Đáp án:
\(\begin{array}{l}
2.\\
a.\\
R = 24\Omega \\
b.\\
{I_A} = 0,25A\\
A = 450J\\
3.\\
a.\\
R = 2,8\Omega \\
b.\\
Q = 58974,9cal
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2.\\
a.\\
{R_{23}} = \frac{{{R_2}.{R_3}}}{{{R_2} + {R_2}}} = \frac{{10.40}}{{10 + 40}} = 8\Omega \\
R = {R_1} + {R_{23}} = 16 + 8 = 24\Omega \\
b.\\
{I_A} = I = \frac{U}{R} = \frac{6}{{24}} = 0,25A\\
A = R{I^2}t = 24.0,{25^2}.5.60 = 450J\\
3.\\
a.\\
R = p\frac{l}{s} = 5,{5.10^{ - 8}}\frac{{40}}{{0,{{785.10}^{ - 6}}}} = 2,8\Omega \\
b.\\
Q = Pt = \frac{{{U^2}}}{R}.t = \frac{{{{24}^2}}}{{2,8}}.20.60 = 246857,143J = 58974,9cal
\end{array}\)