Đáp án:

 a.$36\left( \Omega  \right)$

b.${I_1} = 0,67\left( A \right)$

$\begin{array}{l}a.
{I_2} = 0,27\left( A \right)\\
{I_3} = 0,4\left( A \right)
\end{array}$

c,$4800\left( J \right)$

Giải thích các bước giải:

$\begin{array}{l}
a.{R_1}nt\left( {{R_2}//{R_3}} \right)\\
{R_{23}} = \frac{{{R_2}{R_3}\,}}{{{R_2} + {R_3}}} = \frac{{15.10}}{{15 + 10}} = 6\\
{R_{123}} = {R_{td}} = {R_1} + {R_{23}} = 30 + 6 = 36\left( \Omega  \right)\\
b.I = {I_1} = {I_{23}} = \frac{{{U_{AB}}}}{{{R_{td}}}} = \frac{{24}}{{36}} = 0,67\left( A \right)\\
{U_{23}} = {U_2} = {U_3} = {I_{23}}{R_{23}} = \frac{2}{3}.6 = 4\\
{I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{4}{{15}} = 0,27\left( A \right)\\
{I_3} = \frac{{{U_3}}}{{{R_3}}} = \frac{4}{{10}} = 0,4\left( A \right)\\
c.A = {I^2}{R_{td}}.t = {\left( {\frac{{24}}{{36}}} \right)^2}.36.5.60 = 4800\left( J \right)
\end{array}$