Đáp án:
\(\left\{ \begin{array}{l}
{R_1} = 20\Omega \\
{R_2} = 20\Omega \\
{R_3} = 40\Omega
\end{array} \right.\)
Giải thích các bước giải:
* Trường hợp 1:
\(\dfrac{{{R_3}}}{{{R_1}}} = \dfrac{{{U_3}}}{{{U_1}}} = \dfrac{{2,4}}{{3,6 - 2,4}} = 2 \Rightarrow {R_3} = 2{R_1}\)
* Trường hợp 3:
\(\dfrac{{{R_3}}}{{{R_2}}} = \dfrac{{{U_3}}}{{{U_2}}} = \dfrac{{2,4}}{{3,6 - 2,4}} = 2 \Rightarrow {R_3} = 2{R_2} \Rightarrow {R_1} = {R_2}\)
* Trường hợp 2:
\(\begin{array}{l}
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{2{R_2}}}{3}\\
{R_{123}} = {R_1} + {R_{23}} = {R_2} + \dfrac{{2{R_2}}}{3} = \dfrac{5}{3}{R_2}\\
I = \dfrac{U}{{{R_{123}}}} = \dfrac{{3,6}}{{\dfrac{5}{3}{R_2}}} = \dfrac{{2,16}}{{{R_2}}}\\
{I_A} = \dfrac{{{R_3}}}{{{R_2} + {R_3}}}.I = \dfrac{2}{3}.\dfrac{{2,16}}{{{R_2}}} = \dfrac{{1,44}}{{{R_2}}} = {72.10^{ - 3}}\\
\Rightarrow \left\{ \begin{array}{l}
{R_1} = 20\Omega \\
{R_2} = 20\Omega \\
{R_3} = 40\Omega
\end{array} \right.
\end{array}\)
