Đáp án:
a,${R_{AB}} = 4\left( \Omega \right)$
$3A = {I_1}$
$\begin{array}{l}
{I_2} = 1\left( A \right)\\
{I_3} = {I_4} = 2\left( A \right)
\end{array}$
b.$\begin{array}{l}
+ BD\\
q = C{U_{DB}} = {6.10^{ - 5}}\left( C \right)\\
+ nt{R_3}\\
q = C{U_c} = 0\left( {{U_c} = 0} \right)
\end{array}$
Giải thích các bước giải:
a. mạch điện
$\begin{array}{l}
{R_1}nt\left( {{R_2}//\left( {{R_3}nt{R_4}} \right)} \right)\\
{R_{34}} = {R_3} + {R_4} = 1 + 2 = 3\\
{R_{234}} = \frac{{{R_{34}}{R_2}}}{{{R_{34}} + {R_2}}} = \frac{{3.6}}{{3 + 6}} = 2\\
{R_{AB}} = {R_1} + {R_{234}} = 2 + 2 = 4\left( \Omega \right)\\
I = \frac{{{U_{AB}}}}{{{R_{AB}}}} = \frac{{12}}{4} = 3 = {I_1} = {I_{DB}}\\
{U_{DB}} = {U_{34}} = {U_2} = {I_{DB}}.{R_{234}} = 3.2 = 6\\
{I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{6}{6} = 1\left( A \right)\\
{I_{34}} = {I_3} = {I_4} = \frac{{{U_{34}}}}{{{R_{34}}}} = \frac{6}{3} = 2\left( A \right)
\end{array}$
b,$\begin{array}{l}
+ BD\\
q = C{U_{DB}} = {6.10^{ - 5}}\left( C \right)
\end{array}$
+ mắc nt R3 hdt 2 đầu tụ = 0
$q = C{U_c} = 0\left( {{U_c} = 0} \right)$
