Đáp án:
1.$3\left( \Omega \right)$
2.$6\left( V \right)$
3.$\begin{array}{l}
{I_1} = 1\left( A \right)\\
{P_1} = 6\left( {\rm{W}} \right)
\end{array}$
$\begin{array}{l}
{I_2} = {I_3} = 1\left( A \right)\\
{P_2} = 2\left( {\rm{W}} \right)\\
{P_3} = 4\left( {\rm{W}} \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
1.{R_1}//\left( {{R_2}nt{R_3}} \right)\\
{R_{23}} = {R_2} + {R_3} = 2 + 4 = 6\\
{R_{td}} = {R_{123}} = \frac{{{R_1}.{R_{23}}}}{{{R_1} + {R_{23}}}} = \frac{{6.6}}{{6 + 6}} = 3\left( \Omega \right)\\
2.U = I.{R_{td}} = 2.3 = 6\left( V \right)\\
3.{U_1} = {U_{23}} = U = 6\\
{I_1} = \frac{{{U_1}}}{{{R_1}}} = 1\left( A \right)\\
{P_1} = I_1^2{R_1} = {1^2}.6 = 6\left( {\rm{W}} \right)\\
{I_{23}} = {I_2} = {I_3} = \frac{{{U_{23}}}}{{{R_{23}}}} = \frac{6}{6} = 1\left( A \right)\\
{P_2} = I_2^2{R_2} = {1^2}.2 = 2\left( {\rm{W}} \right)\\
{P_3} = I_3^2{R_3} = {1^1}.4 = 4\left( {\rm{W}} \right)
\end{array}$
