Đáp án:
\(\begin{array}{l}
a.R = \dfrac{{220}}{{71}}\Omega \\
b.\\
{I_A} = \dfrac{{639}}{{181}}A\\
c.\\
{A_{ngoai}} = 38,6197t(J)\\
{P_{ngoai}} = 38,6197{\rm{W}}\\
H = 60,77\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = {R_1} + {R_2} = 6 + 4 = 10\Omega \\
{R_{123}} = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{10.6}}{{10 + 6}} = 3,75\Omega \\
{R_{12345}} = {R_{123}} + {R_4} + {R_5} = 3,75 + 4 + 6 = 13,75\Omega \\
R = \dfrac{{{R_{12345}}{R_6}}}{{{R_{12345}} + {R_6}}} = \dfrac{{13,75.4}}{{13,75 + 4}} = \dfrac{{220}}{{71}}\Omega \\
b.\\
{I_A} = I = \dfrac{E}{{R + r}} = \dfrac{{18}}{{\dfrac{{220}}{{71}} + 2}} = \dfrac{{639}}{{181}}A\\
c.\\
{A_{ngoai}} = R{I^2}t = \dfrac{{220}}{{71}}.{\dfrac{{639}}{{181}}^2}t = 38,6197t(J)\\
{P_{ngoai}} = R{I^2} = \dfrac{{220}}{{71}}.{\dfrac{{639}}{{181}}^2} = 38,6197{\rm{W}}\\
H = \dfrac{R}{{R + r}} = \dfrac{{\dfrac{{220}}{{71}}}}{{\dfrac{{220}}{{71}} + 2}} = 60,77\%
\end{array}\)
