Ai giúp mih câu 1 phần 2 vs cảm ơn trc nha

Đáp án:

 \({R_1} = 1\Omega \)

Giải thích các bước giải:

a) K đóng, mạch gồm: R1 nt (R2 // R4) nt (R3 // R5)

Ta có: 

\(\begin{array}{l}
R = {R_1} + \dfrac{{{R_2}{R_4}}}{{{R_2} + {R_4}}} + \dfrac{{{R_3}{R_5}}}{{{R_3} + {R_5}}}\\
 \Rightarrow R = {R_1} + \dfrac{{12.24}}{{12 + 24}} + \dfrac{{8.24}}{{8 + 24}} = 14 + {R_1}\\
I = \dfrac{U}{R} = \dfrac{{36}}{{14 + {R_1}}}\\
{I_2} = \dfrac{{{R_4}}}{{{R_2} + {R_4}}}.I = \dfrac{{24}}{{12 + 24}}.\dfrac{{36}}{{14 + {R_1}}} = \dfrac{{24}}{{14 + {R_1}}}\\
{I_3} = \dfrac{{{R_5}}}{{{R_3} + {R_5}}}.I = \dfrac{8}{{8 + 24}}.\dfrac{{36}}{{14 + {R_1}}} = \dfrac{9}{{14 + {R_1}}}\\
{I_A} = {I_2} - {I_3} = \dfrac{{24}}{{14 + {R_1}}} - \dfrac{9}{{14 + {R_1}}} = \dfrac{{15}}{{14 + {R_1}}}\\
 \Rightarrow 1 = \dfrac{{15}}{{14 + {R_1}}} \Rightarrow {R_1} = 1\Omega 
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
\dfrac{1}{{{R_{35x}}}} = \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_5}}} + \dfrac{1}{{{R_x}}} = \dfrac{1}{{24}} + \dfrac{1}{8} + \dfrac{1}{{{R_x}}} = \dfrac{1}{6} + \dfrac{1}{{{R_x}}}\\
 \Rightarrow \dfrac{1}{{{R_{35x}}}} = \dfrac{{{R_x} + 6}}{{6{R_x}}} \Rightarrow {R_{35x}} = \dfrac{{6{R_x}}}{{{R_x} + 6}}\\
R = {R_1} + {R_{24}} + {R_{35x}} = 8 + 8 + \dfrac{{6{R_x}}}{{{R_x} + 6}}\\
 \Rightarrow R = \dfrac{{22{R_x} + 96}}{{{R_x} + 6}}\\
I = \dfrac{U}{R} = \dfrac{{36\left( {{R_x} + 6} \right)}}{{22{R_x} + 96}}\\
{I_2} = \dfrac{{{R_4}}}{{{R_2} + {R_4}}}.I = \dfrac{{24}}{{12 + 24}}.\dfrac{{36\left( {{R_x} + 6} \right)}}{{22{R_x} + 96}} = \dfrac{{24\left( {{R_x} + 6} \right)}}{{22{R_x} + 96}}\\
{U_{35x}} = I.{R_{35x}} = \dfrac{{36\left( {{R_x} + 6} \right)}}{{22{R_x} + 96}}.\dfrac{{6{R_x}}}{{{R_x} + 6}} = \dfrac{{216{R_x}}}{{22{R_x} + 96}}\\
{I_3} = \dfrac{{{U_{35x}}}}{{{R_3}}} = \dfrac{{216{R_x}}}{{24\left( {22{R_x} + 96} \right)}} = \dfrac{{9{R_x}}}{{22{R_x} + 96}}\\
{I_A} =  - {I_3} + {I_2} =  - \dfrac{{9{R_x}}}{{22{R_x} + 96}} + \dfrac{{24\left( {{R_x} + 6} \right)}}{{22{R_x} + 96}}\\
 \Rightarrow 0,9 =  - \dfrac{{9{R_x}}}{{22{R_x} + 96}} + \dfrac{{24\left( {{R_x} + 6} \right)}}{{22{R_x} + 96}}\\
 \Rightarrow {R_x} = 12\Omega 
\end{array}\)