Đáp án:
\(\begin{array}{l}
a.\\
\left[ \begin{array}{l}
R = 4\Omega \\
R = 1\Omega
\end{array} \right.\\
b.{P_{N\max }} = 1,125W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
P = R{I^2} = R.{\dfrac{E}{{R + r}}^2}\\
\Rightarrow 1 = R{\dfrac{3}{{R + 2}}^2}\\
\Rightarrow {(R + 2)^2} = 9R\\
\Rightarrow {R^2} - 5R + 4 = 0\\
\Rightarrow \left[ \begin{array}{l}
R = 4\Omega \\
R = 1\Omega
\end{array} \right.\\
b.\\
{P_N} = R{I^2} = R.{\dfrac{E}{{R + r}}^2} = \dfrac{{{E^2}}}{{\dfrac{{{{(R + r)}^2}}}{R}}} = \dfrac{{{E^2}}}{{{{(\sqrt R + \dfrac{r}{{\sqrt R }})}^2}}}\\
{P_{N\max }} \Rightarrow {(\sqrt R + \dfrac{r}{{\sqrt R }})^2}\min
\end{array}\)
Áp dụng bất đẳng thức Cô si:
\(\sqrt R + \dfrac{r}{{\sqrt R }} \ge 2\sqrt {\sqrt R .\dfrac{r}{{\sqrt R }}} = 2\sqrt r \)
Dấu " = " xảy ra khi:
\(\begin{array}{l}
\sqrt R = \dfrac{r}{{\sqrt R }} \Rightarrow r = R = 2\Omega \\
{P_{N\max }} = \dfrac{{{E^2}}}{{{{(2\sqrt r )}^2}}} = \dfrac{{{E^2}}}{{4r}} = \dfrac{{{3^2}}}{{4.2}} = 1,125W
\end{array}\)
