Đáp án:
\(\begin{array}{l}
a.I = \dfrac{2}{3}A\\
b.\\
{I_3} = \dfrac{2}{3}A\\
{I_1} = \dfrac{4}{9}A\\
c.{Q_2} = \dfrac{{3200}}{3}J
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{E_b} = E = 12V\\
{r_b} = \dfrac{r}{2} = \dfrac{4}{2} = 2V\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{12.24}}{{12 + 24}} = 8\Omega \\
R = {R_{12}} + {R_3} = 8 + 8 = 16\Omega \\
I = \dfrac{{{E_b}}}{{{r_b} + R}} = \dfrac{{12}}{{2 + 16}} = \dfrac{2}{3}A\\
b.\\
U = IR = \dfrac{2}{3}.16 = \dfrac{{32}}{3}V\\
{I_3} = I = \dfrac{2}{3}A\\
{U_3} = {I_3}{R_3} = \dfrac{2}{3}.8 = \dfrac{{16}}{3}V\\
{U_1} = {U_2} = U - {U_3} = \dfrac{{32}}{3} - \dfrac{{16}}{3} = \dfrac{{16}}{3}V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{\dfrac{{16}}{3}}}{{12}} = \dfrac{4}{9}A\\
c.\\
{I_2} = I - {I_1} = \dfrac{2}{3} - \dfrac{4}{9} = \dfrac{2}{9}A\\
{Q_2} = {R_2}I_2^2t = 24.{\dfrac{2}{9}^2}.900 = \dfrac{{3200}}{3}J
\end{array}\)
