Đáp án:

a) \(\dfrac{1}{3}\mu F\)

b) \(0,5\mu F\)

Giải thích các bước giải:

a) Mạch gồm: C3 nt C4 nt C5.

Ta có: 

\(\begin{array}{l}
\dfrac{1}{{{C_b}}} = \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}} + \dfrac{1}{{{C_5}}} = 1 + 1 + 1 = 3\\
 \Rightarrow {C_b} = \dfrac{1}{3}\mu F\\
{Q_3} = {Q_4} = {Q_5} = U{C_b} = 15.\dfrac{1}{3} = 5\mu C
\end{array}\)

b) Mạch gồm: ((C1 nt C2) // (C3 nt C4)) nt C5

Ta có: 

\(\begin{array}{l}
{C_{12}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = \dfrac{{1.1}}{{1 + 1}} = 0,5\mu F\\
{C_{34}} = \dfrac{{{C_3}{C_4}}}{{{C_3} + {C_4}}} = \dfrac{{1.1}}{{1 + 1}} = 0,5\mu F\\
{C_{1234}} = {C_{12}} + {C_{34}} = 0,5 + 0,5 = 1\mu F\\
{C_b} = \dfrac{{{C_{1234}}.{C_5}}}{{{C_{1234}} + {C_5}}} = \dfrac{{1.1}}{{1 + 1}} = 0,5\mu F\\
{Q_{1234}} = {Q_5} = U{C_b} = 15.0,5 = 7,5\mu C\\
{U_{12}} = {U_{34}} = {U_{1234}} = \dfrac{{{Q_{1234}}}}{{{C_{1234}}}} = \dfrac{{7,5}}{1} = 7,5V\\
{Q_1} = {Q_2} = {U_{12}}.{C_{12}} = 7,5.0,5 = 3,75V\\
{Q_3} = {Q_4} = {U_{34}}.{C_{34}} = 7,5.0,5 = 3,75V
\end{array}\)