Đáp án:

 D

Giải thích các bước giải:

Ta có: 

\(\begin{array}{l}
{R_{45}} = \dfrac{{{R_4}{R_5}}}{{{R_4} + {R_5}}} = \dfrac{{20.20}}{{20 + 20}} = 10\Omega \\
{R_{345}} = {R_3} + {R_{45}} = 20 + 10 = 30\Omega \\
{R_{2345}} = \dfrac{{{R_2}{R_{345}}}}{{{R_2} + {R_{345}}}} = \dfrac{{60.30}}{{60 + 30}} = 20\Omega \\
R = {R_1} + {R_{2345}} = 20 + 20 = 40\Omega \\
I = \dfrac{U}{R} = \dfrac{U}{{40}}\\
{I_{345}} = \dfrac{{{R_2}}}{{{R_2} + {R_{345}}}}.I = \dfrac{{60}}{{60 + 30}}.\dfrac{U}{{40}} = \dfrac{U}{{60}}\\
{I_4} = \dfrac{{{R_5}}}{{{R_4} + {R_5}}}.{I_{345}} = \dfrac{1}{2}.\dfrac{U}{{60}} = \dfrac{U}{{120}}\\
 \Rightarrow \dfrac{U}{{120}} = 1 \Rightarrow U = 120V\\
{I_2} = \dfrac{{{R_{345}}}}{{{R_2} + {R_{345}}}}.I = \dfrac{{30}}{{60 + 30}}.\dfrac{U}{{40}} = \dfrac{{120}}{{120}} = 1A
\end{array}\)