a/ K mở -->(C1 nt C2) // (C3 nt C4)
Cb = C1.C2/(C1+C2) + C3.C4/(C3+C4)=3,15.10^-6F
Ub=U12=U34 = 20V
Qb = Cb.Ub= 6,3.10^-5C
Q1 =Q2 = Q12 = U12.C12=1,5.10^-5C
Q3 = Q4 = Q34 =U34.C34=4,8.10^-5C
U1 = Q1/C1=15V
U2 = Q2/C2=5V
U3 = 8V
U4 =12V
b/ K đóng --->(C1 // C3) nt (C2 // C4)
C13=C1+C3 = 7 .10^-6F
C24=C2+C4 = 7 .10^-6F
Cb=C13.C24/(C13+C24) =3,5.10^-6F
Qb=Q13=Q24 = Cb.Ub = 7.10^-5(C)
U1=U3=U13=Q13/C13=10V
U2=U4=U24=Q24/C24=10V
Q1=U1.C1=10^-5C
Q2=U2.C2=3.10^-5C
Q3=...=6.10^-5C
Q4=...=4.10^-5C