Giúp mình với mình cần gấp cảm ơn ạ

Đáp án:

\({R_{AC}} = 16\Omega \)

Giải thích các bước giải:

Mạch gồm: \(\left( {{R_{AC}}//\left( {\,nt\,{R_{CB}}} \right)} \right)nt\,{R_d}\)

Ta có:

\(\begin{array}{l}
{R_{den}} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{3} = 12\Omega \\
{R_{dBC}} = {R_{den}} + {R_{BC}} = 12 + \left( {20 - {R_{AC}}} \right) = 32 - {R_{AC}}\\
{R_{dAB}} = \dfrac{{{R_{AC}}.{R_{dBC}}}}{{{R_{AC}} + {R_{dBC}}}} = \dfrac{{{R_{AC}}\left( {32 - {R_{AC}}} \right)}}{{32 - {R_{AC}} + {R_{AC}}}}\\
 \Rightarrow {R_{dAB}} = \dfrac{{{R_{AC}}\left( {32 - {R_{AC}}} \right)}}{{32}}\\
R = {R_{dAB}} + {R_d} = \dfrac{{{R_{AC}}\left( {32 - {R_{AC}}} \right)}}{{32}} + 1\\
 \Rightarrow R = \dfrac{{ - R_{AC}^2 + 32{R_{AC}} + 32}}{{32}}\\
I = \dfrac{U}{R} = \dfrac{{288}}{{ - R_{AC}^2 + 32{R_{AC}} + 32}}\\
{I_d} = \dfrac{{{R_{AC}}}}{{{R_{AC}} + {R_{dBC}}}}.I = \dfrac{{{R_{AC}}}}{{32}}.\dfrac{{288}}{{ - R_{AC}^2 + 32{R_{AC}} + 32}}\\
 \Rightarrow {I_d} = \dfrac{{9{R_{AC}}}}{{ - R_{AC}^2 + 32{R_{AC}} + 32}}\\
 \Rightarrow \dfrac{3}{6} = \dfrac{{9{R_{AC}}}}{{ - R_{AC}^2 + 32{R_{AC}} + 32}}\\
 \Rightarrow {R_{AC}} = 16\Omega 
\end{array}\)