Đáp án:
\(\begin{array}{l}
a.I = 1,8A\\
b.{Q_1} = 5184J\\
c.{R_b} = 14\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{{12}^2}}}{6} = 24\Omega \\
{R_{1d}} = \dfrac{{{R_1}{R_d}}}{{{R_1} + {R_d}}} = \dfrac{{12.24}}{{12 + 24}} = 8\Omega \\
R = {R_b} + {R_{1d}} = 10 + 8 = 18\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{36}}{{18 + 2}} = 1,8A\\
b.\\
{U_d} = {\rm{I}}{{\rm{R}}_{1d}} = 1,8.8 = 14,4V\\
{I_d} = \dfrac{{{U_d}}}{{{R_d}}} = \dfrac{{14,4}}{{24}} = 0,6A\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{6}{{12}} = 0,5A\\
{I_d} > {I_{dm}}
\end{array}\)
Suy ra đèn sáng mạnh.
\({Q_1} = \dfrac{{U_1^2}}{{{R_1}}}t = \dfrac{{14,{4^2}}}{{12}}.300 = 5184J\)
\(\begin{array}{l}
c.\\
{U_{1d}} = {U_d} = 12V\\
I = \dfrac{{{U_{1d}}}}{{{R_{1d}}}} = \dfrac{{12}}{8} = 1,5A\\
R = {R_b} + {R_{1d}} = {R_b} + 8\\
I = \dfrac{E}{{R + r}}\\
\Rightarrow 1,5 = \dfrac{{36}}{{{R_b} + 8 + 2}}\\
\Rightarrow {R_b} = 14\Omega
\end{array}\)
