Đáp án:
\(\begin{array}{l}
a.I = 0,8A\\
b.\\
{I_3} = 0,2A\\
{I_1} = 0,6A\\
{I_2} = 0,2A\\
{I_4} = 0,6A\\
c.\\
{P_1} = 3,6W\\
{P_2} = 1,2W\\
{P_3} = 1,2W\\
{P_4} = 3,6W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
({R_1}//{R_3})nt({R_2}{R_4})\\
a.\\
{R_{13}} = \dfrac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \dfrac{{10.30}}{{10 + 30}} = 7,5\Omega \\
{R_{24}} = \dfrac{{{R_2}{R_4}}}{{{R_2} + {R_4}}} = \dfrac{{10.30}}{{10 + 30}} = 7,5\Omega \\
R = {R_{13}} + {R_{24}} = 7,5 + 7,5 = 15\Omega \\
I = \frac{U}{R} = \frac{{12}}{{15}} = 0,8A\\
b.\\
{U_1} = {U_3} = {U_{13}} = {R_{13}}I = 7,5.0,8 = 6V\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{6}{{30}} = 0,2A\\
{I_1} = I - {I_3} = 0,8 - 0,2 = 0,6A\\
{U_2} = {U_4} = {U_{24}} = {R_{24}}I = 7,5.0,8 = 6V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{6}{{30}} = 0,2A\\
{I_4} = I - {I_2} = 0,8 - 0,2 = 0,6A\\
c.\\
{P_1} = {U_1}{I_1} = 6.0,6 = 3,6W\\
{P_2} = {U_2}{I_2} = 6.0,2 = 1,2W\\
{P_3} = {U_3}{I_3} = 6.0,2 = 1,2W\\
{P_4} = {U_4}{I_4} = 6.0,6 = 3,6W
\end{array}\)
