Đáp án:

 B

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{12{R_3}}}{{12 + {R_3}}}\\
R = {R_1} + {R_{23}} = 3 + \dfrac{{12{R_3}}}{{12 + {R_3}}} = \dfrac{{15{R_3} + 36}}{{12 + {R_3}}}\\
I = \dfrac{U}{R} = \dfrac{{U\left( {12 + {R_3}} \right)}}{{15{R_3} + 36}}\\
{I_3} = \dfrac{{{R_2}}}{{{R_2} + {R_3}}}.I = \dfrac{{12}}{{12 + {R_3}}}.\dfrac{{U\left( {12 + {R_3}} \right)}}{{15{R_3} + 36}} = \dfrac{{12U}}{{15{R_3} + 36}}
\end{array}\)

Công suất tiêu thụ trên R3 là:

\(\begin{array}{l}
{P_3} = I_3^2{R_3} = \dfrac{{{{12}^2}{U^2}}}{{{{\left( {15{R_3} + 36} \right)}^2}}}.{R_3} = \dfrac{{144{U^2}}}{{{{\left( {15\sqrt {{R_3}}  + \dfrac{{36}}{{\sqrt {{R_3}} }}} \right)}^2}}}\\
 \Rightarrow {P_3} \le \dfrac{{144{U^2}}}{{4.15\sqrt {{R_3}} .\dfrac{{36}}{{\sqrt {{R_3}} }}}} = \dfrac{{{U^2}}}{{15}}
\end{array}\)

Dấu = xảy ra khi:

\(15\sqrt {{R_3}}  = \dfrac{{36}}{{\sqrt {{R_3}} }} \Rightarrow {R_3} = 2,4\Omega \)