Đáp án:

\(\begin{array}{l}
2.12\\
R = 1\Omega \\
R = 4\Omega \\
2.14E = 12V
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2.12\\
P = R{I^2} = R\dfrac{{{E^2}}}{{{{(R + r)}^2}}} = \dfrac{{{E^2}}}{{\frac{{{{(R + r)}^2}}}{R}}} = \dfrac{{{E^2}}}{{{{(\sqrt R  + \dfrac{r}{{\sqrt R }})}^2}}}\\
 \Rightarrow 16 = \dfrac{{{{12}^2}}}{{{{(\sqrt R  + \dfrac{2}{{\sqrt R }})}^2}}}\\
 \Rightarrow 16{(\sqrt R  + \dfrac{2}{{\sqrt R }})^2} = 144\\
 \Rightarrow \sqrt R  + \dfrac{2}{{\sqrt R }} = 3\\
 \Rightarrow R - 3\sqrt R  + 2 = 0\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt R  = 1\\
\sqrt R  = 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
R = 1\Omega \\
R = 4\Omega 
\end{array} \right.\\
2.14\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{12^2}}}{8} = 18\Omega \\
{P_d} = {R_d}{I^2}t\\
 \Rightarrow I = \sqrt {\dfrac{{{P_d}}}{{{R_d}t}}}  = \sqrt {\dfrac{{7776}}{{18.1200}}}  = 0,6A\\
E = I({R_d} + r) = 0,6(18 + 2) = 12V
\end{array}\)