Đáp án:
\(\begin{array}{l}
a.\\
{E_b} = 20V\\
{r_b} = 4\Omega \\
b.\\
{I_d} = {I_p} = I = \dfrac{5}{7}A\\
{I_1} = \dfrac{5}{{14}}A\\
{I_2} = \dfrac{5}{{14}}A\\
d.\\
{P_1} = 1,02W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{E_b} = 2E = 2.10 = 20V\\
{r_b} = 2r = 2.2 = 4\Omega \\
b.\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{8.8}}{{8 + 8}} = 4\Omega \\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{3} = 12\Omega \\
R = {R_d} + {R_{12}} + {R_p} = 12 + 4 + 8 = 24\Omega \\
{I_d} = {I_p} = I = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{{20}}{{24 + 4}} = \dfrac{5}{7}A\\
{U_1} = {U_2} = {\rm{I}}{{\rm{R}}_{12}} = \dfrac{5}{7}.4 = \dfrac{{20}}{7}V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{\frac{{20}}{7}}}{8} = \dfrac{5}{{14}}A\\
{I_2} = I - {I_1} = \dfrac{5}{7} - \dfrac{5}{{14}} = \dfrac{5}{{14}}A\\
c.\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{3}{6} = 0,5A\\
{I_d} > {I_{dm}}
\end{array}\)
Suy ra đèn mạnh hơn bình thường.
\(\begin{array}{l}
d.\\
{P_1} = {R_1}I_1^2 = 8.{\dfrac{5}{{14}}^2} = 1,02W
\end{array}\)
