Đáp án:
\(\begin{array}{l}
2.1\\
{E_b} = 3,5V\\
{r_b} = 1\Omega \\
2.2\\
m = 0,783g
\end{array}\)
2.3 đèn sáng yếu
Giải thích các bước giải:
\(\begin{array}{l}
2.1\\
{E_b} = {E_1} + {E_2} = 1,25 + 2,25 = 3,5V\\
{r_b} = {r_1} + {r_2} = 0,5 + 0,5 = 1\Omega \\
2.2\\
{R_{12}} = {R_1} + {R_2} = 13,5 + 6 = 19,5\Omega \\
{R_d} = \frac{{U_{dm}^2}}{{{P_{dm}}}} = \frac{{{3^2}}}{3} = 3\Omega \\
R = \frac{{{R_{12}}{R_d}}}{{{R_{12}} + {R_d}}} = \frac{{19,5.3}}{{19,5 + 3}} = 2,6\Omega \\
I = \frac{{{E_b}}}{{R + {r_b}}} = \frac{{3,5}}{{2,6 + 1}} = \frac{{35}}{{36}}A\\
{I_1}{R_{12}} = {I_d}{R_d}\\
19,5{I_1} = 3{I_d}\\
{I_1} + {I_d} = I = \frac{{35}}{{36}}\\
{I_1} = \frac{7}{{54}}A\\
m = \frac{{A{I_1}t}}{{96500.n}} = \frac{{108.\frac{7}{{54}}.5400}}{{96500.1}} = 0,783g\\
2.3\\
{R_2}//{R_d}\\
R = \frac{{{R_2}{R_d}}}{{{R_2} + {R_d}}} = \frac{{3.6}}{{3 + 6}} = 2\Omega \\
I = \frac{{{E_b}}}{{R + {r_b}}} = \frac{{3,5}}{{2 + 1}} = \frac{7}{6}A\\
{I_2}{R_2} = {I_d}{R_d}\\
6{I_2} = 3{I_d}\\
{I_2} + {I_d} = I = \frac{7}{6}\\
{I_d} = \frac{7}{9}A\\
{I_{dm}} = \frac{{{P_{dm}}}}{{{U_{dm}}}} = \frac{3}{3} = 1A\\
{I_d} < {I_{dm}}
\end{array}\)
đèn sáng yếu