Đáp án:

a) \(\dfrac{{18}}{{11}}A\)

b) \( - \dfrac{{153}}{{11}}V\)

Giải thích các bước giải:

a) Ta có:

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 8 + 10 = 18\Omega \\
{R_{123}} = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{6.18}}{{6 + 18}} = 4,5\Omega \\
{R_{td}} = {R_{123}} + {R_4} = 4,5 + 4,5 = 9\Omega \\
I = \dfrac{E}{{r + {R_{td}}}} = \dfrac{{18}}{{9 + 2}} = \dfrac{{18}}{{11}}A
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{I_4} = I = \dfrac{{18}}{{11}}A\\
{U_4} = {I_4}{R_4} = \dfrac{{18}}{{11}}.6 = \dfrac{{108}}{{11}}V\\
{I_3} = \dfrac{{{R_1}}}{{{R_1} + {R_{23}}}}.I = \dfrac{6}{{6 + 18}}.\dfrac{{18}}{{11}} = \dfrac{9}{{22}}A\\
{U_3} = {I_3}{R_3} = \dfrac{9}{{22}}.10 = \dfrac{{45}}{{11}}V\\
{U_{MN}} =  - {U_4} - {U_3} =  - \dfrac{{108}}{{11}} - \dfrac{{45}}{{11}} =  - \dfrac{{153}}{{11}}V
\end{array}\)