Đáp án:
\(\begin{array}{l}
{R_x} = 3\Omega \\
{P_{x\max }} = 20,25W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{R_x} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{3} = 12\Omega \\
{R_{xd}} = \dfrac{{{R_x}{R_d}}}{{{R_x} + {R_d}}} = \dfrac{{12{R_x}}}{{{R_x} + 12}}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{U}{{{R_1} + {R_{xd}}}} = \dfrac{{12}}{{4 + \dfrac{{12{R_x}}}{{{R_x} + 12}}}} = \dfrac{{12({R_x} + 12)}}{{48 + 16{R_x}}}\\
{U_x} = {U_{xd}} = I{R_{xd}} = \dfrac{{12({R_x} + 12)}}{{48 + 16{R_x}}}.\dfrac{{12{R_x}}}{{{R_x} + 12}} = \dfrac{{144{R_x}}}{{48 + 16{R_x}}}\\
{P_x} = \dfrac{{U_x^2}}{{{R_x}}} = \dfrac{{{{(\dfrac{{144{R_x}}}{{48 + 16{R_x}}})}^2}}}{{{R_x}}} = \dfrac{{20736{R_x}}}{{{{(48 + 16{R_x})}^2}}} = \dfrac{{20736}}{{{{(\dfrac{{48}}{{\sqrt {{R_x}} }} + 16\sqrt {{R_x}} )}^2}}}\\
{P_x}\max \Rightarrow (\dfrac{{48}}{{\sqrt {{R_x}} }} + 16\sqrt {{R_x}} )\min
\end{array}\)
Áp dụng bất đẳng thức cô si:
\(\dfrac{{48}}{{\sqrt {{R_x}} }} + 16\sqrt {{R_x}} \ge 2\sqrt {\dfrac{{48}}{{\sqrt {{R_x}} }}.16\sqrt {{R_x}} } = 32\sqrt 3 \)
Dấu "=" xảy ra khi
\(\frac{{48}}{{\sqrt {{R_x}} }} = 16\sqrt {{R_x}} \Rightarrow {R_x} = 3\Omega \)
Công suất cực đại đó là:
\({P_{x\max }} = \dfrac{{20736{R_x}}}{{{{(48 + 16{R_x})}^2}}} = \dfrac{{20736.3}}{{{{(32\sqrt 3 )}^2}}} = 20,25W\)
