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Đáp án:

a) $2\mu F$

b) $12V$

Giải thích các bước giải:

a) Ta có: 

$\begin{array}{l} \dfrac{1}{{{C_{123}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{2}\\  \Rightarrow {C_{123}} = 2\mu F\\ {C_{1234}} = {C_{123}} + {C_4} = 2 + 2 = 4\mu F\\ {C_b} = \dfrac{{{C_{1234}}.{C_5}}}{{{C_{1234}} + {C_5}}} = \dfrac{{4.4}}{{4 + 4}} = 2\mu F \end{array}$

b) Ta có: 

$\begin{array}{l} {U_4} = \dfrac{{{Q_4}}}{{{C_4}}} = \dfrac{{12}}{2} = 6V\\ {U_{123}} = {U_4} = 6V\\ {Q_1} = {Q_2} = {Q_3} = {C_{123}}{U_{123}} = 2.4 = 12\mu C\\ {U_1} = {U_2} = {U_3} = \dfrac{{{Q_1}}}{{{C_1}}} = \dfrac{{12}}{6} = 2V\\ {Q_5} = {Q_{1234}} = {C_{1234}}.{U_{123}} = 4.6 = 24\mu C\\ {U_5} = \dfrac{{{Q_5}}}{{{C_5}}} = \dfrac{{24}}{4} = 6V\\ U = {U_{123}} + {U_5} = 6 + 6 = 12V \end{array}$