Đáp án:
\(\begin{array}{l}
1.\\
{I_2} = \dfrac{2}{3}A\\
{I_3} = \dfrac{1}{3}A\\
{I_1} = {I_4} = 1A\\
2.\\
{R_4} = 4\Omega \\
{R_1} = 6\Omega \\
3.{I_A} = 1,2A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
{U_V} = {U_2} = {U_3} = 2V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{2}{3}A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{2}{6} = \dfrac{1}{3}A\\
I = {I_1} = {I_4} = {I_2} + {I_3} = \dfrac{2}{3} + \dfrac{1}{3} = 1A\\
2.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
{U_1} + {U_4} = U - {U_2}\\
\Rightarrow {I_1}{R_1} + {I_4}{R_4} = 12 - 2\\
\Rightarrow 1.1,5{R_4} + 1{R_4} = 10\\
\Rightarrow {R_4} = 4\Omega \\
{R_1} = 1,5{R_4} = 6\Omega \\
3.\\
{R_1}nt{R_4}\\
R' = {R_1} + {R_4} = 6 + 4 = 10\Omega \\
{I_A} = I' = \dfrac{U}{{R'}} = \dfrac{{12}}{{10}} = 1,2A
\end{array}\)
