Đáp án:
\(\begin{array}{l}
1.R = 36\Omega \\
2.\\
{I_1} = \dfrac{2}{3}A\\
{I_2} = \dfrac{4}{{15}}A\\
{I_3} = 0,4A\\
3.A = 4800J
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{15.10}}{{15 + 10}} = 6\Omega \\
R = {R_1} + {R_{23}} = 30 + 6 = 36\Omega \\
2.\\
{I_1} = I = \dfrac{U}{R} = \dfrac{{24}}{{36}} = \dfrac{2}{3}A\\
{U_2} = {U_3} = {U_{23}} = {\rm{I}}{{\rm{R}}_{23}} = \dfrac{2}{3}.6 = 4V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{4}{{15}}A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{4}{{10}} = 0,4A\\
3.\\
A = R{I^2}t = 36.{\dfrac{2}{3}^2}.300 = 4800J
\end{array}\)
