(i) Find the equivalent capacitance of the combination shown in the figure (a), when `C_(1)=2.0 muF,C_(2)=4.0 muF and C_(3) = 3.0 muF`
(ii) The input terminals A and B in Fig. (a) are connected to a battery of 12 V. Find the potential and the charge of each capacitor.

Explanation

(a) `C_(1) "and " C_(2)` are in parallel hence their equivalent capacitance is
`C'=C_(1)+C_(2)=2.0 muF+4.0muF=6.0muF`
Figure (b) shows the combination of C' and `C_(3)` in series. The final equivalent capacitance shown in figure ( c) is given by
`(1)/(C_(epsilonq))=(1)/(C')+(1)/(C_(3))=(1)/(6.0muF)+(1)/(3.0muF)=(1)/(2.0 muF)`
or `C_(eq)=2.0muF`
(b) To find the charge or potential difference , we retrace the path to the original in figure (a) The charge supplied by the battery is
`q=C_(eq)`V when the inputs are joined to a V-volt battery
`=(2.0muF) (12.0 V) =24.0 muC`
The charge on each capacitor in series in figure (b ) is
`q_(3)=q=q=24.0 mu C`
So `V_(3)= (q_(3))/(C_(3))=(24.0muC)/(3.0muF) 8 V `
The potential difference across C in figure ( b) is

`V=(q)/(C) =(24.0muC)/(6.0 muF) = 4 V " " `(b) `C_(1) "and" C_(2)` in figure (a) and hence the charges on them are
`q_(1)=C_(1)V_(1)=(2.0muF)(4.0 V) = 8.0 mu C`
and `q_(2)= C_(2)V_(2)= (4.0muF) (4.0V) = 16.muC `
Thus we have `V_(1)=V_(2)=4.0V, V_(3)= 8 V`

and `q_(1)=8.0 mu C , q_(2) =16.0 muC , q_(3)=24.0 muC " " ` (c) C' and `C_(3)` are replaced by `C_(eq)`