Câu 9: 

Ta có: 

\(\begin{array}{l}
{I_1} = I = \dfrac{U}{R} = \dfrac{{78}}{{7,8}} = 10A\\
{U_1} = {I_1}{R_1} = 10.7 = 70V\\
{U_2} = {U_3} = U - {U_1} = 78 - 70 = 8V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{8}{4} = 2A\\
{I_3} = I - {I_1} = 10 - 2 = 8A\\
{R_3} = \dfrac{{{U_3}}}{{{I_3}}} = \dfrac{8}{8} = 1\Omega 
\end{array}\)

Câu 10: 

Ta có: 

\(\begin{array}{l}
I = \dfrac{U}{R} = \dfrac{{12}}{{4,8}} = 2,5A\\
E = U + I.r = 12 + 2,5.0,1 = 12,25V
\end{array}\)

Câu 11: 

Ta có: 

\(E = \dfrac{A}{q} = \dfrac{{{{840.10}^{ - 3}}}}{{{{7.10}^{ - 2}}}} = 12V\)

Câu 12: 

Ta có: 

\(\begin{array}{l}
Q = CU = {6.10^{ - 6}}.3 = {18.10^{ - 6}}C\\
I = \dfrac{Q}{t} = \dfrac{{{{18.10}^{ - 6}}}}{{{{10}^{ - 4}}}} = 0,18A
\end{array}\)