Đáp án:
\(\begin{array}{l}
a.R = 19,5\Omega \\
b.{Q_1} = \dfrac{{16}}{{15}}J\\
c.{P_d} = 5,12W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{{12}^2}}}{8} = 18\Omega \\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{3.3}}{{3 + 3}} = 1,5\Omega \\
R = {R_{12}} + {R_d} = 18 + 1,5 = 19,5\Omega \\
b.\\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{19,5 + 3}} = \dfrac{8}{{15}}A\\
{U_1} = {U_{12}} = {\rm{I}}{{\rm{R}}_{12}} = \dfrac{8}{{15}}.1,5 = 0,8V\\
{Q_1} = \dfrac{{U_1^2}}{{{R_1}}}t = \dfrac{{0,{8^2}}}{3}.5 = \dfrac{{16}}{{15}}J\\
c.\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{8}{{12}} = \dfrac{2}{3}A\\
{I_d} = I = \dfrac{8}{{15}}A\\
{I_d} < {I_{dm}}
\end{array}\)
Suy ra đèn sáng yếu.
\({P_d} = {R_d}I_d^2 = 18.{\dfrac{8}{{15}}^2} = 5,12W\)
