Đáp án:

\(\begin{array}{l}
a.\\
{U_{dm}} = 6V\\
{P_{dm}} = 6W\\
{R_{MC}} = 3\Omega \\
b.\\
{R_{MN}} = 9\Omega \\
c.\\
{U_V} = 15 - \frac{{{U_V}}}{6}{R_X}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{U_{dm}} = {U_V} = 6V\\
I = {I_A} = 1V\\
{P_{dm}} = {U_V}I = 6.1 = 6W\\
{U_{MC}} = U - {U_V} = 9 - 6 = 3V\\
{R_{MC}} = \frac{{{U_{MC}}}}{I} = \frac{3}{1} = 3\Omega \\
b.\\
{R_d} = \frac{{U_{dm}^2}}{{{P_{dm}}}} = \frac{{{6^2}}}{6} = 6\Omega \\
{I_d} = I = \frac{{{U_V}}}{{{R_d}}} = \frac{{3,6}}{6} = 0,6A\\
{U_{MN}} = U - {U_V} = 9 - 3,6 = 5,4V\\
{R_{MN}} = \frac{{{U_{MN}}}}{I} = \frac{{5,4}}{{0,6}} = 9\Omega \\
c.\\
I = \frac{{{U_V}}}{{{R_d}}} = \frac{{{U_V}}}{6}\\
{U_V} = U - {U_X} = 15 - {R_X}I = 15 - \frac{{{U_V}}}{6}{R_X}\\
{U_V} = 15 - \frac{{{U_V}}}{6}{R_X}
\end{array}\)