Đáp án:
\(\begin{array}{l}
a.{I_1} = {I_2} = 1,2A\\
b.\\
{I_1} = \dfrac{{12}}{7}A\\
{U_1} = \dfrac{{48}}{7}V\\
{U_2} = {U_3} = \dfrac{{36}}{7}V\\
{I_2} = \dfrac{6}{7}A\\
{I_3} = \dfrac{6}{7}A
\end{array}\)
\(c.{R_3} = 12\Omega \)
Giải thích các bước giải:
a.
\(\begin{array}{l}
a.\\
R = {R_1} + {R_2} = 4 + 6 = 10\Omega \\
{I_1} = {I_2} = I = \dfrac{U}{R} = \dfrac{{12}}{{10}} = 1,2A\\
b.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{6.6}}{{6 + 6}} = 3\Omega \\
R = {R_1} + {R_{23}} = 4 + 3 = 7\Omega \\
{I_1} = I = \dfrac{U}{R} = \dfrac{{12}}{7}A\\
{U_1} = {I_1}{R_1} = \dfrac{{12}}{7}.4 = \dfrac{{48}}{7}V\\
{U_2} = {U_3} = U - {U_1} = 12 - \dfrac{{48}}{7} = \dfrac{{36}}{7}V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{\dfrac{{36}}{7}}}{6} = \dfrac{6}{7}A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{\dfrac{{36}}{7}}}{6} = \dfrac{6}{7}A\\
c.
\(\begin{array}{l}
{I_1} = {I_2} + {I_3}\\
\Leftrightarrow {I_1} = \dfrac{2}{3}{I_1} + {I_3}\\
\Leftrightarrow {I_3} = \dfrac{1}{3}{I_1}\\
\Rightarrow {I_2} = 2{I_3}
\end{array}\)
Ta có:
\(\begin{array}{l}
{U_2} = {U_3}\\
\Rightarrow {I_2}{R_2} = {I_3}{R_3}\\
\Rightarrow {R_3} = \dfrac{{{I_2}{R_2}}}{{{I_3}}} = \dfrac{{2{I_3}.6}}{{{I_3}}} = 12\Omega
\end{array}\)
