Đáp án:
\(\begin{array}{l}
a.R = 4\Omega \\
b.\\
{U_3} = 12V\\
{I_3} = 2A\\
{I_1} = {I_2} = 1A\\
{U_1} = 8V\\
{U_2} = 4V
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = {R_1} + {R_2} = 8 + 4 = 12\Omega \\
R = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{12.6}}{{12 + 6}} = 4\Omega \\
b.\\
U = {U_3} = {U_{12}} = 12V\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \frac{{12}}{6} = 2A\\
{I_1} = {I_2} = {I_{12}} = \dfrac{{{U_{12}}}}{{{R_{12}}}} = \dfrac{{12}}{{12}} = 1A\\
{U_1} = {I_1}{R_1} = 1.8 = 8V\\
{U_2} = {U_{12}} - {U_1} = 12 - 8 = 4V
\end{array}\)
