Đáp án:

a) 0

b) \(\dfrac{{60}}{{11}}\Omega \)

c) \(\dfrac{{60}}{{11}}V\)

Giải thích các bước giải:

a)

Công suất tỏa nhiệt trên AB là: 

\({P_{AB}} = {I^2}{R_1}\)

Để \({P_{AB}}\) max thì \(I\) max; khi đó R tương đương min nên \({R_{BC}}\) min

Ta có:

\({R_{BC}} = \dfrac{{{R_0}{R_x}}}{{{R_0} + {R_x}}} = \dfrac{{10{R_x}}}{{10 + {R_x}}} = \dfrac{{10}}{{\dfrac{{10}}{{{R_x}}} + 1}}\)

Ta thấy:

\({R_{BC}}\) min khi \({\dfrac{{10}}{{{R_x}}} + 1}\) max nên \({\dfrac{{10}}{{{R_x}}}}\) max nên \({{R_x}}\) min = 0 

b) 

Ta có:

\(\begin{array}{l}
R = {R_1} + \dfrac{{{R_0}{R_x}}}{{{R_0} + {R_x}}} + r = 10 + \dfrac{{10{R_x}}}{{10 + {R_x}}} + 2\\
 \Rightarrow R = 12 + \dfrac{{10{R_x}}}{{10 + {R_x}}} = \dfrac{{22{R_x} + 120}}{{{R_x} + 10}}\\
I = \dfrac{U}{R} = \dfrac{{12\left( {{R_x} + 10} \right)}}{{22{R_x} + 120}}\\
{I_x} = \dfrac{{{R_0}}}{{{R_0} + {R_x}}}.I = \dfrac{{10.12}}{{22{R_x} + 120}} = \dfrac{{60}}{{11{R_x} + 60}}\\
{P_x} = I_x^2.{R_x} = \dfrac{{{{60}^2}{R_x}}}{{{{\left( {11{R_x} + 60} \right)}^2}}}\\
 \Rightarrow {P_x} = \dfrac{{3600{R_x}}}{{{{\left( {11{R_x} + 60} \right)}^2}}} \le \dfrac{{3600{R_x}}}{{4.11{R_x}.60}} = \dfrac{{15}}{{11}}
\end{array}\)

Dấu = xảy ra khi: \(11{R_x} = 60 \Rightarrow {R_x} = \dfrac{{60}}{{11}}\Omega \)

c) 

* Khi không có Rx:

Ta có:

\(\begin{array}{l}
I = \dfrac{U}{{{R_1} + {R_0} + r}} = \dfrac{{12}}{{10 + 10 + 2}} = \dfrac{6}{{11}}A\\
{U_{AB}} = I.{R_1} = \dfrac{6}{{11}}.10 = \dfrac{{60}}{{11}}V
\end{array}\)

* Khi Rx = 1:

Ta có:

\(\begin{array}{l}
I = \dfrac{U}{R} = \dfrac{{12\left( {{R_x} + 10} \right)}}{{22{R_x} + 120}} = \dfrac{{66}}{{71}}A\\
{U_{AB}} = I.{R_1} = \dfrac{{66}}{{71}}.10 = \dfrac{{660}}{{71}}V
\end{array}\)

* Khi Rx = 100:

Ta có:

\(\begin{array}{l}
I = \dfrac{U}{R} = \dfrac{{12\left( {{R_x} + 10} \right)}}{{22{R_x} + 120}} = \dfrac{{33}}{{58}}A\\
{U_{AB}} = I.{R_1} = \dfrac{{33}}{{58}}.10 = \dfrac{{165}}{{29}}V
\end{array}\)