Câu 11:

a) Ta có:

\(\begin{array}{l}
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{6.12}}{{6 + 12}} = 4\Omega \\
{R_{td}} = {R_{12}} + {R_3} = 4 + 8 = 12\Omega 
\end{array}\)

b) Cường độ dòng điện là:

\(\begin{array}{l}
{I_3} = I = \dfrac{U}{{{R_{td}}}} = \dfrac{{24}}{{12}} = 2A\\
{I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}.I = \dfrac{{12}}{{6 + 12}}.2 = \dfrac{4}{3}A\\
{I_2} = I - {I_1} = 2 - \dfrac{4}{3} = \dfrac{2}{3}A\\
{U_1} = {U_2} = {I_1}{R_1} = \dfrac{4}{3}.6 = 8V\\
{U_3} = U - {U_1} = 24 - 8 = 16V
\end{array}\)

Câu 12: 

Mạch gồm: R4 // (R1 nt (R5 // (R2 nt (R3 // R6))))

Ta có:

\(\begin{array}{l}
{R_{36}} = \dfrac{{{R_3}{R_6}}}{{{R_3} + {R_6}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
{R_{236}} = {R_2} + {R_{36}} = 2 + 2 = 4\Omega \\
{R_{2356}} = \dfrac{{{R_{236}}.{R_5}}}{{{R_{236}} + {R_5}}} = \dfrac{{4.4}}{{4 + 4}} = 2\Omega \\
{R_{12356}} = {R_1} + {R_{2356}} = 1 + 2 = 3\Omega \\
{R_{td}} = \dfrac{{{R_4}.{R_{12356}}}}{{{R_4} + {R_{12356}}}} = \dfrac{{6.3}}{{6 + 3}} = 2\Omega 
\end{array}\)