Đáp án:
\(\begin{array}{l}
a.\\
E = 12V\\
r = 2\Omega \\
b.\\
I = 0,48A\\
U = 11,04V\\
{Q_2} = 7776J\\
d.{R_3} = 8\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
E = {E_0} = 12V\\
r = \dfrac{{{r_0}}}{2} = \dfrac{4}{2} = 2\Omega \\
b.\\
{R_d} = \dfrac{{{U_{dm}}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{{4,5}} = 8\Omega \\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{40.24}}{{40 + 24}} = 15\Omega \\
R = {R_{12}} + {R_d} = 8 + 15 = 23\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{23 + 2}} = 0,48A\\
U = IR = 0,48.23 = 11,04V\\
{U_2} = {U_{12}} = I{R_{12}} = 0,48.15 = 7,2V\\
{Q_2} = \dfrac{{U_2^2}}{{{R_2}}}t = \dfrac{{7,{2^3}}}{{24}}.3600 = 7776J\\
c.\\
{I_d} = I = 0,48A\\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{{4,5}}{6} = 0,75A\\
{I_d} < {I_{dm}}
\end{array}\)
Suy ra đèn sáng yếu
\(\begin{array}{l}
d.\\
I = {I_{dm}} = 0,75A\\
E = I(R + r)\\
\Rightarrow R = \dfrac{E}{I} - r = \dfrac{{12}}{{0,75}} - 2 = 14\Omega \\
{R_{23}} = R - {R_d} = 14 - 8 = 6\Omega \\
\frac{1}{{{R_{23}}}} = \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}\\
\Rightarrow \dfrac{1}{6} = \dfrac{1}{{24}} + \dfrac{1}{{{R_3}}}\\
\Rightarrow {R_3} = 8\Omega
\end{array}\)
