Đáp án:

 a. $25\left( \Omega  \right)$

b.50V

c.

$\begin{array}{l}
{U_1} = 50\left( V \right)\\
{I_2} = {I_3} = 1\left( A \right)\\
{U_2} = 20\left( V \right)\\
{U_3} = 30\left( V \right)\\
{I_1} = 1\left( A \right)
\end{array}$

d. 62000J

Giải thích các bước giải:

$\begin{array}{l}
a.\left( {{R_2}nt{R_3}} \right)//{R_1}\\
{R_{23}} = {R_2} + {R_3} = 20 + 30 = 50\\
{R_{td}} = \frac{{{R_{23}}{R_1}}}{{{R_{23}} + {R_1}}} = \frac{{50.50}}{{50 + 50}} = 25\left( \Omega  \right)\\
b.{U_{AB}} = I.{R_{td}} = 2.25 = 50\left( V \right)\\
c.\left( {{R_2}nt{R_3}} \right)//{R_1}\\
 \Rightarrow {U_1} = {U_{23}} = {U_{AB}} = 50\left( V \right)\\
{I_{23}} = {I_2} = {I_3} = \frac{{{U_{23}}}}{{{R_{23}}}} = \frac{{50}}{{50}} = 1\left( A \right)\\
{U_2} = {I_2}{R_2} = 1.20 = 20\left( V \right)\\
{U_3} = {I_3}{R_3} = 1.30 = 30\left( V \right)\\
{I_1} = \frac{{{U_1}}}{{{R_1}}} = \frac{{50}}{{50}} = 1\left( A \right)\\
d.t = 10p20s = 620\left( s \right)\\
A = P.t = \frac{{U_{AB}^2}}{{{R_{td}}}}.t = \frac{{{{50}^2}}}{{25}}.620 = 62000\left( J \right)
\end{array}$