(a) Taking the potential at the point `B` to be zero, find the potential of point `D`.
(b) Find `V_(A)-V_(B)`. Under what condition is it equal to zero.
(c) Consider the situation as shown in diagram. How much charge will flow through cell if the switch `S` is closed?
(d) In the circuit shown in the figure, calculate
(i) Potential difference between the points `X` and `Y`.
(ii) Potential of the point `X` when switch `S` is closed.
(iii) Charge flown through the switch when it was closed.

Explanation

(a)
`V_(1)=((12)/(12+6))xx24=16V`
potential of capacitor means potential difference between its plates.
`V_(1)=V_(D)-V_(B)=16`
`V_(B)=0`
`V_(D)=16V`
(b)
`V_(1)=((C_(2))/(C_(1)+C_(2)))V, V_(3)=((C_(4))/(C_(3)+C_(4)))V`
`V_(1)=V_(X)-V_(A)=(C_(2)V)/(C_(1)+C_(2))` ...`(i)`
`V_(3)=V_(X)-V_(B)=(C_(4)V)/(C_(3)+C_(4))`
Subtracting (i) from (ii), we get
`V_(A)-V_(B)=((C_(4))/(C_(3)+C_(4))-(C_(2))/(C_(1)+C_(2)))V`
If `V_(A)-V_(B)=0`
`C_(4)(C_(1)+C_(2))=C_(2)(C_(3)+C_(4))`
`(C_(1))/(C_(2))=(C_(3))/(C_(4))`
(c) when switch `S` is open:

`C_(eq)=(12Cxx4C)/(12Cxx4C)=3C`
Charge suppplied by battery `Q_(1)=3CV_(0)`
When switch `S` is closed:

Points `D` and `B` are at same potential after closing the switch, the capacitors between `D` and `B` are redundant, remove them.
`C_(eq)=12C`
Charge on capacitors, `Q_(2)=12CV_(0)`
Charge supplied buy battery after closing the switch
`DeltaQ=Q_(2)-Q_(1)=12CV_(0)-3CV_(0)=9CV_(0)`
(d) (i)
since point `B` is connected to earth, `V_(B)=0,V_(A)=60V`
`V_(A)-V_(B)=60V`
`V_(2)=((4)/(4+2))xx60=40V`
`V_(4)=((2)/(2+10))xx60=10V`
`V_(2)=V_(X)-V_(B)=40` implies `V_(x)=40V`
`V_(X)-V_(Y)=40-10=30V`
(ii) Now switch `S` is closed, points `X` and `Y` are are at same potential.

`V_(1)=((12)/(6+12))xx60=40V`, Now return to previous diagram.

`V_(2)=60- V_(X)-V_(B)=20 implies V_(X)=20V V_(1)=20V`
Now return to previous diagram
`V_(X) - V_(B) = 20 rArr V_(X) 20 V`
(iii) Charge on `4 muF,4xx40=160 muC`
Charge on `2muF,2xx20=40muC` Charge flown through switch=`160-40=120 muC`