,2\Omega \\
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = \dfrac{{4.6}}{{4 + 6}} = 2,4\Omega \\
R = {R_{12}} + {R_{34}} = 1,2 + 2,4 = 3,6\Omega \\
b.\\
I = \dfrac{U}{R} = \dfrac{{18}}{{3,6}} = 5A\\
{U_1} = {U_2} = {U_{12}} = I{R_{12}} = 5.1,2 = 6V\\
{U_4} = {U_3} = {U_{34}} = U - {U_{12}} = 18 - 6 = 12V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{6}{2} = 3A\\
{I_2} = I - {I_1} = 5 - 3 = 2A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{12}}{4} = 3A\\
{I_4} = I - {I_3} = 5 - 3 = 2A
\end{array}\)
