Đáp án:

 \(\begin{array}{l}
a.{R_{MN}} = 48\Omega \\
b.I = \dfrac{2}{3}A\\
c.{R_c} = 6\Omega 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{R_{MN}} = p\dfrac{l}{S} = 1,{2.10^{ - 6}}\dfrac{{20}}{{0,{{5.10}^{ - 6}}}} = 48\Omega \\
b.\\
{R_c} = \dfrac{{{R_{MN}}}}{2} = \dfrac{{48}}{2} = 24\Omega \\
{R_{d1}} = {R_{d2}} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{3} = 12\Omega \\
{R_{2d}} = \dfrac{{{R_{d1}}{R_{d2}}}}{{{R_{d1}} + {R_{d2}}}} = \dfrac{{12.12}}{{12.12}} = 6\Omega \\
R = {R_c} + {R_{2d}} = 12 + 6 = 18\Omega \\
I = \dfrac{U}{R} = \dfrac{{12}}{{18}} = \dfrac{2}{3}A\\
c.\\
{U_{2d}} = {U_{dm}} = 6V\\
{U_c} = U - {U_{2d}} = 12 - 6 = 6V\\
{I_{d1}} = {I_{d2}} = {I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{3}{6} = 0,5A\\
{I_c} = I = {I_{d1}} + {I_{d2}} = 0,5 + 0,5 = 1A\\
{R_c} = \dfrac{{{U_c}}}{{{I_c}}} = \dfrac{6}{1} = 6\Omega 
\end{array}\)