Đáp án:
\(\begin{array}{l}
a.\\
R = 9\Omega \\
{I_1} = {I_2} = \dfrac{4}{3}A\\
b.\\
R = 2\Omega \\
{I_1} = 4A\\
{I_2} = 2A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
R = {R_1} + {R_2} = 3 + 6 = 9\Omega \\
{I_1} = {I_2} = I = \dfrac{U}{R} = \dfrac{{12}}{9} = \dfrac{4}{3}A\\
b.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
{U_1} = {U_2} = U = 12V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{12}}{3} = 4A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{12}}{6} = 2A
\end{array}\)