Đáp án:
\(\begin{array}{l}
a.\\
{R_3} = 12\Omega \\
R = 7,5\Omega \\
b.\\
{I_A} = 1,25A\\
m = 0,48g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_3} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{3} = 12\Omega \\
{R_{12}} = {R_1} + {R_2} = 2 + 6 = 8\Omega \\
{R_{123}} = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{8.12}}{{8 + 12}} = 4,8\Omega \\
R = {R_{123}} + {R_4} = 4,8 + 2,7 = 7,5\Omega \\
b.\\
{I_A} = I = \dfrac{E}{{R + r}} = \dfrac{{10}}{{7,5 + 0,5}} = 1,25A\\
{U_{12}} = {U_{123}} = {\rm{I}}{{\rm{R}}_{123}} = 1,25.4,8 = 6V\\
{I_2} = {I_{12}} = \dfrac{{{U_{12}}}}{{{R_{12}}}} = \dfrac{6}{8} = 0,75A\\
m = \dfrac{{AIt}}{{Fn}} = \dfrac{{64.0,75.1930}}{{96500.2}} = 0,48g
\end{array}\)
