Đáp án:
\(\begin{array}{l}
a.I = \dfrac{{24}}{{13}}A\\
b.t = 3126,25s\\
c.{I_A} = \dfrac{6}{7}A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = {R_1} + {R_2} = 5 + 10 = 15\Omega \\
R = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{15.3}}{{15 + 3}} = 2,5\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{6}{{2,5 + 0,75}} = \dfrac{{24}}{{13}}A\\
b.\\
{U_3} = U = {\rm{IR}} = \dfrac{{24}}{{13}}.2,5 = \dfrac{{60}}{{13}}V\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{\dfrac{{60}}{{13}}}}{3} = \dfrac{{20}}{{13}}A\\
m = \dfrac{{A{I_3}t}}{{Fn}}\\
\Rightarrow t = \dfrac{{mFn}}{{A{I_3}}} = \dfrac{{1,6.96500.2}}{{64.\dfrac{{20}}{{13}}}} = 3126,25s\\
c.\\
{R_1}//{R_3}\\
R' = \dfrac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \dfrac{{5.3}}{{5 + 3}} = 1,875\Omega \\
I' = \dfrac{E}{{R' + r}} = \dfrac{6}{{1,875 + 0,75}} = \frac{{16}}{7}A\\
{U_1} = U' = I'R' = \dfrac{{16}}{7}.1,875 = \dfrac{{30}}{7}V\\
{I_A} = {I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{\dfrac{{30}}{7}}}{5} = \dfrac{6}{7}A
\end{array}\)
