Đáp án:

\(\begin{array}{l}
12.\\
a.\\
{U_1} = 4,8V\\
{U_2} = 7,2V\\
13.\\
a.\\
{R_d} = 24\Omega \\
{R_{td}} = 40\Omega 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
12.\\
a.\\
R = {R_1} + {R_2} = 10 + 15 = 25\Omega \\
I = \dfrac{U}{R} = \dfrac{{12}}{{25}} = 0,48A\\
{U_1} = {\rm{I}}{{\rm{R}}_1} = 0,48.10 = 4,8V\\
{U_2} = U - {U_1} = 12 - 4,8 = 7,2V\\
b.\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{3} = 12\Omega \\
R' = {R_2} + {R_d} = 12 + 15 = 27\Omega \\
I' = \dfrac{U}{{R'}} = \dfrac{{12}}{{27}} = 0,44A\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{3}{6} = 0,5A\\
I' < {I_{dm}}
\end{array}\)

Suy ra đèn sáng yếu.

\(\begin{array}{l}
13.\\
a.\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{{12}^2}}}{6} = 24\Omega \\
{R_{td}} = {R_d} + R = 24 + 16 = 40\Omega \\
b.\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{18}}{{40}} = 0,45A\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{6}{{12}} = 0,5A\\
I < {I_{dm}}
\end{array}\)

Suy ra đèn sáng yếu