Đáp án:
\(\begin{array}{l}
7.\\
a.R = \dfrac{{20}}{3}\Omega \\
b.\\
{I_1} = 1,8A\\
{U_1} = 7,2V\\
{U_3} = 4,8V\\
{I_3} = 1,2A\\
{I_2} = {I_3} = 0,6A\\
{U_2} = 2,4V\\
{U_3} = 2,4V
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
7.\\
(({R_2}nt{R_4})//{R_3})nt{R_1}\\
a.\\
{R_{24}} = {R_2} + {R_4} = 4 + 4 = 8\Omega \\
{R_{234}} = \dfrac{{{R_3}{R_{24}}}}{{{R_3} + {R_{24}}}} = \dfrac{{8.4}}{{8 + 4}} = \dfrac{8}{3}\Omega \\
R = {R_1} + {R_{234}} = 4 + \frac{8}{3} = \dfrac{{20}}{3}\Omega \\
b.\\
{I_1} = I = \dfrac{U}{R} = \dfrac{{12}}{{\dfrac{{20}}{3}}} = 1,8A\\
{U_1} = {I_1}{R_1} = 1,8.4 = 7,2V\\
{U_3} = {U_{24}} = {U_{234}} = U - {U_1} = 12 - 7,2 = 4,8V\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{4,8}}{4} = 1,2A\\
{I_2} = {I_3} = {I_{23}} = I - {I_3} = 1,8 - 1,2 = 0,6A\\
{U_2} = {I_2}{R_2} = 0,6.4 = 2,4V\\
{U_3} = {I_3}{R_3} = 0,6.4 = 2,4V
\end{array}\)
