Đáp án:
$r = 12\Omega $
Giải thích các bước giải:
$\begin{array}{l}
{U_{MN}}^2 = {U_r}^2 + {U_L}^2 \Leftrightarrow {U_L} = \sqrt {{U_{MN}}^2 - {U_r}^2} = \sqrt {{{13}^2} - {U_r}^2} (1)\\
{U_{AB}}^2 = {({U_R} + {U_r})^2} + {({U_L} - {U_C})^2}\\
{65^2} = {(13 + {U_r})^2} + {({U_L} - 65)^2}(2)\\
(1)\& (2) \Leftrightarrow \left\{ \begin{array}{l}
{U_r} = 12V\\
{U_L} = 5V
\end{array} \right.\\
P = \frac{{{{({U_R} + {U_r})}^2}}}{{R + r}} \Leftrightarrow R + r = \frac{{{{({U_R} + {U_r})}^2}}}{P} = 25(3)\\
\frac{R}{r} = \frac{{{U_R}}}{{{U_r}}} = \frac{{13}}{{12}}(4)\\
(3)\& (4) \Rightarrow \left\{ \begin{array}{l}
R = 13\Omega \\
r = 12\Omega
\end{array} \right.
\end{array}$