Đáp án:
$\begin{array}{l}
a.{R_x} = 24\Omega \\
b.{R_x} = 3\Omega \\
{P_{{x_{\max }}}} = 3,8W
\end{array}$
Giải thích các bước giải:
a. Khi đèn sáng bình thường:
Điện trở Rx là:
$\begin{array}{l}
{U_x} = {U_D} = 6V\\
\Rightarrow U = {U_{AB}} - {U_D} = 9 - 6 = 3V\\
\Leftrightarrow I = \dfrac{U}{R} = \dfrac{3}{4} = 0,75A\\
{I_D} = \dfrac{{{P_D}}}{{{U_D}}} = \dfrac{3}{6} = 0,5A\\
\Rightarrow {I_x} = I - {I_D} = 0,75 - 0,5 = 0,25A\\
\Rightarrow {R_x} = \dfrac{{{U_x}}}{{{I_x}}} = \dfrac{6}{{0,25}} = 24\Omega
\end{array}$
b. Ta có:
$\begin{array}{l}
{R_D} = \dfrac{{{U_D}}}{{{I_D}}} = \dfrac{6}{{0,5}} = 12\Omega \\
{R_{td}} = \dfrac{{{R_D}.{R_x}}}{{{R_D} + {R_x}}} + R = \dfrac{{12x}}{{12 + x}} + 4 = \dfrac{{16x + 48}}{{12 + x}}\\
\Rightarrow {I_m} = \dfrac{{{U_{AB}}}}{{{R_{td}}}} = \dfrac{{9\left( {12 + x} \right)}}{{16x + 48}}\\
\Rightarrow {I_x} = \dfrac{{{R_D}}}{{{R_x} + {R_D}}}.{I_m} = \dfrac{{12}}{{12 + x}}.\dfrac{{9\left( {12 + x} \right)}}{{16x + 48}} = \dfrac{{27}}{{4x + 12}}\\
\Rightarrow {P_x} = {I_x}^2.{R_x} = \dfrac{{{{27}^2}x}}{{{{\left( {4x + 12} \right)}^2}}} = \dfrac{{729}}{{{{\left( {4\sqrt x + \dfrac{{12}}{{\sqrt x }}} \right)}^2}}}\\
{P_{{x_{\max }}}}khi\\
4\sqrt x = \dfrac{{12}}{{\sqrt x }} \Leftrightarrow x = 3\Omega \Rightarrow {R_x} = 3\Omega \\
{p_{{x_{\max }}}} = \dfrac{{729}}{{{{\left( {4\sqrt 3 + \dfrac{{12}}{{\sqrt 3 }}} \right)}^2}}} = 3,8W
\end{array}$
