Đáp án:

\(\left\{ \begin{array}{l}
{U_1} = 32V\\
{U_2} = 8V\\
{U_3} = 8V\\
{U_4} = 2V
\end{array} \right.\)

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{R_1} = \dfrac{{\rho {l_1}}}{{{S_1}}}\\
{R_2} = \dfrac{{\rho {l_2}}}{{{S_2}}} = \dfrac{{\rho .\dfrac{{{l_1}}}{2}}}{{2{S_1}}} = \dfrac{1}{4}.\dfrac{{\rho {l_1}}}{{{S_1}}} = \dfrac{{{R_1}}}{4}\\
 \Rightarrow {U_2} = \dfrac{{{U_1}}}{4}\\
{R_3} = \dfrac{{\rho {l_3}}}{{{S_3}}} = \dfrac{{\rho .\dfrac{3}{4}{l_1}}}{{3{S_1}}} = \dfrac{1}{4}.\dfrac{{\rho {l_1}}}{{{S_1}}} = \dfrac{{{R_1}}}{4}\\
 \Rightarrow {U_3} = \dfrac{{{U_1}}}{4}\\
{R_4} = \dfrac{{\rho {l_4}}}{{{S_4}}} = \dfrac{{\rho .\dfrac{{{l_1}}}{4}}}{{4{S_1}}} = \dfrac{1}{{16}}.\dfrac{{\rho {l_1}}}{{{S_1}}} = \dfrac{{{R_1}}}{{16}}\\
 \Rightarrow {U_4} = \dfrac{{{U_1}}}{{16}}\\
U = {U_1} + {U_2} + {U_3} + {U_4}\\
 \Rightarrow 50 = {U_1} + \dfrac{{{U_1}}}{4} + \dfrac{{{U_1}}}{4} + \dfrac{{{U_1}}}{{16}}\\
 \Rightarrow \left\{ \begin{array}{l}
{U_1} = 32V\\
{U_2} = 8V\\
{U_3} = 8V\\
{U_4} = 2V
\end{array} \right.
\end{array}\)