The resistance each of `16 Omega` and capacitance of each `100 mu F` are arranged as shown in the figure. A battery of emf `12 V` is joined across `A` and `B`. Find the
(`i`) reading of the ammeter just after key is ciosed and after long time.
(`ii`) charges in each capacitors when steady state is achieved.

Explanation

(`i`) Just after the key is closed
The circuit will be as shown in figure `1`
Then circuit can be simplify as figure `2`

The points `D` and `E` will be at same potential hence the circuit can be further simplified as shown in figure `3` & `4`

Hence equivalent resistance across `A` and `B`, `R_(eq)=((68)/(7)xx16)/((68)/(7)+16)=6.04 Omega`
Hence reading of ammeter `I=(12)/(6.4)~~2A`.
After long time the circuit can be redrawn as

`R_(AB)=(16xx32)/(16+32)=(32)/(3)Omega`
Hence the reading of the ammeter `I=(9)/(8)A`
(`ii`) The current through paths `ADC` and `AEC` will be same and will be equal to
`I_(2)=(1)/(2)xX(9)/(8)xx(16)/(48)=(3)/(16)A`
Hence potential difference across `AD`, `AE`, `DC` and `CE`, `V'=16xx(3)/(16)=3V`
Hence the charge in the `C_(1)`, `C_(2)`, `C_(5)` and `C_(6)q=100xx3=300muC`
Also `V_(DB)=V_(EB)=12-3=9V`
Hence the charge in the capacitor `C_(3)` and `C_(4)=100xx9=900muC`