Đáp án:
B
Giải thích các bước giải:
\(\begin{array}{l}
{R_{2R}} = \frac{{{R_2}R}}{{{R_2} + R}} = \frac{{10R}}{{10 + R}}\\
I = \frac{E}{{{R_1} + {R_{2R}} + r}} = \frac{E}{{5 + {R_{2R}} + 5}} = \frac{E}{{10 + {R_{2R}}}}\\
{P_R} = \frac{{{U_{2R}}^2}}{R} = \frac{{{{(I{R_{2R}})}^2}}}{R} = \frac{{{{(\frac{E}{{10 + {R_{2R}}}}{R_{2R}})}^2}}}{R} = \frac{{{{(\frac{E}{{10 + \frac{{10R}}{{10 + R}}}}.\frac{{10R}}{{10 + R}})}^2}}}{R}\\
{P_R} = \frac{{{E^2}}}{{{{(\frac{{10}}{{\sqrt R }} + 2\sqrt R )}^2}}}\\
{P_{R\max }} \Rightarrow {(\frac{{10}}{{\sqrt R }} + 2\sqrt R )^2}\min \\
khi:\frac{{10}}{{\sqrt R }} = 2\sqrt R \\
R = 5\Omega
\end{array}\)
