Đáp án:
\(\begin{array}{l}
a.I = 1,5A\\
b.{U_3} = 6V\\
c.{U_{MN}} = - 6V
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = {R_1} + {R_2} = 6 + 6 = 12\Omega \\
{R_{123}} = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{12.6}}{{12 + 6}} = 4\Omega \\
\Rightarrow R = {R_{123}} + {R_4} = 4 + 2 = 6\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{6 + 2}} = 1,5A\\
b.\\
U = {\rm{IR}} = 1,5.6 = 9V\\
{U_4} = {I_4}{R_4} = 1,5.2 = 3V\\
{U_3} = {U_{12}} = {U_{123}} = U - {U_4} = 9 - 3 = 6V\\
c.\\
{I_1} = {I_2} = \dfrac{{{U_{12}}}}{{{R_{12}}}} = \dfrac{6}{{12}} = 0,5A\\
{U_2} = {I_2}{R_2} = 0,5.6 = 3V\\
{U_{MN}} = - {U_{NM}} = - ({U_2} + {U_4}) = - (3 + 3) = - 6V
\end{array}\)
