Đáp án:
\(\begin{array}{l}
a.R = 3\Omega \\
b.I = 6A\\
c.{U_{MN}} = 1,5V
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = {R_1} + {R_2} = 1 + 3 = 4\Omega \\
{R_{34}} = {R_3} + {R_4} = 4 + 8 = 12\Omega \\
R = \dfrac{{{R_{12}}{R_{34}}}}{{{R_{12}} + {R_{34}}}} = \dfrac{{4.12}}{{4 + 12}} = 3\Omega \\
b.\\
I = \dfrac{E}{{R + r}} = \dfrac{{24}}{{3 + 1}} = 6A\\
c.\\
{U_{12}} = U = {\rm{IR}} = 6.3 = 18V\\
{I_1} = {I_{12}} = \dfrac{{{U_{12}}}}{{{R_{12}}}} = \dfrac{{18}}{4} = 4,5A\\
{I_3} = {I_{34}} = I - {I_{12}} = 6 - 4,5 = 1,5A\\
{U_{MN}} = {U_{MA}} + {U_{AN}} = - {U_1} + {U_3} = - {I_1}{R_1} + {I_3}{R_3} = - 4,5.1 + 1,5.4 = 1,5V
\end{array}\)
