Đáp án:
\({R_x} = 4\Omega \)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{R_{td}} = \dfrac{{{R_2}\left( {{R_3} + {R_x}} \right)}}{{{R_2} + {R_3} + {R_x}}} + {R_1} = \dfrac{{6\left( {8 + {R_x}} \right)}}{{14 + {R_x}}} + 4 = \dfrac{{10{R_x} + 104}}{{14 + {R_x}}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{12\left( {14 + {R_x}} \right)}}{{10{R_x} + 104}} = \dfrac{{6\left( {14 + {R_x}} \right)}}{{5{R_x} + 52}}\\
{I_x} = \dfrac{{{R_2}}}{{{R_2} + {R_3} + {R_x}}}.I = \dfrac{6}{{14 + {R_x}}}.\dfrac{{6\left( {14 + {R_x}} \right)}}{{5{R_x} + 52}} = \dfrac{{36}}{{5{R_x} + 52}}\\
{P_x} = I_x^2{R_x} \Rightarrow {\left( {\dfrac{{36}}{{5{R_x} + 52}}} \right)^2}.{R_x} = 1\\
\Rightarrow {R_x} = 4\Omega
\end{array}\)
