giúp mik bài này với ạ chiều mik thi r

Đáp án:

a) 20

b) 10

Giải thích các bước giải:

a) Mạch gồm: (R1 // R2) nt (R3 // R4)

Điện trở tương đương là:

\(\begin{array}{l}
R = {R_{12}} + {R_{34}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} + \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}}\\
 \Rightarrow R = \dfrac{{20{R_1}}}{{{R_1} + 20}} + \dfrac{{20{R_4}}}{{{R_4} + 20}}\\
 \Rightarrow R = \dfrac{{20{R_1}\left( {{R_4} + 20} \right) + 20{R_4}\left( {{R_1} + 20} \right)}}{{\left( {{R_1} + 20} \right)\left( {{R_4} + 20} \right)}}\\
 \Rightarrow R = \dfrac{{40{R_1}{R_4} + 400\left( {{R_1} + {R_4}} \right)}}{{{R_1}{R_4} + 20\left( {{R_1} + {R_4}} \right) + 400}}\\
 \Rightarrow R = \dfrac{{40.400 + 400\left( {{R_1} + {R_4}} \right)}}{{400 + 20\left( {{R_1} + {R_4}} \right) + 400}}\\
 \Rightarrow R = \dfrac{{800 + 20\left( {{R_1} + {R_4}} \right)}}{{40 + \left( {{R_1} + {R_4}} \right)}} = 20\Omega 
\end{array}\)

b) Mạch gồm: (R3 // R2) nt (R1 // R4)

Ta có:

\(\begin{array}{l}
{R_{32}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{20.20}}{{20 + 20}} = 10\Omega \\
{R_{14}} = \dfrac{{{R_1}{R_4}}}{{{R_1} + {R_4}}} = \dfrac{{20.20}}{{{R_1} + \dfrac{{400}}{{{R_1}}}}} = \dfrac{{400{R_1}}}{{R_1^2 + 400}}\\
R = {R_{32}} + {R_{14}} = 10 + \dfrac{{400{R_1}}}{{R_1^2 + 400}} = \dfrac{{10R_1^2 + 400{R_1} + 4000}}{{R_1^2 + 400}}\\
I = \dfrac{U}{R} = \dfrac{{18\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}}\\
{I_3} = \dfrac{{{R_2}}}{{{R_2} + {R_3}}}.I = \dfrac{1}{2}.\dfrac{{18\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}} = \dfrac{{9\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}}\\
{I_1} = \dfrac{{{R_4}}}{{{R_1} + {R_4}}}.I = \dfrac{{\dfrac{{400}}{{{R_1}}}}}{{{R_1} + \dfrac{{400}}{{{R_1}}}}}.\dfrac{{18\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}}\\
 \Rightarrow {I_1} = \dfrac{{400}}{{R_1^2 + 400}}.\dfrac{{18\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}} = \dfrac{{7200}}{{10R_1^2 + 400{R_1} + 4000}}\\
{I_A} = {I_1} - {I_3} = \dfrac{{7200}}{{10R_1^2 + 400{R_1} + 4000}} - \dfrac{{9\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}}\\
 \Rightarrow 0,3 = \dfrac{{7200 - 9\left( {R_1^2 + 400} \right)}}{{10R_1^2 + 400{R_1} + 4000}}\\
 \Rightarrow 0,3\left( {10R_1^2 + 400{R_1} + 4000} \right) = 7200 - 9\left( {R_1^2 + 400} \right)\\
 \Rightarrow {R_1} = 10\Omega 
\end{array}\)