Đáp án:
\(\begin{array}{l}
2.\\
a.\\
{E_b} = 6V\\
{r_b} = 0,6\Omega \\
b.\\
I = 2A\\
{I_1} = 1,2A\\
{I_2} = 0,8A\\
c.\\
{P_d} = \dfrac{8}{3}W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2.\\
a.\\
{E_b} = 2E = 2.3 = 6V\\
{r_b} = 2r = 2.0,3 = 0,6\Omega \\
b.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{4.6}}{{4 + 6}} = 2,4\Omega \\
I = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{6}{{2,4 + 0,6}} = 2A\\
{U_1} = {U_2} = U = IR = 2.2,4 = 4,8V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{4,8}}{4} = 1,2A\\
{I_2} = I - {I_1} = 2 - 1,2 = 0,8A\\
c.\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{6} = 6\Omega \\
R' = R + {R_d} = 2,4 + 6 = 8,4\Omega \\
{I_d} = I' = \dfrac{{{E_b}}}{{R' + {r_b}}} = \dfrac{6}{{8,4 + 0,6}} = \dfrac{2}{3}A\\
{P_d} = {R_d}I_d^2 = 6.{\dfrac{2}{3}^2} = \dfrac{8}{3}W\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{6}{6} = 1A\\
{I_d} < {I_{dm}}(\frac{2}{3}A < 1A)
\end{array}\)
Suy ra đèn sáng yếu.
