Đáp án:

\(\begin{array}{l}
a.R = 24\Omega \\
b.\\
{I_1} = 0,2A\\
{I_2} = 0,3A\\
c.{R_3} = 4,8\Omega 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{60.40}}{{60 + 40}} = 24\Omega \\
b.\\
{U_1} = {U_2} = {\rm{IR}} = 0,5.24 = 12V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{12}}{{60}} = 0,2A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{12}}{{40}} = 0,3A\\
c.\\
{U_1} = {U_2}\\
 \Rightarrow {I_1}{R_1} = {U_2}{R_2}\\
 \Rightarrow 60{I_1} = 40{I_2}\\
 \Rightarrow {I_2} = 1,5{I_1}\\
{I_3} = I = {I_1} + {I_2} = {I_1} + 1,5{I_1} = 2,5{I_1}\\
{P_1} = 2{P_3}\\
 \Rightarrow {U_1}{I_1} = 2{U_3}{I_3} = 2.{U_3}.2,5{I_1}\\
\left. \begin{array}{l}
 \Rightarrow {U_1} = 5{U_3}\\
{U_1} + {U_3} = U = 12V
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}
{U_1} = 10V\\
{U_3} = 2V
\end{array} \right.\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{10}}{{60}} = \dfrac{1}{6}A\\
{I_3} = 2,5{I_1} = 2,5.\frac{1}{6} = \dfrac{5}{{12}}A\\
{R_3} = \dfrac{{{U_3}}}{{{I_3}}} = \dfrac{2}{{\dfrac{5}{{12}}}} = 4,8\Omega 
\end{array}\)