Đáp án:
\(\begin{array}{l}
11.A.2,5V\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
R = {R_1} + {R_{23}} = 3 + 2 = 5\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{3}{{5 + 1}} = 0,5A\\
U = {\rm{IR}} = 0,5.5 = 2,5V\\
12.D.13\Omega \\
{R_{34}} = {R_3} + {R_4} = 3 + 3 = 6\Omega \\
{R_{234}} = \dfrac{{{R_2}{R_{34}}}}{{{R_2} + {R_{34}}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
R = {R_1} + {R_{234}} + {R_5} = 10 + 2 + 1 = 13\Omega
\end{array}\)
