Đáp án:

1. 

a) \(4\Omega \)

b) \(2,5A\)

2. \({R_x} = 9\Omega \)

Giải thích các bước giải:

1. K mở, mạch gồm: R5 nt ((R1 nt R3) // (R2 nt R4))

a) Ta có:

\(\begin{array}{l}
{R_{13}} = {R_1} + {R_3} = 3 + 1 = 4\Omega \\
{R_{24}} = {R_2} + {R_4} = 2 + 2 = 4\Omega \\
{R_{1234}} = \dfrac{{{R_{13}}.{R_{24}}}}{{{R_{13}} + {R_{24}}}} = \dfrac{{4.4}}{{4 + 4}} = 2\Omega \\
{R_{td}} = {R_5} + {R_{1234}} = 2 + 2 = 4\Omega 
\end{array}\)

b) Ta có:

\(\begin{array}{l}
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{20}}{4} = 5A\\
{I_A} = \dfrac{{{R_{13}}}}{{{R_{13}} + {R_{24}}}}.I = \dfrac{I}{2} = 2,5A
\end{array}\)

2. K mở, mạch gồm: R5 nt ((R1 nt R3) // (Rx nt Ry))

\(\begin{array}{l}
{R_{13}} = {R_1} + {R_3} = 3 + 1 = 4\Omega \\
{R_{xy}} = {R_x} + {R_y}\\
{R_{13xy}} = \dfrac{{{R_{13}}.{R_{xy}}}}{{{R_{13}} + {R_{xy}}}} = \dfrac{{4\left( {{R_x} + {R_y}} \right)}}{{4 + {R_x} + {R_y}}}\\
{R_{td}} = {R_5} + {R_{1234}} = 2 + \dfrac{{4\left( {{R_x} + {R_y}} \right)}}{{4 + {R_x} + {R_y}}}\\
 \Rightarrow {R_{td}} = \dfrac{{6\left( {{R_x} + {R_y}} \right) + 8}}{{4 + {R_x} + {R_y}}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{20\left( {4 + {R_x} + {R_y}} \right)}}{{6\left( {{R_x} + {R_y}} \right) + 8}}\\
{I_A} = \dfrac{{{R_{13}}}}{{{R_{13}} + {R_{xy}}}}.I = \dfrac{4}{{4 + {R_x} + {R_y}}}.\dfrac{{20\left( {4 + {R_x} + {R_y}} \right)}}{{6\left( {{R_x} + {R_y}} \right) + 8}}\\
 \Rightarrow {I_A} = \dfrac{{80}}{{6\left( {{R_x} + {R_y}} \right) + 8}} = 1\\
 \Rightarrow {R_x} + {R_y} = 12\Omega 
\end{array}\)

* K đóng, mạch gồm: R5 nt (R1 // Rx) nt (R3 // Ry)

\(\begin{array}{l}
{R_{1x}} = \dfrac{{{R_1}{R_x}}}{{{R_1} + {R_x}}} = \dfrac{{3{R_x}}}{{{R_x} + 3}}\\
{R_{3y}} = \dfrac{{{R_3}{R_y}}}{{{R_3} + {R_y}}} = \dfrac{{12 - {R_x}}}{{1 + 12 - {R_x}}} = \dfrac{{12 - {R_x}}}{{13 - {R_x}}}\\
R = {R_5} + {R_{1x}} + {R_{3y}} = 2 + \dfrac{{3{R_x}}}{{{R_x} + 3}} + \dfrac{{12 - {R_x}}}{{13 - {R_x}}}\\
 \Rightarrow R = \dfrac{{2\left( {{R_x} + 3} \right)\left( {13 - {R_x}} \right) + 3{R_x}\left( {13 - {R_x}} \right) + \left( {12 - {R_x}} \right)\left( {{R_x} + 3} \right)}}{{\left( {{R_x} + 3} \right)\left( {13 - {R_x}} \right)}}\\
 \Rightarrow R = \dfrac{{2\left( { - R_x^2 + 10{R_x} + 39} \right) + 3\left( { - R_x^2 + 13{R_x}} \right) + \left( { - R_x^2 + 9{R_x} + 36} \right)}}{{\left( {{R_x} + 3} \right)\left( {13 - {R_x}} \right)}}\\
 \Rightarrow R = \dfrac{{ - 6R_x^2 + 68{R_x} + 114}}{{\left( {{R_x} + 3} \right)\left( {13 - {R_x}} \right)}}\\
I = \dfrac{U}{R} = \dfrac{{20\left( {{R_x} + 3} \right)\left( {13 - {R_x}} \right)}}{{ - 6R_x^2 + 68{R_x} + 114}}\\
{I_A} = {I_x} = \dfrac{{{R_1}}}{{{R_1} + {R_x}}}.I = \dfrac{3}{{{R_x} + 3}}.\dfrac{{20\left( {{R_x} + 3} \right)\left( {13 - {R_x}} \right)}}{{ - 6R_x^2 + 68{R_x} + 114}}\\
 \Rightarrow {I_A} = \dfrac{{60\left( {13 - {R_x}} \right)}}{{ - 6R_x^2 + 68{R_x} + 114}} = 1\\
 \Rightarrow {R_x} = 9\Omega  \Rightarrow {R_y} = 3\Omega 
\end{array}\)