Đáp án:
\(\begin{array}{l}
a.\\
{I_1} = \dfrac{{80}}{{17}}A\\
{I_2} = {I_3} = \dfrac{{40}}{{51}}A\\
b.{Q_3} = 885,813J
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{23}} = {R_2} + {R_3} = 4 + 8 = 12\Omega \\
R = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{2.12}}{{2 + 12}} = \dfrac{{12}}{7}\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{16}}{{\dfrac{{12}}{7} + 1,2}} = \dfrac{{280}}{{51}}A\\
{U_1} = {U_{23}} = U = {\rm{IR}} = \dfrac{{280}}{{51}}.\dfrac{{12}}{7} = \dfrac{{160}}{{17}}V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{\frac{{160}}{{17}}}}{2} = \dfrac{{80}}{{17}}A\\
{I_2} = {I_3} = {I_{23}} = \dfrac{{{U_{23}}}}{{{R_{23}}}} = \dfrac{{\dfrac{{160}}{{17}}}}{{12}} = \dfrac{{40}}{{51}}A\\
b.\\
{Q_3} = {R_3}I_3^2t = 8.{\dfrac{{40}}{{51}}^2}.180 = 885,813J
\end{array}\)
