Đáp án:
\(\begin{array}{l}
5.I = 2A\\
6.\\
a.I = 1A\\
b.U{\rm{ = 8V}}\\
{\rm{c}}{\rm{.}}{P_3} = \dfrac{2}{3}{\rm{W}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5.\\
R = {R_1} + {R_2} = 2 + 3 = 5\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{5 + 1}} = 2A\\
6.\\
a.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
R = {R_1} + {R_{23}} = 6 + 2 = 8\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{10}}{{8 + 2}} = 1A\\
b.\\
U = {\rm{IR = 1}}{\rm{.8 = 8V}}\\
{\rm{c}}{\rm{.}}\\
{{\rm{U}}_3} = {U_{23}} = {\rm{I}}{{\rm{R}}_{23}} = 1.2 = 2V\\
{P_3} = \dfrac{{U_3^2}}{{{R_3}}} = \dfrac{{{2^2}}}{6} = \dfrac{2}{3}{\rm{W}}
\end{array}\)
