Đáp án:
$\begin{array}{l}
a.{R_{td}} = 10\left( \Omega \right)\\
b.I = 4\left( A \right)\\
{I_1} = 2\left( A \right)\\
{I_2} = 1,33\left( A \right)\\
{I_3} = 0,67\left( A \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a.{r_1}//{r_2}//{r_3}\\
\frac{1}{{{R_{td}}}} = \frac{1}{{{r_1}}} + \frac{1}{{{r_2}}} + \frac{1}{{{r_3}}} = \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{60}} \Rightarrow {R_{td}} = 10\left( \Omega \right)\\
b.I = \frac{U}{{{R_{td}}}} = \frac{{40}}{{10}} = 4\left( A \right)\\
{r_1}//{r_2}//{r_3} \Rightarrow {U_1} = {U_2} = {U_3} = U = 40V\\
{I_1} = \frac{{{U_1}}}{{{r_1}}} = \frac{{40}}{{20}} = 2\left( A \right)\\
{I_2} = \frac{{{U_2}}}{{{r_2}}} = \frac{{40}}{{30}} = 1,33\left( A \right)\\
{I_3} = \frac{{{U_3}}}{{{r_3}}} = \frac{{40}}{{60}} = 0,67\left( A \right)
\end{array}$
