Đáp án:
\(\begin{array}{l}
8.R = \dfrac{{20}}{3}\Omega \\
b.\\
{I_1} = 1,8A\\
{U_1} = 7,2V\\
{U_2} = 4,8V\\
{I_2} = 1,2A\\
{I_4} = {I_3} = 0,6A\\
{U_3} = 2,4V\\
{U_4} = 2,4V\\
9.\\
R = 12\Omega \\
{I_1} = {I_4} = 2A\\
{I_5} = 1A\\
{I_2} = {I_3} = 1A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
8.\\
(({R_3}nt{R_4})//{R_2})nt{R_1}\\
a.\\
{R_{34}} = {R_3} + {R_4} = 4 + 4 = 8\Omega \\
{R_{234}} = \dfrac{{{R_2}{R_{34}}}}{{{R_2} + {R_{34}}}} = \dfrac{{4.8}}{{4 + 8}} = \dfrac{8}{3}\Omega \\
R = {R_1} + {R_{234}} = 4 + \dfrac{8}{3} = \dfrac{{20}}{3}\Omega \\
b.\\
{I_1} = I = \dfrac{U}{R} = \dfrac{{12}}{{\dfrac{{20}}{3}}} = 1,8A\\
{U_1} = {I_1}{R_1} = 1,8.4 = 7,2V\\
{U_{34}} = {U_2} = {U_{234}} = U - {U_1} = 12 - 7,2 = 4,8V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{4,8}}{4} = 1,2A\\
{I_4} = {I_3} = {I_{34}} = \dfrac{{{U_{34}}}}{{{R_{34}}}} = \dfrac{{4,8}}{8} = 0,6A\\
{U_3} = {I_3}{R_3} = 0,6.4 = 2,4V\\
{U_4} = {I_4}{R_4} = 0,6.4 = 2,4V\\
9.\\
{R_1}nt(({R_2}nt{R_3})//{R_5})nt{R_4}\\
{R_{23}} = {R_2} + {R_3} = 4 + 6 = 10\Omega \\
{R_{235}} = \dfrac{{{R_{23}}{R_5}}}{{{R_{23}} + {R_5}}} = \dfrac{{10.10}}{{10 + 10}} = 5\Omega \\
R = {R_{235}} + {R_1} + {R_4} = 5 + 4 + 3 = 12\Omega \\
{I_1} = {I_4} = I = \dfrac{U}{R} = \dfrac{{24}}{{12}} = 2A\\
{U_1} = {I_1}{R_1} = 2.4 = 8V\\
{U_4} = {I_4}{R_4} = 2.3 = 6V\\
{U_{23}} = {U_5} = {U_{235}} = U - {U_1} - {U_4} = 24 - 8 - 6 = 10V\\
{I_5} = \dfrac{{{U_5}}}{{{R_5}}} = \dfrac{{10}}{{10}} = 1A\\
{I_2} = {I_3} = {I_{23}} = \dfrac{{{U_{23}}}}{{{R_{23}}}} = \dfrac{{10}}{{10}} = 1A
\end{array}\)
