Đáp án:

Đổi 2mF = $2.10^{-3}$ F, 6mF = $6.10^{-3} $ F, 4mF = $4.10^{-3}$ F

a) 3 tụ mắc song song:

$C_b=C_1+C_2+C_3=2.10^{-3}+4.10^{-3}+6.10^{-3}=0,012(F)$

$U_B=U_1=U_2=U_3=100\ (v)$

$Q_1=C_1.U_1=2.10^{-3}.100=0,2\ (C)$
$Q_2=C_2.U_2=4.10^{-3}.100=0,4\ (C)$
$Q_3=C_3.U_3=6.10^{-3}.100=0,6\ (C)$

b) 3 tụ điện mắc nối tiếp:

$\dfrac{1}{C_b}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}=\dfrac{1}{2.10^{-3}}+\dfrac{1}{4.10^{-3}}+\dfrac{1}{6.10^{-3}}=\dfrac{2750}{3}$
$\Rightarrow C_b=\dfrac{3}{2750}\ (F)$

$Q_b=Q_1=Q_2=Q_3=C_b.U_b=\dfrac{3}{2750}.100=\dfrac{6}{55}\ (C)$

$U_1=\dfrac{Q_1}{C_1}=\dfrac{6}{55.2.10^{-3}}=\dfrac{600}{11}\ (V)$

$U_2=\dfrac{Q_2}{C_2}=\dfrac{6}{55.4.10^{-3}}=\dfrac{300}{11}\ (V)$

$U_3=\dfrac{Q_3}{C_3}=\dfrac{6}{55.6.10^{-3}}=\dfrac{200}{11}\ (V)$

c) ($C_1$ song song với $C_2$) nối tiếp với $C_3$:

$C_{12}=C_1+C_2=6.10^{-3}\ (F)$
$C_b=\dfrac{C_{12}.C_3}{C_{12}+C_3}=\dfrac{6.10^{-3}.6.10^{-3}}{6.10^{-3}+6.10^{-3}}=3.10^{-3}\ (F)$

$Q_b=Q_{12}=Q_3=C_b.U_b=3.10^{-3}.100=0,3\ (C)$

$U_3=\dfrac{Q_3}{C_3}=\dfrac{0,3}{6.10^{-3}}=50\ (V)$

$\Rightarrow U_{12}=U_1=U_2=U_b-U_3=100-50=50\ (V)$

$Q_1=C_1.U_1=2.10^{-3}.50=0,1\ (C)$

$Q_2=C_2.U_2=4.10^{-3}.50=0,2\ (C)$