Đáp án:

a) \(18V\)

b) \(\dfrac{{38}}{3}V\)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{C_{23}} = {C_2} + {C_3} = 5 + 4 = 9\mu F\\
C = \dfrac{{{C_1}{C_{23}}}}{{{C_1} + {C_{23}}}} = \dfrac{{10.9}}{{10 + 9}} = \dfrac{{90}}{{19}}\mu F\\
{Q_1} = {Q_{23}} = Q = CU = \dfrac{{90}}{{19}}.38 = 180\mu C\\
{U_1} = \dfrac{{{Q_1}}}{{{C_1}}} = \dfrac{{180}}{{10}} = 18V\\
{U_2} = {U_3} = \dfrac{{{Q_{23}}}}{{{C_{23}}}} = \dfrac{{180}}{9} = 20V
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
C = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = \dfrac{{10.5}}{{10 + 5}} = \dfrac{{10}}{3}\mu F\\
{Q_1} = {Q_3} = CU = \dfrac{{10}}{3}.38 = \dfrac{{380}}{3}\mu C\\
{U_1} = \dfrac{{{Q_1}}}{{{C_1}}} = \dfrac{{380}}{{3.10}} = \dfrac{{38}}{3}V
\end{array}\)