Đáp án:

 \(x = \dfrac{{20}}{7}\Omega \)

Giải thích các bước giải:

1. Mạch gồm: ((R1 nt R3) // (((R2 nt R4) // R5) nt Rx)

a) Ta có: 

\(\begin{array}{l}
{R_{13}} = {R_1} + {R_3} = 3 + 3 = 6\Omega \\
{R_{24}} = {R_2} + {R_4} = 3 + 3 = 6\Omega \\
{R_{245}} = \dfrac{{{R_{24}}.{R_5}}}{{{R_{24}} + {R_5}}} = \dfrac{{6.3}}{{6 + 3}} = 2\Omega \\
{R_{245x}} = {R_{245}} + {R_x} = 2 + 1 = 3\Omega \\
R = \dfrac{{{R_{13}}.{R_{245x}}}}{{{R_{13}} + {R_{245x}}}} = \dfrac{{6.3}}{{6 + 3}} = 2\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{5,4}}{{2 + r}}\\
{U_V} = I.R = \dfrac{{5,4}}{{2 + r}}.2 = 3,6\\
 \Rightarrow r = 1\Omega 
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
I = \dfrac{{5,4}}{{2 + r}} = \dfrac{{5,4}}{{2 + 1}} = 1,8A\\
{I_1} = {I_{13}} = \dfrac{{{R_{245x}}}}{{{R_{13}} + {R_{245x}}}}.I = \dfrac{3}{{6 + 3}}.1,8 = 0,6A\\
{U_1} = {I_1}{R_1} = 0,6.3 = 1,8V\\
{I_{245x}} = I - {I_1} = 1,8 - 0,6 = 1,2A\\
{I_2} = \dfrac{{{R_5}}}{{{R_{24}} + {R_5}}}.{I_{245x}} = \dfrac{3}{{6 + 3}}.1,2 = 0,4A\\
{U_2} = {I_2}{R_2} = 0,4.3 = 1,2V\\
{U_C} = {U_1} - {U_2} = 1,8 - 1,2 = 0,6V\\
Q = C{U_C} = 0,01.0,6 = {6.10^{ - 3}}\mu C
\end{array}\)

2. Ta có: 

\(\begin{array}{l}
{R_{13}} = {R_1} + {R_3} = 3 + 3 = 6\Omega \\
{R_{24}} = {R_2} + {R_4} = 3 + 3 = 6\Omega \\
{R_{245}} = \dfrac{{{R_{24}}.{R_5}}}{{{R_{24}} + {R_5}}} = \dfrac{{6.3}}{{6 + 3}} = 2\Omega \\
{R_{245x}} = {R_{245}} + {R_x} = 2 + x\\
R = \dfrac{{{R_{13}}.{R_{245x}}}}{{{R_{13}} + {R_{245x}}}} = \dfrac{{6.\left( {2 + x} \right)}}{{6 + 2 + x}} = \dfrac{{6x + 12}}{{x + 8}}\\
I = \dfrac{E}{{R + r}} = \dfrac{{5,4}}{{\dfrac{{6x + 12}}{{x + 8}} + 1}} = \dfrac{{5,4\left( {x + 8} \right)}}{{7x + 20}}\\
{I_x} = \dfrac{{{R_{13}}}}{{{R_{13}} + {R_{245x}}}}.I = \dfrac{6}{{x + 8}}.\dfrac{{5,4\left( {x + 8} \right)}}{{7x + 20}} = \dfrac{{32,4}}{{7x + 20}}\\
{P_x} = I_x^2{R_x} = {\left( {\dfrac{{32,4}}{{7x + 20}}} \right)^2}x = \dfrac{{1049,76x}}{{{{\left( {7x + 20} \right)}^2}}}
\end{array}\)

Áp dụng BĐT Cô si:

\(\begin{array}{l}
{\left( {7x + 20} \right)^2} \ge 4.7x.20 = 560x\\
 \Rightarrow {P_x} \le \dfrac{{1049,76x}}{{560x}} = 1,87W
\end{array}\)

Dấu = xảy ra khi: \(7x = 20 \Rightarrow x = \dfrac{{20}}{7}\Omega \)