Đáp án:
\(\begin{array}{l}
a.R = 15\Omega \\
b.\\
{I_3} = 1A\\
{I_1} = 0,6A\\
{I_2} = 0,4A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{10.15}}{{10 + 15}} = 6\Omega \\
R = {R_{12}} + {R_3} = 6 + 9 = 15\Omega \\
b.\\
{I_3} = {I_{12}} = I = \dfrac{U}{R} = \dfrac{{15}}{{15}} = 1A\\
{U_1} = {U_2} = {U_{12}} = {I_{12}}{R_{12}} = 1.6 = 6V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{6}{{10}} = 0,6A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{6}{{15}} = 0,4A
\end{array}\)