Đáp án:

\(\begin{array}{l}
 + \\
a.R = 60\Omega \\
b.{I_1} = {I_2} = {I_3} = \dfrac{4}{{15}}A\\
 + \\
a.R = 30\Omega \\
b.\\
{I_1} = \dfrac{8}{{15}}A\\
{I_2} = {I_3} = \dfrac{4}{{15}}A\\
 + \\
a.R = \dfrac{{40}}{3}\Omega \\
b.\\
{I_3} = 0,8A\\
{I_1} = {I_2} = 0,4A\\
 + \\
a.R = \dfrac{{20}}{3}\Omega \\
b.\\
{I_1} = 0,8A\\
{I_2} = 0,8A\\
{I_3} = 0,8A
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
 + \\
a.\\
R = {R_1} + {R_2} + {R_3} = 20 + 20 + 20 = 60\Omega \\
b.\\
{I_1} = {I_2} = {I_3} = \dfrac{U}{R} = \dfrac{{16}}{{60}} = \dfrac{4}{{15}}A\\
 + \\
a.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{20.20}}{{20 + 20}} = 10\Omega \\
R = {R_1} + {R_{23}} = 20 + 10 = 30\Omega \\
b.\\
{I_1} = I = \dfrac{U}{R} = \dfrac{{16}}{{30}} = \dfrac{8}{{15}}A\\
{U_2} = {U_3}\\
 \Rightarrow {I_2}{R_2} = {I_3}{R_3}\\
 \Rightarrow 20{I_2} = 20{T_3}\\
\left. \begin{array}{l}
 \Rightarrow {I_2} = {I_3}\\
{I_2} + {I_3} = I = \dfrac{8}{{15}}
\end{array} \right\} \Rightarrow {I_2} = {I_3} = \dfrac{4}{{15}}A\\
 + \\
a.\\
{R_{12}} = {R_1} + {R_2} = 20 + 20 = 40\Omega \\
R = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{40.20}}{{40 + 20}} = \dfrac{{40}}{3}\Omega \\
b.\\
{I_3} = \dfrac{U}{{{R_3}}} = \dfrac{{16}}{{20}} = 0,8A\\
{I_1} = {I_2} = \dfrac{U}{{{R_{12}}}} = \dfrac{{16}}{{40}} = 0,4A\\
 + \\
a.\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{{20}} + \dfrac{1}{{20}} + \dfrac{1}{{20}} = \dfrac{1}{{\dfrac{{20}}{3}}}\\
 \Rightarrow R = \dfrac{{20}}{3}\Omega \\
b.\\
{I_1} = \dfrac{U}{{{R_1}}} = \dfrac{{16}}{{20}} = 0,8A\\
{I_2} = \dfrac{U}{{{R_2}}} = \dfrac{{16}}{{20}} = 0,8A\\
{I_3} = \dfrac{U}{{{R_3}}} = \dfrac{{16}}{{20}} = 0,8A
\end{array}\)