In the circuit shown in the potential difference between the point a and b is `4 V`, Find the emf `epsilon` of the battery. Assume that before connecting the battery in the circuit, all the capacitors were uncharged.
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Explanation

The capacitors `12 muF` and `6muF` are in series arrangement. The charge in each will be equal.
`q_(1)=CV=12xx4=48 muC`.
The potential defference across the `6muF` capacitors is
`(12xx4)/6 V =8 V`


The potential differnce between points `a` and `c` is `V_(a)-V_(c)=4+8=12V`.
The equivalent capacity in portion (1) is `10muF`.
The charge on it is `(10xx12) muC=120 muC`.
Portions (1) and (2) of the circuit are series combination.
Hence, `12xx(V_(d)-V_(a))=120 muC` or `V_(d)-V_(a)=10 V`. The capacitor `5 muF` is also in series with `10 muF`.
Hence charge on it `120 muC`. Thus, `V_(c)V_(e)=120//5=24 V`.
Now `E=(12+10+24) V=46 V`.