Đáp án:
2.1$\left\{ \begin{array}{l}
{e_b} = 8\left( V \right)\\
{r_b} = 1,6\left( \Omega \right)
\end{array} \right.$
2.2 $0,9\left( A \right)$
2.3 $2,72\left( V \right)$
2.4.$H = 18,18\% $
đèn sáng yếu
2.5 $0,4352\left( g \right)$
Giải thích các bước giải:
2.1
4 nguồn ghép nối tiếp
$\begin{array}{l}
\left\{ \begin{array}{l}
{e_b} = 4e = 4.2 = 8\left( V \right)\\
{r_b} = 4r = 4.0,4 = 1,6\left( \Omega \right)
\end{array} \right.\\
2.2\\
{R_D} = \frac{{U_{dm}^2}}{{{P_{dm}}}} = \frac{{{{12}^2}}}{{12}} = 12\\
{R_1}nt\left( {{R_D}//{R_{dp}}} \right)\\
{R_{MN}} = \frac{{{R_D}{R_{dp}}}}{{{R_D} + {R_{dp}}}} = \frac{{12.4}}{{12 + 4}} = 3\\
{R_{td}} = {R_1} + {R_{MN}} = 4,2 + 3 = 7,2\\
I = \frac{{{e_b}}}{{{r_b} + {R_{td}}}} = \frac{8}{{1,6 + 7,2}} = 0,9\left( A \right) = {I_{MN}}\\
2.3.\\
{U_{MN}} = {I_{MN}}{R_{MN}} = 2,72\left( V \right) = {U_D} = {U_{dp}}\\
2.4\\
{U_D} < {U_{dm}}
\end{array}$
nên đèn sáng yếu
$H = \frac{{{r_b}}}{{{r_b} + {R_{td}}}} = \frac{{1,6}}{{1,6 + 7,2}}.100\% = 18,18\% $
$\begin{array}{l}
2.5\\
{I_{dp}} = \frac{{{U_{dp}}}}{{{R_{dp}}}} = \frac{{2,72}}{4} = 0,68\left( A \right)\\
t = 32.60 + 10 = 1930\left( s \right)\\
{m_{Cu}} = \frac{1}{F}\frac{A}{n}{I_{dp}}t = \frac{1}{{96500}}.\frac{{64}}{2}.0,68.1930 = 0,4352\left( g \right)
\end{array}$
