Câu 48: 

Ta có:

\(\begin{array}{l}
{C_{23}} = {C_2} + {C_3} = 2 + 1 = 3\mu F\\
C = \dfrac{{{C_1}{C_{23}}}}{{{C_1} + {C_{23}}}} = \dfrac{{6.3}}{{6 + 3}} = 2\mu F\\
{Q_1} = {Q_{23}} = Q = CU = 2.10 = 20\mu C\\
{U_2} = {U_{23}} = \dfrac{{{Q_{23}}}}{{{C_{23}}}} = \dfrac{{20}}{3}V\\
{Q_2} = {C_2}{U_2} = 2.\dfrac{{20}}{3} = \dfrac{{40}}{3}\mu C
\end{array}\)

Câu 49: 

Ta có: 

\(\begin{array}{l}
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
R = {R_1} + {R_{23}} = 3 + 2 = 5\Omega \\
I = \dfrac{U}{R} = \dfrac{{10}}{5} = 2A\\
{U_2} = {U_3} = {U_{23}} = I.{R_{23}} = 2.2 = 4V\\
{P_2} = \dfrac{{U_2^2}}{{{R_2}}} = \dfrac{{{4^2}}}{3} = \dfrac{{16}}{3}W\\
{P_3} = \dfrac{{U_3^2}}{{{R_3}}} = \dfrac{{{4^2}}}{1} = 16W
\end{array}\)

Câu 50: 

Ta có: 

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 2 + 4 = 6\Omega \\
{I_2} = {I_3} = \dfrac{U}{{{R_{23}}}} = \dfrac{6}{6} = 1A\\
{P_2} = I_2^2{R_2} = {1^2}.2 = 2W\\
{P_3} = I_3^2{R_3} = {1^2}.4 = 4W
\end{array}\)