Đáp án:
a. \({U_{AB}} = 8V\)
b.m=0,4267g
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{23}} = \frac{{{R_2}{R_3}}}{{{R_3} + {R_2}}} = \frac{{4.4}}{{4 + 4}} = 2\Omega \\
{R_{023}} = R + {R_{23}} = 4 + 2 = 6\Omega \\
{R_{td}} = \frac{{{R_{023}}{R_1}}}{{{R_{023}} + {R_1}}} = \frac{{6.3}}{{6 + 3}} = 2\Omega \\
I = \frac{E}{{{R_{td}} + r}} = \frac{{12}}{{2 + 1}} = 4A\\
{U_{AB}} = I{R_{td}} = 4.2 = 8V\\
b.\\
{I_1}{R_1} = {I_0}{R_{023}}\\
3{I_1} = {I_0}6\\
I = {I_1} + {I_0} = 4\\
{I_0} = \frac{4}{3}A\\
m = \frac{{A{I_0}t}}{{96500.n}} = \frac{{64.\frac{4}{3}.965}}{{96500.2}} = 0,4267g
\end{array}\)
