Đáp án:
$\left\{ \begin{align}
& {{I}_{V}}=0,3A \\
& {{R}_{1}}=60\Omega \\
& {{R}_{2}}=30\Omega \\
& {{R}_{V}}=60\Omega \\
\end{align} \right.$
Giải thích các bước giải:
Ta có:
$\begin{align}
& {{I}_{A1}}={{I}_{{{R}_{1}}}}+{{I}_{V}} \\
& {{U}_{V}}={{U}_{R1}}=18V \\
& {{I}_{A2}}={{I}_{R2}}+{{I}_{V}} \\
& {{U}_{V}}={{U}_{{{R}_{2}}}}=18V \\
& {{I}_{A3}}={{I}_{12}}+{{I}_{V}} \\
& {{U}_{V}}={{U}_{R1}}+{{U}_{R2}}=18V \\
\end{align}$
$\left\{ \begin{align}
& 18={{R}_{1}}.{{I}_{R1}} \\
& 18={{R}_{2}}.{{I}_{R2}} \\
& 18=({{R}_{1}}+{{R}_{2}}).{{I}_{R12}} \\
\end{align} \right.$
$\Leftrightarrow \left\{ \begin{align}
& 18={{R}_{1}}.(0,6-{{I}_{V}}) \\
& 18={{R}_{2}}.(0,9-{{I}_{V}}) \\
& 18=(R{}_{1}+{{R}_{2}}).(0,5-{{I}_{V}}) \\
\end{align} \right.$
$\Rightarrow \left\{ \begin{align}
& {{R}_{1}}=\dfrac{18}{0,6-{{I}_{V}}} \\
& {{R}_{1}}=\dfrac{18}{0,9-{{I}_{V}}} \\
& 18=(\dfrac{18}{0,6-{{I}_{V}}}+\dfrac{18}{0,9-{{I}_{V}}}).(0,5-{{I}_{V}}) \\
\end{align} \right.$
Điện trở:
$\left\{ \begin{align}
& {{I}_{V}}=0,3A \\
& {{R}_{1}}=\dfrac{18}{0,6-0,3}=60\Omega \\
& {{R}_{2}}=\dfrac{18}{0,9-0,3}=30\Omega \\
& {{R}_{V}}=\dfrac{18}{0,3}=60\Omega \\
\end{align} \right.$
