Đáp án:

1.

a) \(3,62W\)

b) \(x = 9\Omega \)

2. \(6\Omega \)

Giải thích các bước giải:

1. K ngắt, mạch gồm: \(\left( {{R_d}\parallel \left( {{R_{BC}}\,nt\,{R_2}} \right)} \right)nt\,{R_1}\)

a) Ta có:

\(\begin{array}{l}
{R_d} = \dfrac{{{3^2}}}{3} = 3\Omega \\
{R_{BC}} = {R_x} - {R_{AC}} = 12 - 2 = 10\Omega \\
{R_{NB}} = {R_2} + {R_{BC}} = 3 + 10 = 13\Omega \\
{R_{dNB}} = \dfrac{{{R_d}.{R_{NB}}}}{{{R_d} + {R_{NB}}}} = \dfrac{{3.13}}{{3 + 13}} = \dfrac{{39}}{{16}}\Omega \\
{R_{td}} = {R_1} + {R_{dNB}} = 2 + \dfrac{{39}}{{16}} = \dfrac{{71}}{{16}}\Omega \\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{6}{{\dfrac{{71}}{{16}}}} = \dfrac{{96}}{{71}}A\\
{I_d} = \dfrac{{{R_{NB}}}}{{{R_d} + {R_{NB}}}}.I = \dfrac{{13}}{{3 + 13}}.\dfrac{{96}}{{71}} = \dfrac{{78}}{{71}}A\\
{P_d} = I_d^2.{R_d} = 3,62W
\end{array}\)

b) Ta có:

\(\begin{array}{l}
{R_{BC}} = {R_x} - {R_{AC}} = 12 - x\\
{R_{NB}} = {R_2} + {R_{BC}} = 3 + 12 - x = 15 - x\\
{R_{dNB}} = \dfrac{{{R_d}.{R_{NB}}}}{{{R_d} + {R_{NB}}}} = \dfrac{{3\left( {15 - x} \right)}}{{18 - x}} = \dfrac{{45 - 3x}}{{18 - x}}\\
{R_{td}} = {R_1} + {R_{dNB}} = 2 + \dfrac{{45 - 3x}}{{18 - x}} = \dfrac{{81 - 5x}}{{18 - x}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{6}{{\dfrac{{81 - 5x}}{{18 - x}}}} = \dfrac{{6\left( {18 - x} \right)}}{{81 - 5x}}\\
{I_d} = \dfrac{{{R_{NB}}}}{{{R_d} + {R_{NB}}}}.I = \dfrac{{15 - x}}{{18 - x}}.\dfrac{{6\left( {18 - x} \right)}}{{81 - 5x}} = \dfrac{{6\left( {15 - x} \right)}}{{81 - 5x}}
\end{array}\)

Để đèn sáng bình thường:

\({I_d} = \dfrac{3}{3} = 1A \Rightarrow \dfrac{{6\left( {15 - x} \right)}}{{81 - 5x}} = 1 \Rightarrow x = 9\Omega \)

2. K đóng, mạch gồm: \(\left\{ {{R_d}\parallel \left( {\left( {{R_{AC}}\parallel {R_{BC}}} \right)nt\,{R_2}} \right)} \right\}nt\,{R_1}\)

Ta có:

\(\begin{array}{l}
{R_{AB}} = \dfrac{{{R_{AC}}.{R_{BC}}}}{{{R_{AC}} + {R_{BC}}}} = \dfrac{{x\left( {12 - x} \right)}}{{12}} = \dfrac{{ - {x^2} + 12x}}{{12}}\\
{R_{2AB}} = {R_{AB}} + {R_2} = \dfrac{{ - {x^2} + 12x}}{{12}} + 3 = \dfrac{{ - {x^2} + 12x + 36}}{{12}}\\
{R_{d2AB}} = \dfrac{{{R_d}.{R_{2AB}}}}{{{R_d} + {R_{2AB}}}} = \dfrac{{3.\dfrac{{ - {x^2} + 12x + 36}}{{12}}}}{{3 + \dfrac{{ - {x^2} + 12x + 36}}{{12}}}} = \dfrac{{3\left( { - {x^2} + 12x + 36} \right)}}{{ - {x^2} + 12x + 72}}\\
{R_{td}} = {R_1} + {R_{d2AB}} = 2 + \dfrac{{3\left( { - {x^2} + 12x + 36} \right)}}{{ - {x^2} + 12x + 72}} = \dfrac{{ - 5{x^2} + 60x + 252}}{{ - {x^2} + 12x + 72}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{6\left( { - {x^2} + 12x + 72} \right)}}{{ - 5{x^2} + 60x + 252}}\\
{I_2} = \dfrac{{{R_d}}}{{{R_d} + {R_{2AB}}}}.I = \dfrac{{3.12}}{{ - {x^2} + 12x + 72}}.\dfrac{{6\left( { - {x^2} + 12x + 72} \right)}}{{ - 5{x^2} + 60x + 252}} = \dfrac{{216}}{{ - 5{x^2} + 60x + 252}}\\
{P_2} = I_2^2.{R_2} = 0,75 \Rightarrow {I_2} = 0,5A\\
 \Rightarrow \dfrac{{216}}{{ - 5{x^2} + 60x + 252}} = 0,5\\
 \Rightarrow x = 6\Omega 
\end{array}\)

\(\begin{array}{l}
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{6\left( { - {x^2} + 12x + 72} \right)}}{{ - 5{x^2} + 60x + 252}} = 1,5A\\
{I_{d2AB}} = \dfrac{{{R_d}}}{{{R_d} + {R_{2AB}}}}.I = \dfrac{{216}}{{ - 5{x^2} + 60x + 252}} = 0,5A\\
{I_{AC}} = \dfrac{{{R_{BC}}}}{{{R_{AB}}}}.{I_{d2AB}} = \dfrac{{{I_{d2AB}}}}{2} = \dfrac{{0,5}}{2} = 0,25A\\
{I_A} = I - {I_{AC}} = 1,5 - 0,25 = 1,25A
\end{array}\)