Đáp án:
\(\dfrac{{24}}{7}\Omega \)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{R_{02}} = \dfrac{{R{R_2}}}{{R + {R_2}}} = \dfrac{{6R}}{{R + 6}}\\
R = {R_1} + {R_{02}} = 2 + \dfrac{{6R}}{{R + 6}} = \dfrac{{8R + 12}}{{R + 6}}\\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{\dfrac{{8R + 12}}{{R + 6}} + 6}} = \dfrac{{12\left( {R + 6} \right)}}{{14R + 48}}\\
{I_R} = \dfrac{{{R_2}}}{{R + {R_2}}}.I = \dfrac{6}{{6 + R}}.\dfrac{{12\left( {R + 6} \right)}}{{14R + 48}} = \dfrac{{72}}{{14R + 48}}\\
{P_R} = I_R^2R = \dfrac{{{{72}^2}.R}}{{{{\left( {14R + 48} \right)}^2}}} = \dfrac{{{{72}^2}}}{{{{\left( {14\sqrt R + \dfrac{{48}}{{\sqrt R }}} \right)}^2}}} \le \dfrac{{27}}{{14}}
\end{array}\)
Dấu = xảy ra khi: \(R = \dfrac{{48}}{{14}} = \dfrac{{24}}{7}\Omega \)
