Đáp án:
$\begin{array}{l}
a.{e_b} = 5e = 20\left( V \right)\\
{r_b} = 5r = 2.5\left( \Omega \right)
\end{array}$
b
$\begin{array}{l}
I = {I_1} = 0,83\left( A \right)\\
{I_2} = 0,33\left( A \right)\\
{I_3} = {I_4} = 0,5\left( A \right)
\end{array}$
$\begin{array}{l}
c.{U_{MN}} = 1\left( V \right)\\
{U_{AM}} = 6\left( V \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a.{e_b} = 5e = 20\left( V \right)\\
{r_b} = 5r = 2.5\left( \Omega \right)\\
b.{R_1}nt\left( {\left( {{R_D}nt{R_2}} \right)//\left( {{R_3}nt{R_4}} \right)} \right)\\
{R_{D2}} = {R_D} + {R_2} = 9\\
{R_{34}} = {R_3} + {R_4} = 6\\
{R_{CB}} = \frac{{{R_{D2}}{R_{34}}}}{{{R_{D2}} + {R_{34}}}} = \frac{{9.6}}{{9 + 6}} = 3,6\\
{R_{AB}} = {R_1} + {R_{CB}} = 6 + 3,6 = 9,6\\
I = {I_1} = {I_{CB}} = \frac{{{e_b}}}{{{R_{AB}} + {r_b}}} = \frac{{20}}{{2,5 + 9,6}} = \frac{5}{6} = 0,83\left( A \right)\\
{U_{CB}} = {U_{D2}} = {U_{34}} = {I_{CB}}.{R_{CB}} = 3\\
{I_2} = {I_{D2}} = {I_D} = \frac{{{U_{D2}}}}{{{R_{D2}}}} = \frac{3}{9} = 0,33\left( A \right)\\
{I_{34}} = {I_3} = {I_4} = \frac{{{U_{34}}}}{{{R_{34}}}} = \frac{3}{6} = 0,5\left( A \right)\\
c.{U_{MN}} = {U_{MC}} + {U_{CN}} = - {U_D} + {U_1} = - {I_D}{R_D} + {I_3}{R_3} = - \frac{1}{3}.3 + 0,5.4 = 1\left( V \right)\\
{U_{AM}} = {U_{AC}} + {U_{CM}} = {U_1} + {U_D} = {I_1}{R_1} + {I_D}{R_D} = \frac{5}{6}.6 + \frac{1}{3}.3 = 6\left( V \right)\\
\end{array}$
