Đáp án:
$\begin{align}
& a)I=3,62A \\
& b){{P}_{N}}=121W;{{P}_{Ng}}=148,8\text{W} \\
& c){{U}_{1}}={{U}_{2}}=11,6V;{{U}_{3}}={{U}_{4}}=2,79V \\
& {{I}_{1}}=3,87A;{{I}_{2}}=1,93A;{{I}_{3}}=2,79A;{{I}_{4}}=0,93A \\
\end{align}$
Giải thích các bước giải:
Bộ nguồn
$\begin{align}
& {{E}_{b}}=4E=40V \\
& {{r}_{b}}=4r=2\Omega \\
\end{align}$
điện trở đèn :
${{R}_{d}}=\dfrac{U_{dm}^{2}}{{{P}_{dm}}}=\dfrac{{{6}^{2}}}{6}=6\Omega $
mạch ngoài: ${{D}_{1}}nt\left( {{R}_{1}}//{{R}_{2}} \right)nt\left( {{R}_{3}}//{{R}_{4}} \right)$
Điện trở mạch ngoài:
$\begin{align}
& {{R}_{12}}=\dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\dfrac{3.6}{3+6}=2\Omega \\
& {{R}_{34}}=\dfrac{{{R}_{3}}.{{R}_{4}}}{{{R}_{3}}+{{R}_{4}}}=\dfrac{1.3}{1+3}=0,75\Omega \\
& {{R}_{N}}={{R}_{d}}+{{R}_{12}}+{{R}_{34}}=6+2+0,75=8,75\Omega \\
\end{align}$
Cường độ mạch ngoài: $I=\frac{{{E}_{b}}}{{{R}_{N}}+{{r}_{b}}}=\frac{40}{8,75+2}=3,72A$
b) Công suất
$\begin{align}
& {{P}_{N}}={{I}^{2}}.{{R}_{N}}=3,{{72}^{2}}.8,75=121W \\
& {{P}_{Ng}}={{E}_{b}}.I=40.3,72=148,8\text{W} \\
\end{align}$
c)
$\begin{align}
& {{U}_{12}}=I.{{R}_{12}}=3,72.3=11,6V={{U}_{1}}={{U}_{2}} \\
& {{U}_{34}}=I.{{R}_{34}}=3,72.0,75=2,79V={{U}_{3}}={{U}_{4}} \\
& {{I}_{1}}=\dfrac{{{U}_{1}}}{R{}_{1}}=\dfrac{11,6}{3}=3,87A \\
& {{I}_{2}}=\dfrac{{{U}_{2}}}{{{R}_{2}}}=\dfrac{11,6}{6}=1,93A \\
& {{I}_{3}}=\dfrac{{{U}_{3}}}{{{R}_{3}}}=\dfrac{2,79}{1}=2,79A \\
& {{I}_{4}}=\dfrac{{{U}_{4}}}{{{R}_{4}}}=\dfrac{2,79}{3}=0,93A \\
\end{align}$
