Đáp án:

\({R_4} = 4\Omega \)

Giải thích các bước giải:

Mạch gồm: (R1 // R2) nt (R3 // R4)

Ta có:

\(\begin{array}{l}
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{15.10}}{{15 + 10}} = 6\Omega \\
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = \dfrac{{12{R_4}}}{{12 + {R_4}}}\\
R = {R_{12}} + {R_{34}} = 6 + \dfrac{{12{R_4}}}{{12 + {R_4}}} = \dfrac{{18{R_4} + 72}}{{12 + {R_4}}}\\
I = \dfrac{U}{R} = \dfrac{{12\left( {12 + {R_4}} \right)}}{{18{R_4} + 72}}\\
{I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}.I = \dfrac{{10}}{{25}}.\dfrac{{12\left( {12 + {R_4}} \right)}}{{18{R_4} + 72}} = \dfrac{{4,8\left( {12 + {R_4}} \right)}}{{18{R_4} + 72}}\\
{I_3} = \dfrac{{{R_4}}}{{{R_3} + {R_4}}}.I = \dfrac{{{R_4}}}{{12 + {R_4}}}.\dfrac{{12\left( {12 + {R_4}} \right)}}{{18{R_4} + 72}} = \dfrac{{12{R_4}}}{{18{R_4} + 72}}\\
{I_A} = {I_1} - {I_3} \Rightarrow \dfrac{{4,8\left( {12 + {R_4}} \right)}}{{18{R_4} + 72}} - \dfrac{{12{R_4}}}{{18{R_4} + 72}} = 0,2\\
 \Rightarrow {R_4} = 4\Omega 
\end{array}\)