Giúp mình giải bài 5

Đáp án:

\({R_2} = 12\Omega \)

Giải thích các bước giải:

Ta có: 

\(\begin{array}{l}
{R_{2x}} = \dfrac{{{R_2}{R_x}}}{{{R_2} + {R_x}}}\\
R = {R_1} + {R_{2x}} = 6 + \dfrac{{{R_2}{R_x}}}{{{R_2} + {R_x}}} = \dfrac{{6{R_2} + 6{R_x} + {R_2}{R_x}}}{{{R_2} + {R_x}}}\\
I = \dfrac{U}{R} = \dfrac{{12\left( {{R_2} + {R_x}} \right)}}{{6{R_2} + 6{R_x} + {R_2}{R_x}}}\\
{I_x} = \dfrac{{{R_2}}}{{{R_2} + {R_x}}}.\dfrac{{12\left( {{R_2} + {R_x}} \right)}}{{6{R_2} + 6{R_x} + {R_2}{R_x}}}\\
 \Rightarrow {I_x} = \dfrac{{12{R_2}}}{{6{R_2} + 6{R_x} + {R_2}{R_x}}}\\
{P_x} = I_x^2{R_x} = \dfrac{{144R_2^2{R_x}}}{{{{\left( {6{R_2} + 6{R_x} + {R_2}{R_x}} \right)}^2}}}\\
 \Rightarrow {P_x} = \dfrac{{144R_2^2{R_x}}}{{{{\left( {6{R_2} + {R_x}\left( {6 + {R_2}} \right)} \right)}^2}}}\\
{\left( {6{R_2} + {R_x}\left( {6 + {R_2}} \right)} \right)^2} \ge 4.6{R_2}.{R_x}\left( {6 + {R_2}} \right)\\
 \Rightarrow {\left( {6{R_2} + {R_x}\left( {6 + {R_2}} \right)} \right)^2} \ge 24{R_2}{R_x}\left( {6 + {R_2}} \right)\\
 \Rightarrow {P_x} \le \dfrac{{144R_2^2{R_x}}}{{24{R_2}{R_x}\left( {6 + {R_2}} \right)}}\\
 \Rightarrow {P_x} \le \dfrac{{6{R_2}}}{{6 + {R_2}}} = 4\\
 \Rightarrow {R_2} = 12\Omega 
\end{array}\)