Đáp án:
\(\begin{array}{l}
a.R = 2\Omega \\
b.\\
I = 2A\\
{{\rm{I}}_1} = 0,8A\\
{{\rm{I}}_2} = 0,8A\\
{{\rm{I}}_3} = 0,4A\\
c.\\
H = 66,67\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{{10}} = \dfrac{1}{2}\\
\Rightarrow R = 2\Omega \\
b.\\
I = \frac{E}{{R + r}} = \frac{6}{{2 + 1}} = 2A\\
{U_1} = {U_2} = {U_3} = U = {\rm{IR = 2}}{\rm{.2 = 4V}}\\
{{\rm{I}}_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{4}{5} = 0,8A\\
{{\rm{I}}_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{4}{5} = 0,8A\\
{{\rm{I}}_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{4}{{10}} = 0,4A\\
c.\\
H = \dfrac{R}{{R + r}} = \dfrac{2}{{2 + 1}} = 66,67\%
\end{array}\)
