Giúp mình câu 9,câu 10 ạ. Cần gấp ạ cảm ơn

Bài 9:

a) Mạch gồm: R1 nt ((R2 nt R3) // (R4 nt R5))

Ta có:

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 4 + 6 = 10\Omega \\
{R_{45}} = {R_4} + {R_5} = 6 + 9 = 15\Omega \\
{R_{2345}} = \dfrac{{{R_{23}}.{R_{45}}}}{{{R_{23}} + {R_{45}}}} = \dfrac{{10.15}}{{10 + 15}} = 6\Omega \\
R = {R_1} + {R_{2345}} = 12 + 6 = 18\Omega \\
I = \dfrac{U}{R} = \dfrac{{24}}{{18}} = \dfrac{4}{3}A \Rightarrow {I_1} = \dfrac{4}{3}A\\
{I_2} = {I_3} = \dfrac{{{R_{45}}}}{{{R_{23}} + {R_{45}}}}.I = \dfrac{{15}}{{10 + 15}}.\dfrac{4}{3} = 0,8A\\
{I_4} = {I_5} = I - {I_2} = \dfrac{4}{3} - 0,8 = \dfrac{8}{{15}}A\\
{U_1} = {I_1}{R_1} = \dfrac{4}{3}.12 = 16V\\
{U_2} = {I_2}{R_2} = 0,8.4 = 3,2V\\
{U_4} = {I_4}{R_4} = \dfrac{8}{{15}}.6 = 3,2V\\
{U_{MN}} =  - {U_2} + {U_4} =  - 3,2 + 3,2 = 0V\\
{U_{AN}} = {U_1} + {U_4} = 16 + 3,2 = 19,2V
\end{array}\)

b)

Mạch gồm: R1 nt ((R2 nt R3) // (R4 nt R5))

Ta có:

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 4 + 6 = 10\Omega \\
{R_{45}} = {R_4} + {R_5} = 7 + 8 = 15\Omega \\
{R_{2345}} = \dfrac{{{R_{23}}.{R_{45}}}}{{{R_{23}} + {R_{45}}}} = \dfrac{{10.15}}{{10 + 15}} = 6\Omega \\
R = {R_1} + {R_{2345}} = 12 + 6 = 18\Omega \\
I = \dfrac{U}{R} = \dfrac{{24}}{{18}} = \dfrac{4}{3}A \Rightarrow {I_1} = \dfrac{4}{3}A\\
{I_2} = {I_3} = \dfrac{{{R_{45}}}}{{{R_{23}} + {R_{45}}}}.I = \dfrac{{15}}{{10 + 15}}.\dfrac{4}{3} = 0,8A\\
{I_4} = {I_5} = I - {I_2} = \dfrac{4}{3} - 0,8 = \dfrac{8}{{15}}A\\
{U_1} = {I_1}{R_1} = \dfrac{4}{3}.12 = 16V\\
{U_2} = {I_2}{R_2} = 0,8.4 = 3,2V\\
{U_4} = {I_4}{R_4} = \dfrac{8}{{15}}.7 = \dfrac{{56}}{{15}}V\\
{U_{MN}} =  - {U_2} + {U_4} =  - 3,2 + \dfrac{{56}}{{15}} = \dfrac{8}{{15}}V\\
{U_{AN}} = {U_1} + {U_4} = 16 + \dfrac{{56}}{{15}} = \dfrac{{296}}{{15}}V
\end{array}\)

Bài 10:

Mạch gồm: R1 nt (R2 // R3)

a) Ta có:

\(\begin{array}{l}
{U_2} = {I_2}{R_2} = 2.6 = 12V\\
{U_1} = U - {U_2} = 18 - 12 = 6V\\
{I_3} = \dfrac{{{U_2}}}{{{R_3}}} = \dfrac{{12}}{3} = 4A\\
{I_1} = {I_2} + {I_3} = 2 + 4 = 6A\\
{R_1} = \dfrac{{{U_1}}}{{{I_1}}} = \dfrac{6}{6} = 1\Omega 
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{U_2} = {I_2}{R_2} = 2.1 = 2V\\
{U_1} = U - {U_2} = 18 - 2 = 16V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{16}}{3}A\\
{I_3} = {I_1} - {I_2} = \dfrac{{16}}{3} - 2 = \dfrac{{10}}{3}A\\
{R_3} = \dfrac{{{U_3}}}{{{I_3}}} = \dfrac{2}{{\dfrac{{10}}{3}}} = 0,6\Omega 
\end{array}\)

c) Ta có:

\(\begin{array}{l}
{U_2} = {I_2}{R_2} = 2{R_2}\\
{U_1} = U - {U_2} = 18 - 2{R_2}\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{18 - 2{R_2}}}{5}\\
{I_3} = {I_1} - {I_2} = \dfrac{{18 - 2{R_2}}}{5} - 2\\
{U_3} = {I_3}{R_3} \Rightarrow 2{R_2} = \left( {\dfrac{{18 - 2{R_2}}}{5} - 2} \right).3\\
 \Rightarrow {R_2} = 1,5\Omega 
\end{array}\)